LeetCode in Kotlin

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

@Suppress("NAME_SHADOWING")
class Solution {
    fun largestSumAfterKNegations(nums: IntArray, k: Int): Int {
        var k = k
        nums.sort()
        var minIndex = 0
        var i = 0
        while (i < nums.size && k > 0) {
            if (nums[i] < 0) {
                nums[i] *= -1
                k--
            }
            if (nums[minIndex] > nums[i]) {
                minIndex = i
            }
            i++
        }
        if (k and 1 == 1) {
            nums[minIndex] *= -1
        }
        return nums.sum()
    }
}

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