1658. Minimum Operations to Reduce X to Zero
作者:互联网
package LeetCode_1658 /** * 1658. Minimum Operations to Reduce X to Zero * https://leetcode.com/problems/minimum-operations-to-reduce-x-to-zero/ * 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. 1 <= nums.length <= 105 2. 1 <= nums[i] <= 104 3. 1 <= x <= 109 * */ class Solution { /* * solution: Sliding Window, Time complexity:O(n), Space complexity:O(1) * 1. finding out the longest sub-array sum up to nums.sum-x, * 2. the result is nums.size - sub-array.size, * */ fun minOperations(nums: IntArray, x: Int): Int { val target = nums.sum() - x var lengthOfSubArray = Int.MIN_VALUE var left = 0 var currentSum = 0 for (right in nums.indices) { currentSum += nums[right] while (currentSum > target && left <= right) { currentSum -= nums[left] left++ } if (currentSum == target) { lengthOfSubArray = Math.max(lengthOfSubArray, right - left + 1) } } return if (lengthOfSubArray == Int.MIN_VALUE) -1 else nums.size - lengthOfSubArray } }
标签:Operations,operations,nums,reduce,zero,1658,Minimum,Input,Output 来源: https://www.cnblogs.com/johnnyzhao/p/13977911.html