1851. Minimum Interval to Include Each Query - Explanation

Description

You are given a 2D integer array intervals, where intervals[i] = [left_i, right_i] represents the ith interval starting at left_i and ending at right_i (inclusive).

You are also given an integer array of query points queries. The result of query[j] is the length of the shortest interval i such that left_i <= queries[j] <= right_i. If no such interval exists, the result of this query is -1.

Return an array output where output[j] is the result of query[j].

Note: The length of an interval is calculated as right_i - left_i + 1.

Example 1:

Input: intervals = [[1,3],[2,3],[3,7],[6,6]], queries = [2,3,1,7,6,8]

Output: [2,2,3,5,1,-1]

Explanation:

Query = 2: The interval [2,3] is the smallest one containing 2, it's length is 2.
Query = 3: The interval [2,3] is the smallest one containing 3, it's length is 2.
Query = 1: The interval [1,3] is the smallest one containing 1, it's length is 3.
Query = 7: The interval [3,7] is the smallest one containing 7, it's length is 5.
Query = 6: The interval [6,6] is the smallest one containing 6, it's length is 1.
Query = 8: There is no interval containing 8.

Constraints:

1 <= intervals.length <= 1000
1 <= queries.length <= 1000
1 <= left_i <= right_i <= 10000
1 <= queries[j] <= 10000

Topics

Recommended Time & Space Complexity

You should aim for a solution as good or better than O(nlogn + mlogm) time and O(n + m) space, where m is the size of the array queries and n is the size of the array intervals.

Hint 1

A brute force approach would be to iterate through each query and, for each query, check all intervals to find the result. This would be an O(m * n) solution. Can you think of a better way? Maybe processing the queries in sorted order could help.

Hint 2

We sort the intervals by start value and process the queries in ascending order. Using a pointer i, we add intervals to a min-heap while their start values are less than or equal to the query, storing their end values and sizes.

Hint 3

The min-heap is ordered by interval size. We remove elements from the heap while the top element’s end value is less than the current query. The result for the query is the top element’s size if the heap is non-empty; otherwise, it is -1.

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Prerequisites

Before attempting this problem, you should be comfortable with:

Sorting - Sorting intervals and queries to process them in order
Min Heap / Priority Queue - Efficiently retrieving the smallest interval covering a query
Sweep Line Algorithm - Processing events (interval starts, ends, queries) in sorted order
Segment Trees with Lazy Propagation - Advanced technique for range minimum queries and updates
Coordinate Compression - Mapping large coordinate values to smaller indices

1. Brute Force

Intuition

For each query value q, we want to find the smallest interval length among all intervals [l, r] that contain q (meaning l <= q <= r).
If no interval contains q, we return -1.

The brute force idea is very direct:

handle queries one by one
for a query, scan through every interval
whenever an interval covers the query, compute its length r - l + 1
keep the minimum length seen

Algorithm

Initialize an empty list res.
For each query q in queries:
- set cur = -1 to represent "no covering interval found yet"
Iterate through every interval [l, r]:
- if l <= q <= r, then the interval covers q
- compute its length len = r - l + 1
- update cur if:
  - cur == -1 (first valid interval), or
  - len < cur (found a smaller covering interval)
Append cur to res for this query.
After all queries are processed, return res.

class Solution:
    def minInterval(self, intervals: List[List[int]], queries: List[int]) -> List[int]:
        n = len(intervals)
        res = []
        for q in queries:
            cur = -1
            for l, r in intervals:
                if l <= q <= r:
                    if cur == -1 or (r - l + 1) < cur:
                        cur = r - l + 1
            res.append(cur)
        return res

public class Solution {
    public int[] minInterval(int[][] intervals, int[] queries) {
        int[] res = new int[queries.length];
        int idx = 0;
        for (int q : queries) {
            int cur = -1;
            for (int[] interval : intervals) {
                int l = interval[0], r = interval[1];
                if (l <= q && q <= r) {
                    if (cur == -1 || (r - l + 1) < cur) {
                        cur = r - l + 1;
                    }
                }
            }
            res[idx++] = cur;
        }
        return res;
    }
}

class Solution {
public:
    vector<int> minInterval(vector<vector<int>>& intervals, vector<int>& queries) {
        vector<int> res;
        for (int q : queries) {
            int cur = -1;
            for (auto& interval : intervals) {
                int l = interval[0], r = interval[1];
                if (l <= q && q <= r) {
                    if (cur == -1 || (r - l + 1) < cur) {
                        cur = r - l + 1;
                    }
                }
            }
            res.push_back(cur);
        }
        return res;
    }
};

class Solution {
    /**
     * @param {number[][]} intervals
     * @param {number[]} queries
     * @return {number[]}
     */
    minInterval(intervals, queries) {
        const res = [];
        for (const q of queries) {
            let cur = -1;
            for (const [l, r] of intervals) {
                if (l <= q && q <= r) {
                    if (cur === -1 || r - l + 1 < cur) {
                        cur = r - l + 1;
                    }
                }
            }
            res.push(cur);
        }
        return res;
    }
}

public class Solution {
    public int[] MinInterval(int[][] intervals, int[] queries) {
        int[] res = new int[queries.Length];
        int idx = 0;
        foreach (int q in queries) {
            int cur = -1;
            foreach (var interval in intervals) {
                int l = interval[0], r = interval[1];
                if (l <= q && q <= r) {
                    if (cur == -1 || (r - l + 1) < cur) {
                        cur = r - l + 1;
                    }
                }
            }
            res[idx++] = cur;
        }
        return res;
    }
}

func minInterval(intervals [][]int, queries []int) []int {
    res := make([]int, len(queries))
    for i, q := range queries {
        cur := -1
        for _, interval := range intervals {
            l, r := interval[0], interval[1]
            if l <= q && q <= r {
                if cur == -1 || (r - l + 1) < cur {
                    cur = r - l + 1
                }
            }
        }
        res[i] = cur
    }
    return res
}

class Solution {
    fun minInterval(intervals: Array<IntArray>, queries: IntArray): IntArray {
        val res = IntArray(queries.size)
        for ((i, q) in queries.withIndex()) {
            var cur = -1
            for (interval in intervals) {
                val l = interval[0]
                val r = interval[1]
                if (l <= q && q <= r) {
                    if (cur == -1 || (r - l + 1) < cur) {
                        cur = r - l + 1
                    }
                }
            }
            res[i] = cur
        }
        return res
    }
}

class Solution {
    func minInterval(_ intervals: [[Int]], _ queries: [Int]) -> [Int] {
        var res = [Int]()

        for q in queries {
            var cur = -1
            for interval in intervals {
                let l = interval[0]
                let r = interval[1]

                if l <= q && q <= r {
                    if cur == -1 || (r - l + 1) < cur {
                        cur = r - l + 1
                    }
                }
            }
            res.append(cur)
        }

        return res
    }
}

Time & Space Complexity

Time complexity: $O(m * n)$
Space complexity:
- $O(1)$ extra space.
- $O(m)$ space for the output array.

