646. Maximum Length of Pair Chain - Explanation

Prerequisites

Before attempting this problem, you should be comfortable with:

Sorting - Pairs must be sorted by end value for greedy approach or by start value for binary search
Greedy Algorithms - The optimal solution uses interval scheduling greedy strategy
Dynamic Programming - Both top-down and bottom-up DP approaches are presented
Binary Search - Used in the optimized DP solution similar to Longest Increasing Subsequence

1. Recursion

Intuition

A pair [a, b] can follow pair [c, d] in a chain only if d < a. To build the longest chain, we can use recursion to explore all possibilities: for each pair, we either include it in the chain (if valid) or skip it. Sorting by the second element helps us process pairs in an order that makes chain-building more intuitive.

Algorithm

Sort pairs by their second element (end value).
Use recursion with parameters i (current index) and j (index of the last pair added to the chain, or -1 if none).
At each step, try skipping pair i. If j is -1 or the last pair's end is less than the current pair's start, also try including pair i.
Return the maximum chain length found.

class Solution:
    def findLongestChain(self, pairs: List[List[int]]) -> int:
        n = len(pairs)
        pairs.sort(key=lambda x: x[1])

        def dfs(i, j):
            if i == n:
                return 0

            res = dfs(i + 1, j)
            if j == -1 or pairs[j][1] < pairs[i][0]:
                res = max(res, 1 + dfs(i + 1, i))

            return res

        return dfs(0, -1)

public class Solution {
    public int findLongestChain(int[][] pairs) {
        int n = pairs.length;
        Arrays.sort(pairs, (a, b) -> Integer.compare(a[1], b[1]));
        return dfs(pairs, 0, -1, n);
    }

    private int dfs(int[][] pairs, int i, int j, int n) {
        if (i == n) {
            return 0;
        }

        int res = dfs(pairs, i + 1, j, n);
        if (j == -1 || pairs[j][1] < pairs[i][0]) {
            res = Math.max(res, 1 + dfs(pairs, i + 1, i, n));
        }

        return res;
    }
}

class Solution {
public:
    int findLongestChain(vector<vector<int>>& pairs) {
        int n = pairs.size();
        sort(pairs.begin(), pairs.end(), [](const auto& a, const auto& b) {
            return a[1] < b[1];
        });

        return dfs(pairs, 0, -1, n);
    }

private:
    int dfs(vector<vector<int>>& pairs, int i, int j, int n) {
        if (i == n) {
            return 0;
        }

        int res = dfs(pairs, i + 1, j, n);
        if (j == -1 || pairs[j][1] < pairs[i][0]) {
            res = max(res, 1 + dfs(pairs, i + 1, i, n));
        }

        return res;
    }
};

class Solution {
    /**
     * @param {number[][]} pairs
     * @return {number}
     */
    findLongestChain(pairs) {
        pairs.sort((a, b) => a[1] - b[1]);
        let n = pairs.length;

        const dfs = (i, j) => {
            if (i === n) {
                return 0;
            }

            let res = dfs(i + 1, j);
            if (j === -1 || pairs[j][1] < pairs[i][0]) {
                res = Math.max(res, 1 + dfs(i + 1, i));
            }

            return res;
        };

        return dfs(0, -1);
    }
}

public class Solution {
    public int FindLongestChain(int[][] pairs) {
        int n = pairs.Length;
        Array.Sort(pairs, (a, b) => a[1].CompareTo(b[1]));
        return Dfs(pairs, 0, -1, n);
    }

    private int Dfs(int[][] pairs, int i, int j, int n) {
        if (i == n) {
            return 0;
        }

        int res = Dfs(pairs, i + 1, j, n);
        if (j == -1 || pairs[j][1] < pairs[i][0]) {
            res = Math.Max(res, 1 + Dfs(pairs, i + 1, i, n));
        }

        return res;
    }
}

func findLongestChain(pairs [][]int) int {
    n := len(pairs)
    sort.Slice(pairs, func(i, j int) bool {
        return pairs[i][1] < pairs[j][1]
    })

    var dfs func(i, j int) int
    dfs = func(i, j int) int {
        if i == n {
            return 0
        }

        res := dfs(i+1, j)
        if j == -1 || pairs[j][1] < pairs[i][0] {
            take := 1 + dfs(i+1, i)
            if take > res {
                res = take
            }
        }

        return res
    }

    return dfs(0, -1)
}

class Solution {
    fun findLongestChain(pairs: Array<IntArray>): Int {
        val n = pairs.size
        pairs.sortBy { it[1] }

