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We are given with an array of positive and negative integers and a number K. The task is to find the maximum sum of the array after K changes in it’s elements. The single change operation here multiplies the single element by -1.

Approach used will be to convert every negative number to positive. If there are N negative numbers then, for this we will sort the array −

If N<K, after N operations every element will be positive, and we are left with K-N operations

Now if K-N is even then for remaining K-N operations changing signs will have no effect, do nothing.

If K-N is odd then for remaining K-N operations change the sign of least number (which will become negative) but the overall sum will be maximum.

Or

If N>K then change the sign of K negative numbers and add the array. Sum will be maximum.

Arr[]= { 0,-2,6,4,8,2,-3 } K=4

Maximum array sum is : 25

**Explanation** − The 4 changes in elements are

1. 0,2,6,4,8,2,-3 -2 changed to 2 2. 0,2,6,4,8,2,3 -3 changed to 3 3. 0,-2,6,4,8,2,3 2 changed to -2 4. 0,2,6,4,8,2,3 -2 changed to 2 Maximum sum is 25

Arr[]= { -1,-2,-3,-4,-5,-6,-7 } K=4

Maximum array sum is : 16

**Explanation** − The 4 changes in elements are

1. -1,-2,-3,-4,-5,-6,7 -7 changed to 7 2. -1,-2,-3,-4,-5,6,7 -6 changed to 6 3. -1,-2,-3,-4,5,6,7 -5 changed to 5 4. -1,-2,-3,4,5,6,7 -4 changed to 4 Maximum sum is 16

The integer array Arr[] is used to store the integers.

Integer ‘size’ stores the length of the array and K is initialized.

Function returnSum( int arr[], int n, int k) takes an array , its size and k as input and returns the maximum sum of it’s elements after exactly k operations.

First of all we will sort the array using sort(arr,arr+n)

Now we will apply the operation arr[i]*-1 to all negative elements until either last index is reached or k becomes 0.

If k is smaller than the number of negative elements, then the above step will change k - ve elements.

If k is larger and the remaining value of k will be checked if it is odd or even.

If remaining k is odd then we will change the value of minimum element by applying operation arr[i]*-1 once,where arr[i] is found at least. ( multiplying by -1 odd times is same as doing it once)

If remaining k is even then arr[i]*-1 will have no effect. Do nothing.

Calculate the sum of the whole array and return the result.

#include <bits/stdc++.h> using namespace std; int returnSum(int arr[], int n, int k){ // Sort the array elements sort(arr, arr + n); // Change signs of the negative elements // starting from the smallest //this loop will change the sign of -ve elements //for each k one -ve element is turned positive for(i=0;i<n;i++) if(k>0 && arr[i]<0){ arr[i]=arr[i]*-1; k--; } //if k is non zero and odd change sign of minimum element //once as it is same as changing its sign odd times if (k % 2 == 1) { int min = arr[0]; int pos=0; //index of minimum element for (i = 1; i < n; i++) if (arr[i]<min){ min = arr[i]; pos=i; } arr[pos] *= -1; } int sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; return sum; } int main(){ int Arr[] = { -3, 4, -3, 6, 8 }; int size =5; int K = 4; cout <<"Maximum array sum that can be obtained after exactly k changes" returnSum(Arr, size, K) << endl; return 0; }

Maximum array sum that can be obtained after exactly k changes : 24

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