Where $m$ is the length of the array $queries$ and $n$ is the length of the array $intervals$ .

2. Sweep Line Algorithm

Intuition

For each query value q, we want the length of the smallest interval [l, r] that contains q (l <= q <= r). If none exists, the answer is -1.

A sweep line approach processes everything (interval starts, interval ends, and queries) in sorted order of time.
While sweeping from left to right, we maintain a data structure of “currently active intervals” (intervals that have started but have not ended yet).
For a query time q, the answer is simply the smallest interval length among the active intervals.

To get that smallest length quickly, we use a min-heap ordered by interval size.

Because intervals can end later, we also need a way to remove expired intervals:

when an interval ends, we mark it as inactive
when answering a query, we pop inactive intervals from the heap until the top is active

Algorithm

Build a list of events:
- For each interval [start, end]:
  - Add a start event at start containing its length and index
  - Add an end event at end containing its index
- For each query q:
  - Add a query event at q containing the query's original position
Sort all events by (time, type) so they are processed in time order.
Use:
- a min-heap sizes storing (interval_length, interval_index) for active intervals
- an array inactive to mark intervals that have ended
- an output array ans initialized with -1
Sweep through events in sorted order:
- If it is an interval start:
  - push (length, idx) into the heap
- If it is an interval end:
  - mark that interval index as inactive
- If it is a query:
  - pop from the heap while the top interval is inactive
  - if heap is not empty, the top length is the smallest covering interval store it in ans
  - otherwise leave -1
Return ans in the original query order.

class Solution:
    def minInterval(self, intervals: List[List[int]], queries: List[int]) -> List[int]:
        events = []
        # Create events for intervals
        for idx, (start, end) in enumerate(intervals):
            events.append((start, 0, end - start + 1, idx))
            events.append((end, 2, end - start + 1, idx))

        # Create events for queries
        for i, q in enumerate(queries):
            events.append((q, 1, i))

        # Sort by time and type (end before query)
        events.sort(key=lambda x: (x[0], x[1]))

        # Min heap storing [size, index]
        sizes = []
        ans = [-1] * len(queries)
        inactive = [False] * len(intervals)

        for time, type, *rest in events:
            if type == 0:  # Interval start
                interval_size, idx = rest
                heapq.heappush(sizes, (interval_size, idx))
            elif type == 2: #Interval end
                idx = rest[1]
                inactive[idx] = True
            else: # Query
                query_idx = rest[0]
                while sizes and inactive[sizes[0][1]]:
                    heapq.heappop(sizes)
                if sizes:
                    ans[query_idx] = sizes[0][0]

        return ans

public class Solution {
   public int[] minInterval(int[][] intervals, int[] queries) {
       List<int[]> events = new ArrayList<>();
       for (int i = 0; i < intervals.length; i++) {
           events.add(new int[]{intervals[i][0], 0, intervals[i][1] - intervals[i][0] + 1, i});
           events.add(new int[]{intervals[i][1], 2, intervals[i][1] - intervals[i][0] + 1, i});
       }

       for (int i = 0; i < queries.length; i++) {
           events.add(new int[]{queries[i], 1, i});
       }

       // Sort by time and type (end before query)
       events.sort((a, b) -> a[0] != b[0] ?
                  Integer.compare(a[0], b[0]) :
                  Integer.compare(a[1], b[1]));

       int[] ans = new int[queries.length];
       Arrays.fill(ans, -1);

       // Min heap storing [size, index]
       PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> Integer.compare(a[0], b[0]));
       boolean[] inactive = new boolean[intervals.length];

       for (int[] event : events) {
           if (event[1] == 0) { // Interval start
               pq.offer(new int[]{event[2], event[3]});
           }
           else if (event[1] == 2) { // Interval end
               inactive[event[3]] = true;
           }
           else { // Query
               while (!pq.isEmpty() && inactive[pq.peek()[1]]) {
                   pq.poll();
               }
               if (!pq.isEmpty()) {
                   ans[event[2]] = pq.peek()[0];
               }
           }
       }

       return ans;
   }
}

class Solution {
public:
    vector<int> minInterval(vector<vector<int>>& intervals, vector<int>& queries) {
        vector<vector<int>> events;
        // Create events for intervals
        for (int i = 0; i < intervals.size(); i++) {
            events.push_back({intervals[i][0], 0, intervals[i][1] - intervals[i][0] + 1, i});
            events.push_back({intervals[i][1], 2, intervals[i][1] - intervals[i][0] + 1, i});
        }

        // Create events for queries
        for (int i = 0; i < queries.size(); i++) {
            events.push_back({queries[i], 1, i});
        }

        // Sort by time and type (end before query)
        sort(events.begin(), events.end(), [](const vector<int>& a, const vector<int>& b) {
            return a[0] == b[0] ? a[1] < b[1] : a[0] < b[0];
        });

        vector<int> ans(queries.size(), -1);
        // Min heap storing [size, index]
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
        vector<bool> inactive(intervals.size(), false);

        for (const auto& event : events) {
            if (event[1] == 0) { // Interval start
                pq.push({event[2], event[3]});
            } else if (event[1] == 2) { // Interval end
                inactive[event[3]] = true;
            } else { // Query
                int queryIdx = event[2];
                while (!pq.empty() && inactive[pq.top().second]) {
                    pq.pop();
                }
                if (!pq.empty()) {
                    ans[queryIdx] = pq.top().first;
                }
            }
        }

        return ans;
    }
};

class Solution {
    /**
     * @param {number[][]} intervals
     * @param {number[]} queries
     * @return {number[]}
     */
    minInterval(intervals, queries) {
        let events = [];
        // Create events for intervals
        for (let i = 0; i < intervals.length; i++) {
            const [start, end] = intervals[i];
            events.push([start, 0, end - start + 1, i]);
            events.push([end, 2, end - start + 1, i]);
        }

        // Create events for queries
        queries.forEach((q, i) => {
            events.push([q, 1, i]);
        });
        // Sort by time and type (end before query)
        events.sort((a, b) => a[0] - b[0] || a[1] - b[1]);

        const ans = Array(queries.length).fill(-1);
        // Min heap storing [size, index]
        const pq = new PriorityQueue((a, b) => a[0] - b[0]);
        const inactive = Array(intervals.length).fill(false);

        for (const [time, type, ...rest] of events) {
            if (type === 0) {
                // Interval start
                pq.push([rest[0], rest[1]]);
            } else if (type === 2) {
                // Interval end
                inactive[rest[1]] = true;
            } else {
                // Query
                while (!pq.isEmpty() && inactive[pq.front()[1]]) {
                    pq.pop();
                }
                if (!pq.isEmpty()) {
                    ans[rest[0]] = pq.front()[0];
                }
            }
        }

        return ans;
    }
}

public class Solution {
    public int[] MinInterval(int[][] intervals, int[] queries) {
        var events = new List<int[]>();
        // Create events for intervals
        for (int i = 0; i < intervals.Length; i++) {
            events.Add(new int[] { intervals[i][0], 0, intervals[i][1] - intervals[i][0] + 1, i });
            events.Add(new int[] { intervals[i][1], 2, intervals[i][1] - intervals[i][0] + 1, i });
        }

