1658. Minimum Operations to Reduce X to Zero (M)

Minimum Operations to Reduce X to Zero (M)

题目

You are given an integer array nums and an integer x. In one operation, you can either remove the leftmost or the rightmost element from the array nums and subtract its value from x. Note that this modifies the array for future operations.

Return the minimum number of operations to reduce x to exactly 0 if it's possible**, otherwise, return -1.

Example 1:

Input: nums = [1,1,4,2,3], x = 5
Output: 2
Explanation: The optimal solution is to remove the last two elements to reduce x to zero.

Example 2:

Input: nums = [5,6,7,8,9], x = 4
Output: -1

Example 3:

Input: nums = [3,2,20,1,1,3], x = 10
Output: 5
Explanation: The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero. 

Constraints:

• 1 <= nums.length <= 10^5
• 1 <= nums[i] <= 10^4
• 1 <= x <= 10^9

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题意

给定一个数组，从头部或尾部取出连续的若干数，使其和为指定值，求这样的若干数的最小数量。

思路

求头部或尾部连续的最少的数，可以转化为求数组中一个最长的子数组，使其和正好为数组之和减去指定值。而求这样的最长子数组可以用滑动窗口来解决。

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代码实现

Java

class Solution {
    public int minOperations(int[] nums, int x) {
        int target = -x;
        for (int num : nums) {
            target += num;
        }
        if (target <= 0) return target == 0 ? nums.length : -1;

        int sum = 0, maxLen = -1;
        int left = 0, right = 0;
        while (left < nums.length) {
            if (sum < target) {
                if (right == nums.length) break;
                sum += nums[right++];
            } else if (sum > target) {
                sum -= nums[left++];
            } else {
                maxLen = Math.max(maxLen, right - left);
                if (right == nums.length) break;
                sum += nums[right++];
                sum -= nums[left++];
            }
        }

        return maxLen == -1 ? -1 : nums.length - maxLen;
    }
}