        fun dfs(i: Int, j: Int): Int {
            if (i == n) {
                return 0
            }

            var res = dfs(i + 1, j)
            if (j == -1 || pairs[j][1] < pairs[i][0]) {
                res = maxOf(res, 1 + dfs(i + 1, i))
            }

            return res
        }

        return dfs(0, -1)
    }
}

class Solution {
    func findLongestChain(_ pairs: [[Int]]) -> Int {
        var pairs = pairs.sorted { $0[1] < $1[1] }
        let n = pairs.count

        func dfs(_ i: Int, _ j: Int) -> Int {
            if i == n {
                return 0
            }

            var res = dfs(i + 1, j)
            if j == -1 || pairs[j][1] < pairs[i][0] {
                res = max(res, 1 + dfs(i + 1, i))
            }

            return res
        }

        return dfs(0, -1)
    }
}

Time & Space Complexity

Time complexity: $O(2 ^ n)$
Space complexity: $O(n)$ for recursion stack.

2. Dynamic Programming (Top-Down)

Intuition

The recursive solution has overlapping subproblems since we may reach the same state (i, j) through different paths. By caching results in a 2D array, we avoid redundant computation. Each state represents the maximum chain length achievable starting from index i when the last included pair was at index j.

Algorithm

Sort pairs by their second element.
Initialize a 2D memoization array dp with -1 values.
Use the same recursive logic as before, but check dp[i][j+1] before computing. Store results before returning.
The offset j+1 handles the case where j = -1 (no previous pair selected).

class Solution:
    def findLongestChain(self, pairs: List[List[int]]) -> int:
        n = len(pairs)
        pairs.sort(key=lambda x: x[1])
        dp = [[-1] * (n + 1) for _ in range(n)]

        def dfs(i, j):
            if i == n:
                return 0
            if dp[i][j + 1] != -1:
                return dp[i][j + 1]

            res = dfs(i + 1, j)
            if j == -1 or pairs[j][1] < pairs[i][0]:
                res = max(res, 1 + dfs(i + 1, i))

            dp[i][j + 1] = res
            return res

        return dfs(0, -1)

public class Solution {
    private int[][] dp;

    public int findLongestChain(int[][] pairs) {
        int n = pairs.length;
        Arrays.sort(pairs, (a, b) -> Integer.compare(a[1], b[1]));
        dp = new int[n][n + 1];

        for (int[] row : dp) {
            Arrays.fill(row, -1);
        }

        return dfs(pairs, 0, -1, n);
    }

    private int dfs(int[][] pairs, int i, int j, int n) {
        if (i == n) {
            return 0;
        }
        if (dp[i][j + 1] != -1) {
            return dp[i][j + 1];
        }

        int res = dfs(pairs, i + 1, j, n);
        if (j == -1 || pairs[j][1] < pairs[i][0]) {
            res = Math.max(res, 1 + dfs(pairs, i + 1, i, n));
        }

        dp[i][j + 1] = res;
        return res;
    }
}

class Solution {
public:
    vector<vector<int>> dp;

    int findLongestChain(vector<vector<int>>& pairs) {
        int n = pairs.size();
        sort(pairs.begin(), pairs.end(), [](const auto& a, const auto& b) {
            return a[1] < b[1];
        });

        dp = vector<vector<int>>(n, vector<int>(n + 1, -1));
        return dfs(pairs, 0, -1, n);
    }

private:
    int dfs(vector<vector<int>>& pairs, int i, int j, int n) {
        if (i == n) {
            return 0;
        }
        if (dp[i][j + 1] != -1) {
            return dp[i][j + 1];
        }

        int res = dfs(pairs, i + 1, j, n);
        if (j == -1 || pairs[j][1] < pairs[i][0]) {
            res = max(res, 1 + dfs(pairs, i + 1, i, n));
        }

        dp[i][j + 1] = res;
        return res;
    }
};

class Solution {
    /**
     * @param {number[][]} pairs
     * @return {number}
     */
    findLongestChain(pairs) {
        pairs.sort((a, b) => a[1] - b[1]);
        let n = pairs.length;
        let dp = Array.from({ length: n }, () => new Array(n + 1).fill(-1));

        const dfs = (i, j) => {
            if (i === n) {
                return 0;
            }
            if (dp[i][j + 1] !== -1) {
                return dp[i][j + 1];
            }