        // Create events for queries
        for (int i = 0; i < queries.Length; i++) {
            events.Add(new int[] { queries[i], 1, i });
        }
        // Sort by time and type (end before query)
        events.Sort((a, b) => a[0] == b[0] ? a[1].CompareTo(b[1]) : a[0].CompareTo(b[0]));

        int[] ans = new int[queries.Length];
        Array.Fill(ans, -1);
        // Min heap storing [size, index]
        var pq = new PriorityQueue<(int size, int idx), int>();
        var inactive = new bool[intervals.Length];

        foreach (var e in events) {
            if (e[1] == 0) { // Interval start
                pq.Enqueue((e[2], e[3]), e[2]);
            } else if (e[1] == 2) { // Interval end
                inactive[e[3]] = true;
            } else {
                int queryIdx = e[2];
                while (pq.Count > 0 && inactive[pq.Peek().idx]) {
                    pq.Dequeue();
                }
                if (pq.Count > 0) {
                    ans[queryIdx] = pq.Peek().size;
                }
            }
        }

        return ans;
    }
}

type Event struct {
	time int
	typ  int // 0: interval start, 1: query, 2: interval end
	val  int // interval size or query index
	idx  int // interval index
}

func minInterval(intervals [][]int, queries []int) []int {
	events := []Event{}

	// Create events for intervals
	for idx, interval := range intervals {
		start, end := interval[0], interval[1]
		size := end - start + 1
		events = append(events, Event{start, 0, size, idx})
		events = append(events, Event{end, 2, size, idx})
	}

	// Create events for queries
	for i, q := range queries {
		events = append(events, Event{q, 1, i, -1})
	}

	// Sort events by time and type
	sort.Slice(events, func(i, j int) bool {
		if events[i].time == events[j].time {
			return events[i].typ < events[j].typ
		}
		return events[i].time < events[j].time
	})

	// Priority queue to store intervals with the smallest size on top
	sizes := priorityqueue.NewWith(func(a, b interface{}) int {
		return utils.IntComparator(a.([]int)[0], b.([]int)[0])
	})
	ans := make([]int, len(queries))
	for i := range ans {
		ans[i] = -1
	}
	inactive := make([]bool, len(intervals))

	for _, event := range events {
		switch event.typ {
		case 0: // Interval start
			sizes.Enqueue([]int{event.val, event.idx})
		case 2: // Interval end
			inactive[event.idx] = true
		case 1: // Query
			queryIdx := event.val
			for !sizes.Empty() {
				top, _ := sizes.Peek()
				if inactive[top.([]int)[1]] {
					sizes.Dequeue()
				} else {
					break
				}
			}
			if !sizes.Empty() {
				top, _ := sizes.Peek()
				ans[queryIdx] = top.([]int)[0]
			}
		}
	}
	return ans
}

data class Event(val time: Int, val type: Int, val sizeOrQueryIndex: Int, val idx: Int)

class Solution {
    fun minInterval(intervals: Array<IntArray>, queries: IntArray): IntArray {
        val events = mutableListOf<Event>()

        // Create events for intervals
        for ((idx, interval) in intervals.withIndex()) {
            val (start, end) = interval
            val size = end - start + 1
            events.add(Event(start, 0, size, idx))
            events.add(Event(end, 2, size, idx))
        }

        // Create events for queries
        for ((i, q) in queries.withIndex()) {
            events.add(Event(q, 1, i, -1))
        }

        // Sort events by time and type
        events.sortWith(compareBy({ it.time }, { it.type }))

        val sizes = PriorityQueue<Pair<Int, Int>>(compareBy { it.first })
        val ans = IntArray(queries.size) { -1 }
        val inactive = BooleanArray(intervals.size)

        for (event in events) {
            when (event.type) {
                0 -> { // Interval start
                    sizes.add(Pair(event.sizeOrQueryIndex, event.idx))
                }
                2 -> { // Interval end
                    inactive[event.idx] = true
                }
                1 -> { // Query
                    val queryIdx = event.sizeOrQueryIndex
                    while (sizes.isNotEmpty() && inactive[sizes.peek().second]) {
                        sizes.poll()
                    }
                    if (sizes.isNotEmpty()) {
                        ans[queryIdx] = sizes.peek().first
                    }
                }
            }
        }

        return ans
    }
}

struct Event {
    let time: Int
    let type: Int    // 0: interval start, 1: query, 2: interval end
    let size: Int?   // interval size for start/end events
    let idx: Int?    // index for interval (for start/end) or query index (for query)
}

struct Item: Comparable {
    let size: Int
    let idx: Int

    static func < (lhs: Item, rhs: Item) -> Bool {
        return lhs.size < rhs.size
    }
}

class Solution {
    func minInterval(_ intervals: [[Int]], _ queries: [Int]) -> [Int] {
        var events = [Event]()

        // Create events for intervals
        for (idx, interval) in intervals.enumerated() {
            let start = interval[0]
            let end = interval[1]
            let intervalSize = end - start + 1
            events.append(Event(time: start, type: 0, size: intervalSize, idx: idx))
            events.append(Event(time: end, type: 2, size: intervalSize, idx: idx))
        }

        // Create events for queries
        for (i, q) in queries.enumerated() {
            events.append(Event(time: q, type: 1, size: nil, idx: i))
        }

        // Sort by time and type (end before query)
        events.sort { (a, b) in
            if a.time != b.time {
                return a.time < b.time
            }
            return a.type < b.type
        }

        // Min heap storing [size, index]
        var sizes = Heap<Item>()
        var ans = Array(repeating: -1, count: queries.count)
        var inactive = Array(repeating: false, count: intervals.count)

        for event in events {
            if event.type == 0 {  // Interval start
                let intervalSize = event.size!
                let idx = event.idx!
                sizes.insert(Item(size: intervalSize, idx: idx))
            } else if event.type == 2 { // Interval end
                let idx = event.idx!
                inactive[idx] = true
            } else { // Query
                let queryIdx = event.idx!
                while !sizes.isEmpty, inactive[sizes.min!.idx] {
                    sizes.removeMin()
                }
                if !sizes.isEmpty {
                    ans[queryIdx] = sizes.min!.size
                }
            }
        }

        return ans
    }
}

Time & Space Complexity

Time complexity: $O((n + m) \log (n + m))$
Space complexity: $O(n + m)$

Where $m$ is the length of the array $queries$ and $n$ is the length of the array $intervals$ .

