1658. Minimum Operations to Reduce X to Zero

Problem Description

Given an integer array nums and an integer x, you can remove elements from either the left or right end of the array and subtract their values from x. The goal is to find the minimum number of operations needed to reduce x to exactly 0. If it's impossible, return -1.

Solution Approach: Two Pointers (Optimal)

This approach leverages the observation that we're essentially looking for a subarray within nums whose sum, when subtracted from the total sum of nums, equals x. Finding the longest such subarray minimizes the number of operations (elements removed from both ends).

1. Calculate Target Sum: Compute the target sum s = sum(nums) - x. If s is negative, it's impossible to reach 0, so we return -1.

2. Two Pointers: Use two pointers, i (left) and j (right), to maintain a sliding window. i iterates through the array, adding each element to a running sum t.

3. Window Adjustment: If t exceeds s, we shrink the window by incrementing j and subtracting nums[j] from t. This ensures t always remains less than or equal to s.

4. Longest Subarray: When t equals s, we've found a subarray with the target sum. We update mx (maximum length) with the current window size (i - j + 1).

5. Result: The minimum number of operations is len(nums) - mx. If mx remains -1 (no subarray found), we return -1.

Time and Space Complexity

• Time Complexity: O(N), where N is the length of nums. We traverse the array once.
• Space Complexity: O(1). The algorithm uses a constant amount of extra space.

Code Implementation (Python)

class Solution:
    def minOperations(self, nums: List[int], x: int) -> int:
        s = sum(nums) - x
        if s < 0:
            return -1
        mx, t = -1, 0
        j = 0
        for i, v in enumerate(nums):
            t += v
            while j <= i and t > s:
                t -= nums[j]
                j += 1
            if t == s:
                mx = max(mx, i - j + 1)
        return len(nums) - mx if mx != -1 else -1
 

Code Implementation (Java)

class Solution {
    public int minOperations(int[] nums, int x) {
        int s = 0;
        for (int num : nums) {
            s += num;
        }
        s -= x;
        if (s < 0) {
            return -1;
        }
        int maxLen = -1;
        int sum = 0;
        int j = 0;
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            while (j <= i && sum > s) {
                sum -= nums[j++];
            }
            if (sum == s) {
                maxLen = Math.max(maxLen, i - j + 1);
            }
        }
        return maxLen == -1 ? -1 : nums.length - maxLen;
    }
}

Code Implementation (C++)

class Solution {
public:
    int minOperations(vector<int>& nums, int x) {
        int s = accumulate(nums.begin(), nums.end(), 0) - x;
        if (s < 0) return -1;
        int mx = -1, t = 0;
        int j = 0;
        for (int i = 0; i < nums.size(); ++i) {
            t += nums[i];
            while (j <= i && t > s) {
                t -= nums[j++];
            }
            if (t == s) {
                mx = max(mx, i - j + 1);
            }
        }
        return mx == -1 ? -1 : nums.size() - mx;
    }
};

(Similar implementations can be easily adapted for other languages like JavaScript, Go, etc.)

This two-pointer approach provides an efficient solution with optimal time complexity, making it a preferred method for solving this problem. The other approach using Hash Tables and Prefix Sums, while functionally correct, has a slightly higher space complexity due to the hash table.