Naive Logic

1. Define a recursive function solve(left_index, right_index, current_x).
2. Base case: If current_x == 0, return 0 (success).
3. Base case: If current_x < 0 or indices cross, return infinity (failure).
4. Recursive step: Try removing nums[left_index] and nums[right_index].
5. Return 1 + min(left_branch, right_branch).

Alternatively, an iterative brute force approach would involve checking every possible prefix sum combined with every possible suffix sum to see if they add up to x.

python
# Pseudo-code for iterative brute force
min_ops = infinity
for i from 0 to n:
    for j from 0 to n:
        if prefix_sum[i] + suffix_sum[j] == x:
             # Ensure prefix and suffix don't overlap
             if i + j <= n:
                 min_ops = min(min_ops, i + j)

Analysis

• Time Complexity: The recursive approach has a time complexity of $O(2^N)$ without memoization. The iterative prefix/suffix check is $O(N^2)$ because we iterate through all pairs of split points.
• Why it fails: The constraint nums.length <= 100,000 means an $O(N^2)$ or exponential solution will immediately trigger a Time Limit Exceeded (TLE) error. We need a linear $O(N)$ solution.

Core Insight for Minimum Operations to Reduce X to Zero

The key to solving "Minimum Operations to Reduce X to Zero" efficiently lies in inverting the problem statement.

Directly finding the minimum prefix and suffix is difficult because the two parts are disjoint (separated by the middle of the array). However, if we remove a prefix and a suffix that sum to x, the remaining subarray in the middle must sum to total_sum - x.

Let target = sum(nums) - x.

1. If we find a subarray with sum equal to target, the elements outside this subarray sum to x.
2. To minimize the operations (elements outside), we must maximize the length of this internal subarray.

The Algorithm Visualized: Imagine a window defined by left and right pointers. As right moves forward, the window expands and the window sum increases. If the sum exceeds the target, the left pointer moves forward to shrink the window until the sum is less than or equal to the target. We record the length whenever the window sum exactly matches the target.

Execution Flow

Consider nums = [1, 1, 4, 2, 3] and x = 5. Total Sum = 11. Target Subarray Sum = $11 - 5 = 6$.

1. Init: left=0, current_sum=0, max_len=-1.
2. right=0 (Val=1): current_sum becomes 1. $1 < 6$. No shrink.
3. right=1 (Val=1): current_sum becomes 2. $2 < 6$. No shrink.
4. right=2 (Val=4): current_sum becomes 6. $6 == 6$. Match found!
   • max_len = max(-1, 2 - 0 + 1) = 3. (Subarray: [1, 1, 4])
5. right=3 (Val=2): current_sum becomes 8. $8 > 6$. Shrink left.
   • Remove nums[0] (1). current_sum = 7. left = 1.
   • Remove nums[1] (1). current_sum = 6. left = 2.
   • $6 == 6$. Match found!
6. right=4 (Val=3): current_sum becomes 9. $9 > 6$. Shrink left.
   • Remove nums[2] (4). current_sum = 5. left = 3.
   • $5 < 6$. Stop shrinking.
7. End: max_len is 3.
8. Result: nums.length - max_len = $5 - 3 = 2$.

Correct Answer: 2 operations (remove the last two elements, 2 and 3, which sum to 5). Wait, looking at the example logic provided in the problem description: "remove the last two elements". In our array [1,1,4,2,3], the last two are 2, 3 summing to 5. The middle subarray is [1,1,4] summing to 6. Our algorithm correctly identified the middle subarray of length 3.

Proof of Correctness

The algorithm relies on the property that nums contains only positive integers. This ensures that current_sum is monotonically non-decreasing with respect to the right pointer and monotonically non-increasing with respect to the left pointer.

For any specific right index, there is at most one valid left index such that the sum of nums[left...right] equals target. By expanding right and shrinking left only when necessary (when sum exceeds target), we strictly visit every possible window that could sum to target. Since we track the maximum length among all valid windows, the complement (total length minus max window length) guarantees the minimum number of operations.

Reference Implementation

C++ Solution for LeetCode 1658

cpp
class Solution {
public:
    int minOperations(vector<int>& nums, int x) {
        int totalSum = 0;
        for (int num : nums) totalSum += num;
        
        int target = totalSum - x;
        
        // If x is greater than total sum, impossible
        if (target < 0) return -1;
        // If x equals total sum, remove all elements
        if (target == 0) return nums.size();
        
        int n = nums.size();
        int maxLength = -1;
        int currentSum = 0;
        int left = 0;
        
        for (int right = 0; right < n; ++right) {
            currentSum += nums[right];
            
            // Shrink window if sum exceeds target
            while (currentSum > target && left <= right) {
                currentSum -= nums[left];
                left++;
            }
            
            // Check if current window matches target
            if (currentSum == target) {
                maxLength = max(maxLength, right - left + 1);
            }
        }
        
        return maxLength == -1 ? -1 : n - maxLength;
    }
};

Java Solution for LeetCode 1658

java
class Solution {
    public int minOperations(int[] nums, int x) {
        int totalSum = 0;
        for (int num : nums) {
            totalSum += num;
        }
        
        int target = totalSum - x;
        
        // If target is negative, x is larger than the sum of all elements
        if (target < 0) return -1;
        // If target is 0, we need to remove all elements (x == totalSum)
        if (target == 0) return nums.length;
        
        int maxLength = -1;
        int currentSum = 0;
        int left = 0;
        
        for (int right = 0; right < nums.length; right++) {
            currentSum += nums[right];
            
            // Shrink window while sum is too large
            while (currentSum > target && left <= right) {
                currentSum -= nums[left];
                left++;
            }
            
            // If valid window found, update max length
            if (currentSum == target) {
                maxLength = Math.max(maxLength, right - left + 1);
            }
        }
        
        return maxLength == -1 ? -1 : nums.length - maxLength;
    }
}

Python Solution for LeetCode 1658

python
class Solution:
    def minOperations(self, nums: List[int], x: int) -> int:
        total_sum = sum(nums)
        target = total_sum - x
        
        # If x is larger than the total array sum
        if target < 0:
            return -1
        # If x equals the total sum, we remove everything
        if target == 0:
            return len(nums)
            
        max_len = -1
        current_sum = 0
        left = 0
        
        for right in range(len(nums)):
            current_sum += nums[right]
            
            # Shrink window while sum is too large
            while current_sum > target and left <= right:
                current_sum -= nums[left]
                left += 1
            
            # Update max_len if we found the target sum
            if current_sum == target:
                max_len = max(max_len, right - left + 1)
                
        return len(nums) - max_len if max_len != -1 else -1