3. Min Heap

Intuition

For each query q, we want the length of the smallest interval [l, r] such that
l ≤ q ≤ r. If no interval covers q, the answer is -1.

A very efficient way to do this is:

process queries in sorted order
as queries increase, we add intervals whose start ≤ q
among those active intervals, remove any that end before q
the smallest valid interval is always at the top of a min heap

The heap is ordered by interval length, so the smallest covering interval is easy to find.

Algorithm

Sort intervals by their start value.
Sort the queries, but keep their original order for the final answer.
Initialize:
- a min heap minHeap storing (interval_length, interval_end)
- an index i = 0 to iterate through intervals
- a map res to store answers for each query
For each query q in sorted order:
- While there are intervals whose start ≤ q:
  - push (r - l + 1, r) into the heap
  - move i forward
- While the heap is not empty and the top interval ends before q:
  - pop it from the heap
- If the heap is not empty:
  - the top element's length is the answer for q
- Otherwise:
  - the answer for q is -1
- Store the result for q
Return the answers in the original query order.

class Solution:
    def minInterval(self, intervals: List[List[int]], queries: List[int]) -> List[int]:
        intervals.sort()
        minHeap = []
        res = {}
        i = 0
        for q in sorted(queries):
            while i < len(intervals) and intervals[i][0] <= q:
                l, r = intervals[i]
                heapq.heappush(minHeap, (r - l + 1, r))
                i += 1

            while minHeap and minHeap[0][1] < q:
                heapq.heappop(minHeap)
            res[q] = minHeap[0][0] if minHeap else -1
        return [res[q] for q in queries]

public class Solution {
    public int[] minInterval(int[][] intervals, int[] queries) {
        Arrays.sort(intervals, Comparator.comparingInt(a -> a[0]));
        PriorityQueue<int[]> minHeap = new PriorityQueue<>((a, b) -> Integer.compare(a[0], b[0]));
        Map<Integer, Integer> res = new HashMap<>();
        int i = 0;
        for (int q : Arrays.stream(queries).sorted().toArray()) {
            while (i < intervals.length && intervals[i][0] <= q) {
                int l = intervals[i][0];
                int r = intervals[i][1];
                minHeap.offer(new int[]{r - l + 1, r});
                i++;
            }

            while (!minHeap.isEmpty() && minHeap.peek()[1] < q) {
                minHeap.poll();
            }
            res.put(q, minHeap.isEmpty() ? -1 : minHeap.peek()[0]);
        }
        int[] result = new int[queries.length];
        for (int j = 0; j < queries.length; j++) {
            result[j] = res.get(queries[j]);
        }
        return result;
    }
}

class Solution {
public:
    vector<int> minInterval(vector<vector<int>>& intervals, vector<int>& queries) {
        // Sort intervals based on the start value
        sort(intervals.begin(), intervals.end(), [](auto& a, auto& b) {
            return a[0] < b[0];
        });

        vector<int> sortedQueries = queries;
        sort(sortedQueries.begin(), sortedQueries.end());
        map<int, int> res;

        auto cmp = [](const vector<int>& a, const vector<int>& b) {
            return a[0] > b[0] || (a[0] == b[0] && a[1] > b[1]);
        };
        priority_queue<vector<int>, vector<vector<int>>, decltype(cmp)> minHeap(cmp);

        int i = 0;
        for (int q : sortedQueries) {
            while (i < intervals.size() && intervals[i][0] <= q) {
                int l = intervals[i][0];
                int r = intervals[i][1];
                minHeap.push({r - l + 1, r});
                i++;
            }

            while (!minHeap.empty() && minHeap.top()[1] < q) {
                minHeap.pop();
            }

            res[q] = minHeap.empty() ? -1 : minHeap.top()[0];
        }

        vector<int> result(queries.size());
        for (int j = 0; j < queries.size(); j++) {
            result[j] = res[queries[j]];
        }
        return result;
    }
};

/**
 * const { MinPriorityQueue } = require('@datastructures-js/priority-queue');
 */

class Solution {
    /**
     * @param {number[][]} intervals
     * @param {number[]} queries
     * @return {number[]}
     */
    minInterval(intervals, queries) {
        intervals.sort((a, b) => a[0] - b[0]);
        const minHeap = new MinPriorityQueue((entry) => entry[0]);
        const res = {};
        let i = 0;

        const sortedQueries = [...queries].sort((a, b) => a - b);

        for (const q of sortedQueries) {
            while (i < intervals.length && intervals[i][0] <= q) {
                const [l, r] = intervals[i];
                minHeap.enqueue([r - l + 1, r]);
                i += 1;
            }

            while (!minHeap.isEmpty() && minHeap.front()[1] < q) {
                minHeap.dequeue();
            }

            res[q] = !minHeap.isEmpty() ? minHeap.front()[0] : -1;
        }

        return queries.map((q) => res[q]);
    }
}

public class Solution {
    public int[] MinInterval(int[][] intervals, int[] queries) {
        Array.Sort(intervals, (a, b) => a[0].CompareTo(b[0]));

        var minHeap = new PriorityQueue<(int Size, int End), int>();
        var res = new Dictionary<int, int>();
        int i = 0;

        int[] sortedQueries = queries.OrderBy(q => q).ToArray();
        foreach (int q in sortedQueries) {
            while (i < intervals.Length && intervals[i][0] <= q) {
                int l = intervals[i][0];
                int r = intervals[i][1];
                minHeap.Enqueue((r - l + 1, r), r - l + 1);
                i++;
            }

            while (minHeap.Count > 0 && minHeap.Peek().End < q) {
                minHeap.Dequeue();
            }

            res[q] = minHeap.Count == 0 ? -1 : minHeap.Peek().Size;
        }

        int[] result = new int[queries.Length];
        for (int j = 0; j < queries.Length; j++) {
            result[j] = res[queries[j]];
        }

        return result;
    }
}

func minInterval(intervals [][]int, queries []int) []int {
    sort.Slice(intervals, func(i, j int) bool {
        return intervals[i][0] < intervals[j][0]
    })

    queriesWithIdx := make([][2]int, len(queries))
    for i, q := range queries {
        queriesWithIdx[i] = [2]int{q, i}
    }
    sort.Slice(queriesWithIdx, func(i, j int) bool {
        return queriesWithIdx[i][0] < queriesWithIdx[j][0]
    })

    comparator := func(a, b interface{}) int {
        pair1 := a.([2]int)
        pair2 := b.([2]int)
        if pair1[0] != pair2[0] {
            if pair1[0] < pair2[0] {
                return -1
            }
            return 1
        }
        return 0
    }

    pq := priorityqueue.NewWith(comparator)
    res := make([]int, len(queries))
    i := 0

    for _, qPair := range queriesWithIdx {
        q, originalIdx := qPair[0], qPair[1]

