Introduction Link to heading
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 Link to heading
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 Link to heading
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.