1658. Minimum Operations to Reduce X to Zero - Explanation

1. Brute Force

Intuition

We can only remove elements from the left or right ends of the array. The total sum we remove must equal x. We try every possible combination: take some elements from the left (prefix) and some from the right (suffix), checking if their combined sum equals x. For each valid combination, we track the min number of elements removed.

Algorithm

First, check all suffix-only cases by iterating from the right and accumulating a suffixSum. If suffixSum == x, update res.
Then, for each prefix length (iterating from left), calculate prefixSum. If prefixSum == x, update res.
For each prefix, try combining it with suffixes by iterating from the right. If prefixSum + suffixSum == x, update res with the combined count.
Return -1 if no valid combination is found, otherwise return res.

class Solution:
    def minOperations(self, nums: List[int], x: int) -> int:
        n = len(nums)
        res = n + 1
        suffixSum = prefixSum = 0

        for i in range(n - 1, -1, -1):
            suffixSum += nums[i]
            if suffixSum == x:
                res = min(res, n - i)

        for i in range(n):
            prefixSum += nums[i]
            suffixSum = 0
            if prefixSum == x:
                res = min(res, i + 1)

            for j in range(n - 1, i, -1):
                suffixSum += nums[j]
                if prefixSum + suffixSum == x:
                    res = min(res, i + 1 + n - j)

        return -1 if res == n + 1 else res

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(1)$ extra space.

2. Prefix Sum + Binary Search

Intuition

Instead of recalculating prefix sums repeatedly, we precompute them. For each suffixSum we consider, we need to find a prefixSum that equals x - suffixSum. Since prefix sums are sorted in increasing order (all elements are positive), we can use binary search to quickly find if such a prefix exists. This eliminates the inner loop from the brute force approach.

Algorithm

Build a prefix sum array where prefixSum[i] represents the sum of the first i elements.
If the total sum is less than x, return -1 immediately.
Use binary search to find if any prefix alone equals x.
Iterate from the right, building suffixSum. For each suffixSum:
- If suffixSum == x, update res.
- If suffixSum > x, break early.
- Binary search in the prefix array for x - suffixSum, ensuring the prefix and suffix do not overlap.
Return -1 if no valid combination exists, otherwise return res.

class Solution:
    def minOperations(self, nums: List[int], x: int) -> int:
        n = len(nums)
        prefixSum = [0] * (n + 1)
        for i in range(n):
            prefixSum[i + 1] = prefixSum[i] + nums[i]

        if x > prefixSum[n]:
            return -1

        def binarySearch(target, m):
            l, r = 1, m
            index = n + 1

            while l <= r:
                mid = (l + r) >> 1
                if prefixSum[mid] >= target:
                    if prefixSum[mid] == target:
                        index = mid
                    r = mid - 1
                else:
                    l = mid + 1

            return index

        res = binarySearch(x, n)
        suffixSum = 0
        for i in range(n - 1, 0, -1):
            suffixSum += nums[i]
            if suffixSum == x:
                res = min(res, n - i)
                break
            if suffixSum > x: break
            res = min(res, binarySearch(x - suffixSum, i) + n - i)

        return -1 if res == n + 1 else res

Time & Space Complexity

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

3. Prefix Sum + Hash Map

Intuition

Here is a clever reframing: removing elements from both ends that sum to x is the same as keeping a contiguous subarray in the middle that sums to total - x. The problem becomes finding the longest subarray with sum equal to target = total - x. A hash map storing prefix sums lets us check in O(1) whether the required complement exists.

Algorithm

Calculate target = total - x. If target is negative, return -1. If target is 0, return the array length.
Use a hash map to store each prefixSum and its index. Initialize with {0: -1} to handle subarrays starting at index 0.
Iterate through the array, maintaining a running prefixSum:
- Check if prefixSum - target exists in the map. If so, we found a subarray with sum target.
- Track the max length of such subarrays.
- Store the current prefixSum in the map.
Return n - maxLength if a valid subarray was found, otherwise -1.

class Solution:
    def minOperations(self, nums: List[int], x: int) -> int:
        total = sum(nums)
        if total == x:
            return len(nums)

        target = total - x
        if target < 0:
            return -1

        res = -1
        prefixSum = 0
        prefixMap = {0: -1}  # prefixSum -> index

        for i, num in enumerate(nums):
            prefixSum += num
            if prefixSum - target in prefixMap:
                res = max(res, i - prefixMap[prefixSum - target])
            prefixMap[prefixSum] = i

        return len(nums) - res if res != -1 else -1

Time & Space Complexity

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

4. Sliding Window

Intuition

Building on the same insight as the hash map approach, we want the longest subarray summing to target = total - x. Since all elements are positive, the subarray sum increases as we expand and decreases as we shrink. This monotonic property allows us to use a sliding window: expand the right boundary to include more elements, and shrink from the left when the sum exceeds the target.

Algorithm

Calculate target = total - x. This is the sum we want the middle subarray to have.
Use two pointers l and r to define the current window, with curSum tracking the window sum.
Expand r to include elements. If curSum exceeds target, shrink from l until curSum <= target.
When curSum == target, record the window length if it is the max seen.
The answer is n - maxWindow. Return -1 if no valid window was found.

class Solution:
    def minOperations(self, nums: List[int], x: int) -> int:
        target = sum(nums) - x
        cur_sum = 0
        max_window = -1
        l = 0

        for r in range(len(nums)):
            cur_sum += nums[r]

            while l <= r and cur_sum > target:
                cur_sum -= nums[l]
                l += 1

            if cur_sum == target:
                max_window = max(max_window, r - l + 1)

        return -1 if max_window == -1 else len(nums) - max_window

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$ extra space.

Common Pitfalls

Not Recognizing the Problem Transformation

The key insight is that removing elements from both ends that sum to x is equivalent to finding the longest contiguous subarray that sums to total - x. Many solutions fail because they try to directly simulate removing from both ends, leading to inefficient or incorrect approaches.

Forgetting to Handle Edge Cases

When total == x, the answer is n (remove all elements). When total < x, the answer is -1 (impossible). When target = total - x is negative, return -1. Missing any of these edge cases causes wrong answers on specific test cases.

Overlapping Prefix and Suffix in Brute Force

In approaches that combine prefix and suffix sums, you must ensure the prefix and suffix do not overlap (i.e., the suffix must start after the prefix ends). A common bug is allowing j <= i instead of j > i, which double-counts elements and produces incorrect results.