        for i < len(intervals) && intervals[i][0] <= q {
            size := intervals[i][1] - intervals[i][0] + 1
            pq.Enqueue([2]int{size, intervals[i][1]})
            i++
        }

        for !pq.Empty() {
            if top, _ := pq.Peek(); top.([2]int)[1] < q {
                pq.Dequeue()
            } else {
                break
            }
        }

        if !pq.Empty() {
            if top, _ := pq.Peek(); true {
                res[originalIdx] = top.([2]int)[0]
            }
        } else {
            res[originalIdx] = -1
        }
    }

    return res
}

class Solution {
    fun minInterval(intervals: Array<IntArray>, queries: IntArray): IntArray {
        intervals.sortBy { it[0] }

        val queriesWithIndex = queries.withIndex()
            .map { it.value to it.index }
            .sortedBy { it.first }

        val minHeap = PriorityQueue<Pair<Int, Int>>(compareBy { it.first })
        val result = IntArray(queries.size)
        var i = 0

        for ((q, originalIdx) in queriesWithIndex) {
            while (i < intervals.size && intervals[i][0] <= q) {
                val size = intervals[i][1] - intervals[i][0] + 1
                minHeap.offer(size to intervals[i][1])
                i++
            }

            while (minHeap.isNotEmpty() && minHeap.peek().second < q) {
                minHeap.poll()
            }

            result[originalIdx] = if (minHeap.isNotEmpty()) minHeap.peek().first else -1
        }

        return result
    }
}

struct HeapItem: Comparable {
    let size: Int
    let end: Int
    static func < (lhs: HeapItem, rhs: HeapItem) -> Bool {
        if lhs.size == rhs.size {
            return lhs.end < rhs.end
        }
        return lhs.size < rhs.size
    }
}

class Solution {
    func minInterval(_ intervals: [[Int]], _ queries: [Int]) -> [Int] {
        let sortedIntervals = intervals.sorted { $0[0] < $1[0] }
        var minHeap = Heap<HeapItem>()
        var res = [Int: Int]()
        var i = 0

        for q in queries.sorted() {
            while i < sortedIntervals.count && sortedIntervals[i][0] <= q {
                let l = sortedIntervals[i][0]
                let r = sortedIntervals[i][1]
                minHeap.insert(HeapItem(size: r - l + 1, end: r))
                i += 1
            }

            while !minHeap.isEmpty, minHeap.min!.end < q {
                minHeap.removeMin()
            }
            res[q] = minHeap.isEmpty ? -1 : minHeap.min!.size
        }

        return queries.map { res[$0] ?? -1 }
    }
}

Time & Space Complexity

Time complexity: $O(n \log n + m \log m)$
Space complexity: $O(n + m)$

Where $m$ is the length of the array $queries$ and $n$ is the length of the array $intervals$ .

4. Min Segment Tree (Lazy Propagation)

Intuition

For each query q, we need the length of the smallest interval [l, r] such that l <= q <= r.
If no interval covers q, the answer is -1.

This solution treats the number line as a set of discrete important points (all interval endpoints and all queries).
Then it uses a segment tree that supports:

Range update: “For every point covered by this interval, the best (minimum) interval length might become smaller.”
Point query: “At this query point, what is the minimum interval length that covers it?”

Since many intervals update large ranges, we use lazy propagation to apply updates efficiently without touching every point in the range.

Because coordinates can be large, we first do coordinate compression:

map every unique coordinate to a compact index 0..M-1
this allows the segment tree to work on a small index range

Algorithm

Collect all important points
- Add every interval start and end
- Add every query value
Coordinate compress
- Sort and deduplicate points
- Create a map compress[value] -> index
Build a min segment tree with lazy propagation
- Each node stores the minimum interval length applied to that segment
- Lazy values store “pending minimum updates” that still need to be pushed down
Apply each interval as a range update
- For interval [l, r]:
  - convert to compressed indices [L, R]
  - compute its length len = r - l + 1
  - update the segment tree on range [L, R] with len using:
    - tree[node] = min(tree[node], len)
  - lazy propagation ensures this is efficient
Answer each query with a point query
- Convert query q to compressed index idx
- Query the segment tree at idx to get the minimum covering length
- If the value is still infinity, return -1 (no interval covers it)
Return answers in the original query order.

class SegmentTree:
    def __init__(self, N):
        self.n = N
        self.tree = [float('inf')] * (4 * N)
        self.lazy = [float('inf')] * (4 * N)

    def propagate(self, treeidx, lo, hi):
        if self.lazy[treeidx] != float('inf'):
            self.tree[treeidx] = min(self.tree[treeidx], self.lazy[treeidx])
            if lo != hi:
                self.lazy[2 * treeidx + 1] = min(self.lazy[2 * treeidx + 1], self.lazy[treeidx])
                self.lazy[2 * treeidx + 2] = min(self.lazy[2 * treeidx + 2], self.lazy[treeidx])
            self.lazy[treeidx] = float('inf')

    def update(self, treeidx, lo, hi, left, right, val):
        self.propagate(treeidx, lo, hi)
        if lo > right or hi < left:
            return
        if lo >= left and hi <= right:
            self.lazy[treeidx] = min(self.lazy[treeidx], val)
            self.propagate(treeidx, lo, hi)
            return
        mid = (lo + hi) // 2
        self.update(2 * treeidx + 1, lo, mid, left, right, val)
        self.update(2 * treeidx + 2, mid + 1, hi, left, right, val)
        self.tree[treeidx] = min(self.tree[2 * treeidx + 1], self.tree[2 * treeidx + 2])

    def query(self, treeidx, lo, hi, idx):
        self.propagate(treeidx, lo, hi)
        if lo == hi:
            return self.tree[treeidx]
        mid = (lo + hi) // 2
        if idx <= mid:
            return self.query(2 * treeidx + 1, lo, mid, idx)
        else:
            return self.query(2 * treeidx + 2, mid + 1, hi, idx)

    def update_range(self, left, right, val):
        self.update(0, 0, self.n - 1, left, right, val)

    def query_point(self, idx):
        return self.query(0, 0, self.n - 1, idx)

class Solution:
    def minInterval(self, intervals: List[List[int]], queries: List[int]) -> List[int]:
        points = []
        for interval in intervals:
            points.append(interval[0])
            points.append(interval[1])
        for q in queries:
            points.append(q)

        # Compress the coordinates
        points = sorted(set(points))
        compress = {points[i]: i for i in range(len(points))}

        # Lazy Segment Tree
        segTree = SegmentTree(len(points))

        for interval in intervals:
            start = compress[interval[0]]
            end = compress[interval[1]]
            length = interval[1] - interval[0] + 1
            segTree.update_range(start, end, length)

        ans = []
        for q in queries:
            idx = compress[q]

            # query for minSize
            res = segTree.query_point(idx)
            ans.append(res if res != float('inf') else -1)
        return ans

class SegmentTree {
    int n;
    int[] tree;
    int[] lazy;

