将 x 减到 0 的最小操作数

/*
 * Java Stream Solution:
 * This problem doesn't particularly lend itself to a pure Stream-based solution
 * because it involves finding subarrays, which are more naturally handled with loops.
 * However, we can still utilize streams for array processing such as calculating sums.
 */

import java.util.Arrays;

public class Solution {
    public int minOperations(int[] nums, int x) {
        int totalSum = Arrays.stream(nums).sum();
        int target = totalSum - x;
        if (target < 0) return -1;
        if (target == 0) return nums.length;

        int n = nums.length;
        int maxLength = -1, currentSum = 0;
        int left = 0;
        for (int right = 0; right < n; right++) {
            currentSum += nums[right];
            while (currentSum > target) {
                currentSum -= nums[left];
                left++;
            }
            if (currentSum == target) {
                maxLength = Math.max(maxLength, right - left + 1);
            }
        }
        return maxLength == -1 ? -1 : n - maxLength;
    }
}