Maximize Sum Of Array After K Negations

May 17, 2019

Introduction

User Accepted: 2691 User Tried: 2945 Total Accepted: 2760 Total Submissions: 6124 Difficulty: Easy Given an array A of integers, we must modify the array in the following way: we choose an i and replace A[i] with -A[i], and we repeat this process K times in total. (We 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: A = [4,2,3], K = 1
Output: 5
Explanation: Choose indices (1,) and A becomes [4,-2,3].

Example 2:

Input: A = [3,-1,0,2], K = 3
Output: 6
Explanation: Choose indices (1, 2, 2) and A becomes [3,1,0,2].

Example 3:

Input: A = [2,-3,-1,5,-4], K = 2
Output: 13
Explanation: Choose indices (1, 4) and A becomes [2,3,-1,5,4].

Note:

1 <= A.length <= 10000
1 <= K <= 10000
-100 <= A[i] <= 100

Solution

Golang:

func largestSumAfterKNegations(A []int, K int) int {
    n := len(A)
    if n == 0 {
        return 0
    }
    sort.Ints(A)
    i := 0
    for ;i < n && A[i] <= 0; i++ {
        if K == 0 || A[i] == 0 {
            return sumOf(A)
        }
        A[i] = -A[i]
        K--
    }
    if K % 2 == 1 {
        if i == n {
            A[n-1] = -A[n-1]
        } else if i - 1 >= 0 && A[i-1] < A[i] {
            A[i-1] = -A[i-1]
        } else {
            A[i] = -A[i]
        }
    }
    return sumOf(A)  
}

func sumOf(A []int) int {
    s := 0
    for _, a := range A {
        s += a
    }
    return s
}

Explanation

Lets sort array first, convert negatives to positives, stop on zero, if no zero let’s take min positive value and for even left K convert it. Result is the sum of array.