    SegmentTree(int N) {
        this.n = N;
        tree = new int[4 * N];
        lazy = new int[4 * N];
        Arrays.fill(tree, Integer.MAX_VALUE);
        Arrays.fill(lazy, Integer.MAX_VALUE);
    }

    void propagate(int treeidx, int lo, int hi) {
        if (lazy[treeidx] != Integer.MAX_VALUE) {
            tree[treeidx] = Math.min(tree[treeidx], lazy[treeidx]);
            if (lo != hi) {
                lazy[2 * treeidx + 1] = Math.min(lazy[2 * treeidx + 1], lazy[treeidx]);
                lazy[2 * treeidx + 2] = Math.min(lazy[2 * treeidx + 2], lazy[treeidx]);
            }
            lazy[treeidx] = Integer.MAX_VALUE;
        }
    }

    void update(int treeidx, int lo, int hi, int left, int right, int val) {
        propagate(treeidx, lo, hi);
        if (lo > right || hi < left) return;
        if (lo >= left && hi <= right) {
            lazy[treeidx] = Math.min(lazy[treeidx], val);
            propagate(treeidx, lo, hi);
            return;
        }
        int mid = (lo + hi) / 2;
        update(2 * treeidx + 1, lo, mid, left, right, val);
        update(2 * treeidx + 2, mid + 1, hi, left, right, val);
        tree[treeidx] = Math.min(tree[2 * treeidx + 1], tree[2 * treeidx + 2]);
    }

    int query(int treeidx, int lo, int hi, int idx) {
        propagate(treeidx, lo, hi);
        if (lo == hi) return tree[treeidx];
        int mid = (lo + hi) / 2;
        if (idx <= mid) return query(2 * treeidx + 1, lo, mid, idx);
        else return query(2 * treeidx + 2, mid + 1, hi, idx);
    }

    void update(int left, int right, int val) {
        update(0, 0, n - 1, left, right, val);
    }

    int query(int idx) {
        return query(0, 0, n - 1, idx);
    }
}

public class Solution {
    public int[] minInterval(int[][] intervals, int[] queries) {
        List<Integer> points = new ArrayList<>();
        for (int[] interval : intervals) {
            points.add(interval[0]);
            points.add(interval[1]);
        }
        for (int q : queries) {
            points.add(q);
        }
        points = new ArrayList<>(new HashSet<>(points));
        Collections.sort(points);

        // Compress the points to indices
        Map<Integer, Integer> compress = new HashMap<>();
        for (int i = 0; i < points.size(); i++) {
            compress.put(points.get(i), i);
        }

        // Create the segment tree
        SegmentTree segTree = new SegmentTree(points.size());

        // Update the segment tree with intervals
        for (int[] interval : intervals) {
            int start = compress.get(interval[0]);
            int end = compress.get(interval[1]);
            int length = interval[1] - interval[0] + 1;
            segTree.update(start, end, length);
        }

        // Query the segment tree for each query
        int[] ans = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            int idx = compress.get(queries[i]);
            int res = segTree.query(idx);
            ans[i] = (res == Integer.MAX_VALUE) ? -1 : res;
        }

        return ans;
    }
}

class SegmentTree {
public:
    int n;
    vector<int> tree;
    vector<int> lazy;

    SegmentTree(int N) {
        this->n = N;
        tree.assign(4 * N, INT_MAX);
        lazy.assign(4 * N, INT_MAX);
    }

    void propagate(int treeidx, int lo, int hi) {
        if (lazy[treeidx] != INT_MAX) {
            tree[treeidx] = min(tree[treeidx], lazy[treeidx]);
            if (lo != hi) {
                lazy[2 * treeidx + 1] = min(lazy[2 * treeidx + 1], lazy[treeidx]);
                lazy[2 * treeidx + 2] = min(lazy[2 * treeidx + 2], lazy[treeidx]);
            }
            lazy[treeidx] = INT_MAX;
        }
    }

    void update(int treeidx, int lo, int hi, int left, int right, int val) {
        propagate(treeidx, lo, hi);

        if (lo > right || hi < left) return;

        if (lo >= left && hi <= right) {
            lazy[treeidx] = min(lazy[treeidx], val);
            propagate(treeidx, lo, hi);
            return;
        }

        int mid = (lo + hi) / 2;
        update(2 * treeidx + 1, lo, mid, left, right, val);
        update(2 * treeidx + 2, mid + 1, hi, left, right, val);

        tree[treeidx] = min(tree[2 * treeidx + 1], tree[2 * treeidx + 2]);
    }

    int query(int treeidx, int lo, int hi, int idx) {
        propagate(treeidx, lo, hi);
        if (lo == hi) return tree[treeidx];

        int mid = (lo + hi) / 2;
        if (idx <= mid) return query(2 * treeidx + 1, lo, mid, idx);
        else return query(2 * treeidx + 2, mid + 1, hi, idx);
    }

    void update(int left, int right, int val) {
        update(0, 0, n - 1, left, right, val);
    }

    int query(int idx) {
        return query(0, 0, n - 1, idx);
    }
};

class Solution {
public:
    vector<int> minInterval(vector<vector<int>>& intervals, vector<int>& queries) {
        vector<int> points;

        for (const auto& interval : intervals) {
            points.push_back(interval[0]);
            points.push_back(interval[1]);
        }
        for (int q : queries) {
            points.push_back(q);
        }

        sort(points.begin(), points.end());
        points.erase(unique(points.begin(), points.end()), points.end());

        // compress the coordinates
        unordered_map<int, int> compress;
        for (int i = 0; i < points.size(); ++i) {
            compress[points[i]] = i;
        }

        SegmentTree segTree(points.size());

        for (const auto& interval : intervals) {
            int start = compress[interval[0]];
            int end = compress[interval[1]];
            int len = interval[1] - interval[0] + 1;
            segTree.update(start, end, len);
        }

        vector<int> ans;
        for (int q : queries) {
            int idx = compress[q];
            int res = segTree.query(idx);
            ans.push_back(res == INT_MAX ? -1 : res);
        }

        return ans;
    }
};

class SegmentTree {
    constructor(N) {
        this.n = N;
        this.tree = new Array(4 * N).fill(Infinity);
        this.lazy = new Array(4 * N).fill(Infinity);
    }

    /**
     * @param {number} treeidx
     * @param {number} lo
     * @param {number} hi
     * @return {void}
     */
    propagate(treeidx, lo, hi) {
        if (this.lazy[treeidx] !== Infinity) {
            this.tree[treeidx] = Math.min(
                this.tree[treeidx],
                this.lazy[treeidx],
            );
            if (lo !== hi) {
                this.lazy[2 * treeidx + 1] = Math.min(
                    this.lazy[2 * treeidx + 1],
                    this.lazy[treeidx],
                );
                this.lazy[2 * treeidx + 2] = Math.min(
                    this.lazy[2 * treeidx + 2],
                    this.lazy[treeidx],
                );
            }
            this.lazy[treeidx] = Infinity;
        }
    }