            let res = dfs(i + 1, j);
            if (j === -1 || pairs[j][1] < pairs[i][0]) {
                res = Math.max(res, 1 + dfs(i + 1, i));
            }

            dp[i][j + 1] = res;
            return res;
        };

        return dfs(0, -1);
    }
}

public class Solution {
    private int[,] dp;

    public int FindLongestChain(int[][] pairs) {
        int n = pairs.Length;
        Array.Sort(pairs, (a, b) => a[1].CompareTo(b[1]));
        dp = new int[n, n + 1];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= n; j++) {
                dp[i, j] = -1;
            }
        }
        return Dfs(pairs, 0, -1, n);
    }

    private int Dfs(int[][] pairs, int i, int j, int n) {
        if (i == n) {
            return 0;
        }
        if (dp[i, j + 1] != -1) {
            return dp[i, j + 1];
        }

        int res = Dfs(pairs, i + 1, j, n);
        if (j == -1 || pairs[j][1] < pairs[i][0]) {
            res = Math.Max(res, 1 + Dfs(pairs, i + 1, i, n));
        }

        dp[i, j + 1] = res;
        return res;
    }
}

func findLongestChain(pairs [][]int) int {
    n := len(pairs)
    sort.Slice(pairs, func(i, j int) bool {
        return pairs[i][1] < pairs[j][1]
    })

    dp := make([][]int, n)
    for i := range dp {
        dp[i] = make([]int, n+1)
        for j := range dp[i] {
            dp[i][j] = -1
        }
    }

    var dfs func(i, j int) int
    dfs = func(i, j int) int {
        if i == n {
            return 0
        }
        if dp[i][j+1] != -1 {
            return dp[i][j+1]
        }

        res := dfs(i+1, j)
        if j == -1 || pairs[j][1] < pairs[i][0] {
            take := 1 + dfs(i+1, i)
            if take > res {
                res = take
            }
        }

        dp[i][j+1] = res
        return res
    }

    return dfs(0, -1)
}

class Solution {
    fun findLongestChain(pairs: Array<IntArray>): Int {
        val n = pairs.size
        pairs.sortBy { it[1] }
        val dp = Array(n) { IntArray(n + 1) { -1 } }

        fun dfs(i: Int, j: Int): Int {
            if (i == n) {
                return 0
            }
            if (dp[i][j + 1] != -1) {
                return dp[i][j + 1]
            }

            var res = dfs(i + 1, j)
            if (j == -1 || pairs[j][1] < pairs[i][0]) {
                res = maxOf(res, 1 + dfs(i + 1, i))
            }

            dp[i][j + 1] = res
            return res
        }

        return dfs(0, -1)
    }
}

class Solution {
    func findLongestChain(_ pairs: [[Int]]) -> Int {
        var pairs = pairs.sorted { $0[1] < $1[1] }
        let n = pairs.count
        var dp = [[Int]](repeating: [Int](repeating: -1, count: n + 1), count: n)

        func dfs(_ i: Int, _ j: Int) -> Int {
            if i == n {
                return 0
            }
            if dp[i][j + 1] != -1 {
                return dp[i][j + 1]
            }

            var res = dfs(i + 1, j)
            if j == -1 || pairs[j][1] < pairs[i][0] {
                res = max(res, 1 + dfs(i + 1, i))
            }

            dp[i][j + 1] = res
            return res
        }

        return dfs(0, -1)
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(n ^ 2)$

3. Dynamic Programming (Bottom-Up)

Intuition

We can build the solution iteratively. For each pair, we look at all previous pairs and find the longest chain that can be extended by the current pair. After sorting by end values, dp[i] represents the longest chain ending at pair i.

Algorithm

Sort pairs by their second element.
Initialize a DP array where dp[i] = 1 (each pair alone forms a chain of length 1).
For each pair i, check all previous pairs j. If pairs[j][1] < pairs[i][0], then pair i can extend the chain ending at j, so update dp[i] = max(dp[i], dp[j] + 1).
Return the maximum value in the DP array.

class Solution:
    def findLongestChain(self, pairs: List[List[int]]) -> int:
        n = len(pairs)
        pairs.sort(key=lambda x: x[1])
        dp = [1] * n

        for i in range(n):
            for j in range(i):
                if pairs[j][1] < pairs[i][0]:
                    dp[i] = max(dp[i], dp[j] + 1)

        return max(dp)

public class Solution {
    public int findLongestChain(int[][] pairs) {
        int n = pairs.length;
        Arrays.sort(pairs, (a, b) -> Integer.compare(a[1], b[1]));
        int[] dp = new int[n];
        Arrays.fill(dp, 1);

