Thursday, 6 April 2017

Geeksforgeeks Solution For " Maximize sum after K negations "

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Problem :- Given an array of size n and a number k. We must modify array K number of times. Here modify array means in each operation we can replace any array element arr[i] by -arr[i]. We need to perform this operation in such a way that after K operations, sum of array must be maximum?

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Solution :- 

#include<bits/stdc++.h>
using namespace std;

int maximumSum(int arr[], int n, int k)
{
    for (int i=1; i<=k; i++)
    {
        int min = INT_MAX;
        int index = -1;
        for (int j=0; j<n; j++)
        {
            if (arr[j] < min)
            {
                min = arr[j];
                index = j;
            }
        }
        if (min == 0)
            break;

        arr[index] = -arr[index];
    }
    int sum = 0;
    for (int i=0; i<n; i++)
        sum += arr[i];
    return sum;
}

int main()
{
int t;
    cin>>t;
    while(t--)
    {
        int n,i,k,res=0;
        cin>>n;
        int arr[n];

        for(i=0;i<n;i++)
        cin>>arr[i];
        cin>>k;
    cout << maximumSum(arr, n, k);
    cout<<endl;
    }
return 0;
}

Output:-



Geeksforgeeks Solution For " Maximize sum after K negations "

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