    /**
     * @param {number} treeidx
     * @param {number} lo
     * @param {number} hi
     * @param {number} left
     * @param {number} right
     * @param {number} val
     * @return {void}
     */
    update(treeidx, lo, hi, left, right, val) {
        this.propagate(treeidx, lo, hi);
        if (lo > right || hi < left) return;
        if (lo >= left && hi <= right) {
            this.lazy[treeidx] = Math.min(this.lazy[treeidx], val);
            this.propagate(treeidx, lo, hi);
            return;
        }
        const mid = Math.floor((lo + hi) / 2);
        this.update(2 * treeidx + 1, lo, mid, left, right, val);
        this.update(2 * treeidx + 2, mid + 1, hi, left, right, val);
        this.tree[treeidx] = Math.min(
            this.tree[2 * treeidx + 1],
            this.tree[2 * treeidx + 2],
        );
    }

    /**
     * @param {number} treeidx
     * @param {number} lo
     * @param {number} hi
     * @param {number} idx
     * @return {number}
     */
    query(treeidx, lo, hi, idx) {
        this.propagate(treeidx, lo, hi);
        if (lo === hi) return this.tree[treeidx];
        const mid = Math.floor((lo + hi) / 2);
        if (idx <= mid) return this.query(2 * treeidx + 1, lo, mid, idx);
        else return this.query(2 * treeidx + 2, mid + 1, hi, idx);
    }

    /**
     * @param {number} left
     * @param {number} right
     * @param {number} val
     * @return {number}
     */
    updateRange(left, right, val) {
        this.update(0, 0, this.n - 1, left, right, val);
    }

    /**
     * @param {number} idx
     * @return {number}
     */
    queryPoint(idx) {
        return this.query(0, 0, this.n - 1, idx);
    }
}

class Solution {
    /**
     * @param {number[][]} intervals
     * @param {number[]} queries
     * @return {number[]}
     */
    minInterval(intervals, queries) {
        const points = [];
        for (const interval of intervals) {
            points.push(interval[0]);
            points.push(interval[1]);
        }
        for (const q of queries) {
            points.push(q);
        }
        const uniquePoints = [...new Set(points)].sort((a, b) => a - b);
        const compress = new Map();
        uniquePoints.forEach((point, idx) => {
            compress.set(point, idx);
        });
        const segTree = new SegmentTree(uniquePoints.length);
        for (const interval of intervals) {
            const start = compress.get(interval[0]);
            const end = compress.get(interval[1]);
            const length = interval[1] - interval[0] + 1;
            segTree.updateRange(start, end, length);
        }
        const ans = [];
        for (const q of queries) {
            const idx = compress.get(q);
            const res = segTree.queryPoint(idx);
            ans.push(res === Infinity ? -1 : res);
        }
        return ans;
    }
}

public class SegmentTree {
    public int n;
    public int[] tree;
    public int[] lazy;

    public SegmentTree(int N) {
        this.n = N;
        tree = new int[4 * N];
        lazy = new int[4 * N];
        Array.Fill(tree, int.MaxValue);
        Array.Fill(lazy, int.MaxValue);
    }

    public void Propagate(int treeidx, int lo, int hi) {
        if (lazy[treeidx] != int.MaxValue) {
            tree[treeidx] = Math.Min(tree[treeidx], lazy[treeidx]);
            if (lo != hi) {
                lazy[2 * treeidx + 1] = Math.Min(lazy[2 * treeidx + 1], lazy[treeidx]);
                lazy[2 * treeidx + 2] = Math.Min(lazy[2 * treeidx + 2], lazy[treeidx]);
            }
            lazy[treeidx] = int.MaxValue;
        }
    }

    public void Update(int treeidx, int lo, int hi, int left, int right, int val) {
        Propagate(treeidx, lo, hi);
        if (lo > right || hi < left) return;
        if (lo >= left && hi <= right) {
            lazy[treeidx] = Math.Min(lazy[treeidx], val);
            Propagate(treeidx, lo, hi);
            return;
        }
        int mid = (lo + hi) / 2;
        Update(2 * treeidx + 1, lo, mid, left, right, val);
        Update(2 * treeidx + 2, mid + 1, hi, left, right, val);
        tree[treeidx] = Math.Min(tree[2 * treeidx + 1], tree[2 * treeidx + 2]);
    }

    public int Query(int treeidx, int lo, int hi, int idx) {
        Propagate(treeidx, lo, hi);
        if (lo == hi) return tree[treeidx];
        int mid = (lo + hi) / 2;
        if (idx <= mid) return Query(2 * treeidx + 1, lo, mid, idx);
        else return Query(2 * treeidx + 2, mid + 1, hi, idx);
    }

    public void Update(int left, int right, int val) {
        Update(0, 0, n - 1, left, right, val);
    }

    public int Query(int idx) {
        return Query(0, 0, n - 1, idx);
    }
}

public class Solution {
    public int[] MinInterval(int[][] intervals, int[] queries) {
        List<int> points = new List<int>();
        foreach (var interval in intervals) {
            points.Add(interval[0]);
            points.Add(interval[1]);
        }
        foreach (var q in queries) {
            points.Add(q);
        }
        points.Sort();
        points = new List<int>(new HashSet<int>(points));
        Dictionary<int, int> compress = new Dictionary<int, int>();
        for (int i = 0; i < points.Count; i++) {
            compress[points[i]] = i;
        }
        SegmentTree segTree = new SegmentTree(points.Count);
        foreach (var interval in intervals) {
            int start = compress[interval[0]];
            int end = compress[interval[1]];
            int length = interval[1] - interval[0] + 1;
            segTree.Update(start, end, length);
        }
        int[] ans = new int[queries.Length];
        for (int i = 0; i < queries.Length; i++) {
            int idx = compress[queries[i]];
            int res = segTree.Query(idx);
            ans[i] = (res == int.MaxValue) ? -1 : res;
        }
        return ans;
    }
}

type SegmentTree struct {
    n    int
    tree []int
    lazy []int
}

func NewSegmentTree(n int) *SegmentTree {
    tree := make([]int, 4*n)
    lazy := make([]int, 4*n)
    for i := range tree {
        tree[i] = math.MaxInt32
        lazy[i] = math.MaxInt32
    }
    return &SegmentTree{n: n, tree: tree, lazy: lazy}
}

func (st *SegmentTree) propagate(treeIdx, lo, hi int) {
    if st.lazy[treeIdx] != math.MaxInt32 {
        st.tree[treeIdx] = min(st.tree[treeIdx], st.lazy[treeIdx])
        if lo != hi {
            st.lazy[2*treeIdx+1] = min(st.lazy[2*treeIdx+1], st.lazy[treeIdx])
            st.lazy[2*treeIdx+2] = min(st.lazy[2*treeIdx+2], st.lazy[treeIdx])
        }
        st.lazy[treeIdx] = math.MaxInt32
    }
}