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (pairs[j][1] < pairs[i][0]) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
        }

        return Arrays.stream(dp).max().getAsInt();
    }
}

class Solution {
public:
    int findLongestChain(vector<vector<int>>& pairs) {
        int n = pairs.size();
        sort(pairs.begin(), pairs.end(), [](const auto& a, const auto& b) {
            return a[1] < b[1];
        });

        vector<int> dp(n, 1);

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (pairs[j][1] < pairs[i][0]) {
                    dp[i] = max(dp[i], dp[j] + 1);
                }
            }
        }

        return *max_element(dp.begin(), dp.end());
    }
};

class Solution {
    /**
     * @param {number[][]} pairs
     * @return {number}
     */
    findLongestChain(pairs) {
        let n = pairs.length;
        pairs.sort((a, b) => a[1] - b[1]);
        let dp = new Array(n).fill(1);

        for (let i = 0; i < n; i++) {
            for (let j = 0; j < i; j++) {
                if (pairs[j][1] < pairs[i][0]) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
        }

        return Math.max(...dp);
    }
}

public class Solution {
    public int FindLongestChain(int[][] pairs) {
        int n = pairs.Length;
        Array.Sort(pairs, (a, b) => a[1].CompareTo(b[1]));
        int[] dp = new int[n];
        Array.Fill(dp, 1);

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (pairs[j][1] < pairs[i][0]) {
                    dp[i] = Math.Max(dp[i], dp[j] + 1);
                }
            }
        }

        return dp.Max();
    }
}

func findLongestChain(pairs [][]int) int {
    n := len(pairs)
    sort.Slice(pairs, func(i, j int) bool {
        return pairs[i][1] < pairs[j][1]
    })

    dp := make([]int, n)
    for i := range dp {
        dp[i] = 1
    }

    for i := 0; i < n; i++ {
        for j := 0; j < i; j++ {
            if pairs[j][1] < pairs[i][0] {
                if dp[j]+1 > dp[i] {
                    dp[i] = dp[j] + 1
                }
            }
        }
    }

    maxVal := dp[0]
    for _, v := range dp {
        if v > maxVal {
            maxVal = v
        }
    }
    return maxVal
}

class Solution {
    fun findLongestChain(pairs: Array<IntArray>): Int {
        val n = pairs.size
        pairs.sortBy { it[1] }
        val dp = IntArray(n) { 1 }

        for (i in 0 until n) {
            for (j in 0 until i) {
                if (pairs[j][1] < pairs[i][0]) {
                    dp[i] = maxOf(dp[i], dp[j] + 1)
                }
            }
        }

        return dp.max()
    }
}

class Solution {
    func findLongestChain(_ pairs: [[Int]]) -> Int {
        let n = pairs.count
        var pairs = pairs.sorted { $0[1] < $1[1] }
        var dp = [Int](repeating: 1, count: n)

        for i in 0..<n {
            for j in 0..<i {
                if pairs[j][1] < pairs[i][0] {
                    dp[i] = max(dp[i], dp[j] + 1)
                }
            }
        }

        return dp.max()!
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(n)$

4. Dynamic Programming (Binary Search)

Intuition

This approach is similar to the Longest Increasing Subsequence optimization. We maintain a list where dp[i] stores the smallest end value of any chain of length i+1. When processing a new pair, we binary search to find where it can extend an existing chain. By keeping track of the smallest possible end values, we maximize opportunities for future extensions.

Algorithm

Sort pairs by their first element (start value).
Maintain a list dp where dp[i] is the smallest end value among all chains of length i+1.
For each pair [a, b], use binary search to find the position where a would be inserted (first position where dp[pos] >= a).
If pos equals dp.length, append b. Otherwise, update dp[pos] = min(dp[pos], b) to keep the smallest end value.
Return the length of dp.

