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Maximize sum after K negations
Easy Accuracy: 48.5% Submissions: 22034 Points: 2

Given an array of integers of size N and a number K., Your must modify array arr[] exactly K number of times. Here modify array means in each operation you can replace any array element either arr[i] by -arr[i] or -arr[i] by arr[i]. You need to perform this operation in such a way that after K operations, the sum of the array must be maximum.

Example 1:

Input:
N = 5, K = 1
arr[] = {1, 2, -3, 4, 5}
Output:
15
Explanation:
We have k=1 so we can change -3 to 3 and
sum all the elements to produce 15 as output.

Example 2:

Input:
N = 10, K = 5
arr[] = {5, -2, 5, -4, 5, -12, 5, 5, 5, 20}
Output:
68
Explanation:
Here  we have k=5 so we turn -2, -4, -12 to
2, 4, and 12 respectively. Since we have
performed 3 operations so k is now 2. To get
maximum sum of array we can turn positive
turned 2 into negative and then positive
again so k is 0. Now sum is
5+5+4+5+12+5+5+5+20+2 = 68

You don't have to print anything, print ting is done by the driver code itself. You have to complete the function maximizeSum() which takes the array A[], its size N, and an integer K as inputs and returns the maximum possible sum.

Expected Time Complexity: O(N*logN)
Expected Auxiliary Space: O(1)

Constraints:
1 ≤ N,K ≤ 105
-109 ≤ Ai ≤ 109

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