func (st *SegmentTree) updateRange(treeIdx, lo, hi, left, right, val int) {
    st.propagate(treeIdx, lo, hi)
    if lo > right || hi < left {
        return
    }
    if lo >= left && hi <= right {
        st.lazy[treeIdx] = min(st.lazy[treeIdx], val)
        st.propagate(treeIdx, lo, hi)
        return
    }
    mid := (lo + hi) / 2
    st.updateRange(2*treeIdx+1, lo, mid, left, right, val)
    st.updateRange(2*treeIdx+2, mid+1, hi, left, right, val)
    st.tree[treeIdx] = min(st.tree[2*treeIdx+1], st.tree[2*treeIdx+2])
}

func (st *SegmentTree) queryPoint(treeIdx, lo, hi, idx int) int {
    st.propagate(treeIdx, lo, hi)
    if lo == hi {
        return st.tree[treeIdx]
    }
    mid := (lo + hi) / 2
    if idx <= mid {
        return st.queryPoint(2*treeIdx+1, lo, mid, idx)
    }
    return st.queryPoint(2*treeIdx+2, mid+1, hi, idx)
}

func (st *SegmentTree) Update(left, right, val int) {
    st.updateRange(0, 0, st.n-1, left, right, val)
}

func (st *SegmentTree) Query(idx int) int {
    return st.queryPoint(0, 0, st.n-1, idx)
}

func minInterval(intervals [][]int, queries []int) []int {
    points := make(map[int]bool)
    for _, interval := range intervals {
        points[interval[0]] = true
        points[interval[1]] = true
    }
    for _, q := range queries {
        points[q] = true
    }

    pointsList := make([]int, 0, len(points))
    for point := range points {
        pointsList = append(pointsList, point)
    }
    sort.Ints(pointsList)

    compress := make(map[int]int)
    for i, point := range pointsList {
        compress[point] = i
    }

    segTree := NewSegmentTree(len(pointsList))

    for _, interval := range intervals {
        start := compress[interval[0]]
        end := compress[interval[1]]
        length := interval[1] - interval[0] + 1
        segTree.Update(start, end, length)
    }

    ans := make([]int, len(queries))
    for i, q := range queries {
        idx := compress[q]
        res := segTree.Query(idx)
        if res == math.MaxInt32 {
            ans[i] = -1
        } else {
            ans[i] = res
        }
    }

    return ans
}

class SegmentTree(private val n: Int) {
    private val tree = IntArray(4 * n) { Int.MAX_VALUE }
    private val lazy = IntArray(4 * n) { Int.MAX_VALUE }

    private fun propagate(treeIdx: Int, lo: Int, hi: Int) {
        if (lazy[treeIdx] != Int.MAX_VALUE) {
            tree[treeIdx] = minOf(tree[treeIdx], lazy[treeIdx])
            if (lo != hi) {
                lazy[2 * treeIdx + 1] = minOf(lazy[2 * treeIdx + 1], lazy[treeIdx])
                lazy[2 * treeIdx + 2] = minOf(lazy[2 * treeIdx + 2], lazy[treeIdx])
            }
            lazy[treeIdx] = Int.MAX_VALUE
        }
    }

    private fun updateRange(treeIdx: Int, lo: Int, hi: Int, left: Int, right: Int, value: Int) {
        propagate(treeIdx, lo, hi)
        if (lo > right || hi < left) return
        if (lo >= left && hi <= right) {
            lazy[treeIdx] = minOf(lazy[treeIdx], value)
            propagate(treeIdx, lo, hi)
            return
        }
        val mid = (lo + hi) / 2
        updateRange(2 * treeIdx + 1, lo, mid, left, right, value)
        updateRange(2 * treeIdx + 2, mid + 1, hi, left, right, value)
        tree[treeIdx] = minOf(tree[2 * treeIdx + 1], tree[2 * treeIdx + 2])
    }

    private fun queryPoint(treeIdx: Int, lo: Int, hi: Int, idx: Int): Int {
        propagate(treeIdx, lo, hi)
        if (lo == hi) return tree[treeIdx]
        val mid = (lo + hi) / 2
        return if (idx <= mid) {
            queryPoint(2 * treeIdx + 1, lo, mid, idx)
        } else {
            queryPoint(2 * treeIdx + 2, mid + 1, hi, idx)
        }
    }

    fun update(left: Int, right: Int, value: Int) {
        updateRange(0, 0, n - 1, left, right, value)
    }

    fun query(idx: Int): Int {
        return queryPoint(0, 0, n - 1, idx)
    }
}

class Solution {
    fun minInterval(intervals: Array<IntArray>, queries: IntArray): IntArray {
        val points = mutableSetOf<Int>()
        intervals.forEach { interval ->
            points.add(interval[0])
            points.add(interval[1])
        }
        queries.forEach { points.add(it) }

        val pointsList = points.sorted()
        val compress = pointsList.withIndex().associate { it.value to it.index }

        val segTree = SegmentTree(pointsList.size)

        intervals.forEach { interval ->
            val start = compress[interval[0]]!!
            val end = compress[interval[1]]!!
            val length = interval[1] - interval[0] + 1
            segTree.update(start, end, length)
        }

        return queries.map { q ->
            val idx = compress[q]!!
            val res = segTree.query(idx)
            if (res == Int.MAX_VALUE) -1 else res
        }.toIntArray()
    }
}

Time & Space Complexity

Time complexity: $O((n + m)\log k)$
Space complexity:
- $O(k)$ extra space.
- $O(m)$ space for the output array.

Where $m$ is the length of the array $queries$ , $n$ is the length of the array $intervals$ and $k$ is the number of unique points.

Common Pitfalls

Losing Original Query Order After Sorting

When sorting queries for efficient processing, the original indices must be preserved. Returning results in sorted query order instead of the original order produces incorrect output. Always pair each query with its original index before sorting.

Forgetting to Remove Expired Intervals from the Heap

Intervals that have ended (their right endpoint is less than the current query) must be removed from the heap before answering. Failing to pop expired intervals means the heap may return an interval that does not actually contain the query point.

Incorrect Event Ordering in Sweep Line Approach

When multiple events occur at the same coordinate, processing order matters. Interval end events should typically be processed before query events at the same point to ensure intervals ending exactly at the query point are still considered active.

Using Wrong Interval Length Calculation

The length of interval [l, r] is r - l + 1, not r - l. This off-by-one error affects which interval is considered smallest and leads to incorrect results when intervals differ by exactly one unit.

Not Handling Queries Outside All Intervals

When no interval covers a query, the answer must be -1. Forgetting to initialize the result array with -1 or not checking for an empty heap after removing expired intervals causes undefined or incorrect values to be returned.