class Solution:
    def findLongestChain(self, pairs: List[List[int]]) -> int:
        pairs.sort(key=lambda x: x[0])
        dp = []

        for a, b in pairs:
            pos = bisect_left(dp, a)
            if pos == len(dp):
                dp.append(b)
            else:
                dp[pos] = min(dp[pos], b)

        return len(dp)

public class Solution {
    public int findLongestChain(int[][] pairs) {
        Arrays.sort(pairs, Comparator.comparingInt(a -> a[0]));
        List<Integer> dp = new ArrayList<>();

        for (int[] pair : pairs) {
            int pos = binarySearch(dp, pair[0]);
            if (pos == dp.size()) {
                dp.add(pair[1]);
            } else {
                dp.set(pos, Math.min(dp.get(pos), pair[1]));
            }
        }

        return dp.size();
    }

    private int binarySearch(List<Integer> dp, int target) {
        int left = 0, right = dp.size() - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (dp.get(mid) < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }
}

class Solution {
public:
    int findLongestChain(vector<vector<int>>& pairs) {
        sort(pairs.begin(), pairs.end());
        vector<int> dp;

        for (auto& pair : pairs) {
            auto it = lower_bound(dp.begin(), dp.end(), pair[0]);
            if (it == dp.end()) {
                dp.push_back(pair[1]);
            } else {
                *it = min(*it, pair[1]);
            }
        }

        return dp.size();
    }
};

class Solution {
    /**
     * @param {number[][]} pairs
     * @return {number}
     */
    findLongestChain(pairs) {
        pairs.sort((a, b) => a[0] - b[0]);
        let dp = [];

        const binarySearch = (target) => {
            let left = 0,
                right = dp.length - 1;
            while (left <= right) {
                let mid = Math.floor((left + right) / 2);
                if (dp[mid] < target) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            }
            return left;
        };

        for (let i = 0; i < pairs.length; i++) {
            let pos = binarySearch(pairs[i][0]);
            if (pos === dp.length) {
                dp.push(pairs[i][1]);
            } else {
                dp[pos] = Math.min(dp[pos], pairs[i][1]);
            }
        }

        return dp.length;
    }
}

public class Solution {
    public int FindLongestChain(int[][] pairs) {
        Array.Sort(pairs, (a, b) => a[0].CompareTo(b[0]));
        List<int> dp = new List<int>();

        foreach (var pair in pairs) {
            int pos = BinarySearch(dp, pair[0]);
            if (pos == dp.Count) {
                dp.Add(pair[1]);
            } else {
                dp[pos] = Math.Min(dp[pos], pair[1]);
            }
        }

        return dp.Count;
    }

    private int BinarySearch(List<int> dp, int target) {
        int left = 0, right = dp.Count - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (dp[mid] < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }
}

func findLongestChain(pairs [][]int) int {
    sort.Slice(pairs, func(i, j int) bool {
        return pairs[i][0] < pairs[j][0]
    })
    dp := []int{}

    binarySearch := func(target int) int {
        left, right := 0, len(dp)-1
        for left <= right {
            mid := left + (right-left)/2
            if dp[mid] < target {
                left = mid + 1
            } else {
                right = mid - 1
            }
        }
        return left
    }

    for _, pair := range pairs {
        pos := binarySearch(pair[0])
        if pos == len(dp) {
            dp = append(dp, pair[1])
        } else {
            if pair[1] < dp[pos] {
                dp[pos] = pair[1]
            }
        }
    }

    return len(dp)
}

class Solution {
    fun findLongestChain(pairs: Array<IntArray>): Int {
        pairs.sortBy { it[0] }
        val dp = mutableListOf<Int>()

        fun binarySearch(target: Int): Int {
            var left = 0
            var right = dp.size - 1
            while (left <= right) {
                val mid = left + (right - left) / 2
                if (dp[mid] < target) {
                    left = mid + 1
                } else {
                    right = mid - 1
                }
            }
            return left
        }

        for (pair in pairs) {
            val pos = binarySearch(pair[0])
            if (pos == dp.size) {
                dp.add(pair[1])
            } else {
                dp[pos] = minOf(dp[pos], pair[1])
            }
        }

        return dp.size
    }
}

class Solution {
    func findLongestChain(_ pairs: [[Int]]) -> Int {
        let pairs = pairs.sorted { $0[0] < $1[0] }
        var dp = [Int]()

        func binarySearch(_ target: Int) -> Int {
            var left = 0
            var right = dp.count - 1
            while left <= right {
                let mid = left + (right - left) / 2
                if dp[mid] < target {
                    left = mid + 1
                } else {
                    right = mid - 1
                }
            }
            return left
        }

        for pair in pairs {
            let pos = binarySearch(pair[0])
            if pos == dp.count {
                dp.append(pair[1])
            } else {
                dp[pos] = min(dp[pos], pair[1])
            }
        }

        return dp.count
    }
}

Time & Space Complexity

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

5. Greedy

Intuition

This problem is identical to the classic interval scheduling problem. By sorting pairs by their end values, we can greedily select pairs. The key insight is that choosing the pair with the smallest end value leaves the most room for subsequent pairs. Whenever we find a pair whose start is greater than our current chain's end, we add it to the chain.

