LeetCode-in-Java

1005. Maximize Sum Of Array After K Negations

Easy

Given an integer array nums and an integer k, modify the array in the following way:

• choose an index i and replace nums[i] with -nums[i].

You should apply this process exactly k times. You may choose the same index i multiple times.

Return the largest possible sum of the array after modifying it in this way.

Example 1:

Input: nums = [4,2,3], k = 1

Output: 5

Explanation: Choose index 1 and nums becomes [4,-2,3].

Example 2:

Input: nums = [3,-1,0,2], k = 3

Output: 6

Explanation: Choose indices (1, 2, 2) and nums becomes [3,1,0,2].

Example 3:

Input: nums = [2,-3,-1,5,-4], k = 2

Output: 13

Explanation: Choose indices (1, 4) and nums becomes [2,3,-1,5,4].

Constraints:

• 1 <= nums.length <= 104
• -100 <= nums[i] <= 100
• 1 <= k <= 104

Solution

import java.util.Arrays;

public class Solution {
    public int largestSumAfterKNegations(int[] nums, int k) {
        Arrays.sort(nums);
        int minIndex = 0;
        for (int i = 0; i < nums.length && k > 0; i++) {
            if (nums[i] < 0) {
                nums[i] *= -1;
                k--;
            }
            if (nums[minIndex] > nums[i]) {
                minIndex = i;
            }
        }
        if ((k & 1) == 1) {
            nums[minIndex] *= -1;
        }
        return Arrays.stream(nums).sum();
    }
}

This site is open source. Improve this page.