Algorithm

Sort pairs by their second element (end value).
Initialize length = 1 and end = pairs[0][1] (start with the first pair).
Iterate through remaining pairs. If a pair's start is greater than end, increment length and update end to this pair's end value.
Return length.

class Solution:
    def findLongestChain(self, pairs: List[List[int]]) -> int:
        pairs.sort(key=lambda p: p[1])
        length = 1
        end = pairs[0][1]

        for i in range(1, len(pairs)):
            if end < pairs[i][0]:
                length += 1
                end = pairs[i][1]

        return length

public class Solution {
    public int findLongestChain(int[][] pairs) {
        Arrays.sort(pairs, (a, b) -> Integer.compare(a[1], b[1]));
        int length = 1;
        int end = pairs[0][1];

        for (int i = 1; i < pairs.length; i++) {
            if (end < pairs[i][0]) {
                length++;
                end = pairs[i][1];
            }
        }

        return length;
    }
}

class Solution {
public:
    int findLongestChain(vector<vector<int>>& pairs) {
        sort(pairs.begin(), pairs.end(), [](const auto& a, const auto& b) {
            return a[1] < b[1];
        });

        int length = 1, end = pairs[0][1];

        for (int i = 1; i < pairs.size(); i++) {
            if (end < pairs[i][0]) {
                length++;
                end = pairs[i][1];
            }
        }

        return length;
    }
};

class Solution {
    /**
     * @param {number[][]} pairs
     * @return {number}
     */
    findLongestChain(pairs) {
        pairs.sort((a, b) => a[1] - b[1]);
        let length = 1;
        let end = pairs[0][1];

        for (let i = 1; i < pairs.length; i++) {
            if (end < pairs[i][0]) {
                length++;
                end = pairs[i][1];
            }
        }

        return length;
    }
}

public class Solution {
    public int FindLongestChain(int[][] pairs) {
        Array.Sort(pairs, (a, b) => a[1].CompareTo(b[1]));
        int length = 1;
        int end = pairs[0][1];

        for (int i = 1; i < pairs.Length; i++) {
            if (end < pairs[i][0]) {
                length++;
                end = pairs[i][1];
            }
        }

        return length;
    }
}

func findLongestChain(pairs [][]int) int {
    sort.Slice(pairs, func(i, j int) bool {
        return pairs[i][1] < pairs[j][1]
    })

    length := 1
    end := pairs[0][1]

    for i := 1; i < len(pairs); i++ {
        if end < pairs[i][0] {
            length++
            end = pairs[i][1]
        }
    }

    return length
}

class Solution {
    fun findLongestChain(pairs: Array<IntArray>): Int {
        pairs.sortBy { it[1] }
        var length = 1
        var end = pairs[0][1]

        for (i in 1 until pairs.size) {
            if (end < pairs[i][0]) {
                length++
                end = pairs[i][1]
            }
        }

        return length
    }
}

class Solution {
    func findLongestChain(_ pairs: [[Int]]) -> Int {
        let pairs = pairs.sorted { $0[1] < $1[1] }
        var length = 1
        var end = pairs[0][1]

        for i in 1..<pairs.count {
            if end < pairs[i][0] {
                length += 1
                end = pairs[i][1]
            }
        }

        return length
    }
}

Time & Space Complexity

Time complexity: $O(n\log n)$
Space complexity: $O(1)$ or $O(n)$ depending on the sorting algorithm.

Common Pitfalls

Sorting by the Wrong Element

A common mistake is sorting pairs by their first element (start value) instead of the second element (end value) when using the greedy approach. Sorting by end values is crucial because it ensures we always pick the pair that leaves the most room for subsequent pairs. Sorting by start values can lead to suboptimal chains.

Using Non-Strict Inequality

The problem requires that for pair [c, d] to follow pair [a, b], we need b < c (strictly less than). Using b <= c instead will incorrectly allow pairs where the end of one equals the start of another, violating the chain condition.

Confusing with Longest Increasing Subsequence

While this problem resembles LIS, the chain condition b < c is different from the standard LIS comparison. Directly applying LIS logic without adapting for the pair structure will produce incorrect results. The binary search approach requires careful handling of which element to compare and which to store.