# Maximize array sum after K negations | Set 1

• Difficulty Level : Easy
• Last Updated : 13 Jul, 2021

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?

Examples :

```Input : arr[] = {-2, 0, 5, -1, 2}
K = 4
Output: 10
Explanation:
1. Replace (-2) by -(-2), array becomes {2, 0, 5, -1, 2}
2. Replace (-1) by -(-1), array becomes {2, 0, 5, 1, 2}
3. Replace (0) by -(0), array becomes {2, 0, 5, 1, 2}
4. Replace (0) by -(0), array becomes {2, 0, 5, 1, 2}

Input : arr[] = {9, 8, 8, 5}
K = 3
Output: 20```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

This problem has very simple solution, we just have to replace the minimum element arr[i] in array by -arr[i] for current operation. In this way we can make sum of array maximum after K operations. One interesting case is, once the minimum element becomes 0, we don’t need to make any more changes.

## C++

 `// C++ program to maximize array sum after``// k operations.``#include ``using` `namespace` `std;` `// This function does k operations on array``// in a way that maximize the array sum.``// index --> stores the index of current minimum``// element for j'th operation``int` `maximumSum(``int` `arr[], ``int` `n, ``int` `k)``{``    ``// Modify array K number of times``    ``for` `(``int` `i = 1; i <= k; i++)``    ``{``        ``int` `min = INT_MAX;``        ``int` `index = -1;` `        ``// Find minimum element in array for``        ``// current operation and modify it``        ``// i.e; arr[j] --> -arr[j]``        ``for` `(``int` `j = 0; j < n; j++)``        ``{``            ``if` `(arr[j] < min) {``                ``min = arr[j];``                ``index = j;``            ``}``        ``}` `        ``// this the condition if we find 0 as``        ``// minimum element, so it will useless to``        ``// replace 0 by -(0) for remaining operations``        ``if` `(min == 0)``            ``break``;` `        ``// Modify element of array``        ``arr[index] = -arr[index];``    ``}` `    ``// Calculate sum of array``    ``int` `sum = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``sum += arr[i];``    ``return` `sum;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { -2, 0, 5, -1, 2 };``    ``int` `k = 4;``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << maximumSum(arr, n, k);``    ``return` `0;``}`

## Java

 `// Java program to maximize array``// sum after k operations.` `class` `GFG {``    ``// This function does k operations``    ``// on array in a way that maximize``    ``// the array sum. index --> stores``    ``// the index of current minimum``    ``// element for j'th operation``    ``static` `int` `maximumSum(``int` `arr[], ``int` `n, ``int` `k)``    ``{``        ``// Modify array K number of times``        ``for` `(``int` `i = ``1``; i <= k; i++)``        ``{``            ``int` `min = +``2147483647``;``            ``int` `index = -``1``;` `            ``// Find minimum element in array for``            ``// current operation and modify it``            ``// i.e; arr[j] --> -arr[j]``            ``for` `(``int` `j = ``0``; j < n; j++)``            ``{``                ``if` `(arr[j] < min)``                ``{``                    ``min = arr[j];``                    ``index = j;``                ``}``            ``}` `            ``// this the condition if we find 0 as``            ``// minimum element, so it will useless to``            ``// replace 0 by -(0) for remaining operations``            ``if` `(min == ``0``)``                ``break``;` `            ``// Modify element of array``            ``arr[index] = -arr[index];``        ``}` `        ``// Calculate sum of array``        ``int` `sum = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``sum += arr[i];``        ``return` `sum;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String arg[])``    ``{``        ``int` `arr[] = { -``2``, ``0``, ``5``, -``1``, ``2` `};``        ``int` `k = ``4``;``        ``int` `n = arr.length;``        ``System.out.print(maximumSum(arr, n, k));``    ``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python3 program to maximize``# array sum after k operations.` `# This function does k operations on array``# in a way that maximize the array sum.``# index --> stores the index of current``# minimum element for j'th operation`  `def` `maximumSum(arr, n, k):` `    ``# Modify array K number of times``    ``for` `i ``in` `range``(``1``, k ``+` `1``):` `        ``min` `=` `+``2147483647``        ``index ``=` `-``1` `        ``# Find minimum element in array for``        ``# current operation and modify it``        ``# i.e; arr[j] --> -arr[j]``        ``for` `j ``in` `range``(n):` `            ``if` `(arr[j] < ``min``):` `                ``min` `=` `arr[j]``                ``index ``=` `j` `        ``# this the condition if we find 0 as``        ``# minimum element, so it will useless to``        ``# replace 0 by -(0) for remaining operations``        ``if` `(``min` `=``=` `0``):``            ``break` `        ``# Modify element of array``        ``arr[index] ``=` `-``arr[index]` `    ``# Calculate sum of array``    ``sum` `=` `0``    ``for` `i ``in` `range``(n):``        ``sum` `+``=` `arr[i]``    ``return` `sum`  `# Driver code``arr ``=` `[``-``2``, ``0``, ``5``, ``-``1``, ``2``]``k ``=` `4``n ``=` `len``(arr)``print``(maximumSum(arr, n, k))` `# This code is contributed by Anant Agarwal.`

## C#

 `// C# program to maximize array``// sum after k operations.``using` `System;` `class` `GFG {` `    ``// This function does k operations``    ``// on array in a way that maximize``    ``// the array sum. index --> stores``    ``// the index of current minimum``    ``// element for j'th operation``    ``static` `int` `maximumSum(``int``[] arr, ``int` `n, ``int` `k)``    ``{` `        ``// Modify array K number of times``        ``for` `(``int` `i = 1; i <= k; i++)``        ``{``            ``int` `min = +2147483647;``            ``int` `index = -1;` `            ``// Find minimum element in array for``            ``// current operation and modify it``            ``// i.e; arr[j] --> -arr[j]``            ``for` `(``int` `j = 0; j < n; j++)``            ``{``                ``if` `(arr[j] < min)``                ``{``                    ``min = arr[j];``                    ``index = j;``                ``}``            ``}` `            ``// this the condition if we find``            ``// 0 as minimum element, so it``            ``// will useless to replace 0 by -(0)``            ``// for remaining operations``            ``if` `(min == 0)``                ``break``;` `            ``// Modify element of array``            ``arr[index] = -arr[index];``        ``}` `        ``// Calculate sum of array``        ``int` `sum = 0;``        ``for` `(``int` `i = 0; i < n; i++)``            ``sum += arr[i];``        ``return` `sum;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] arr = { -2, 0, 5, -1, 2 };``        ``int` `k = 4;``        ``int` `n = arr.Length;``        ``Console.Write(maximumSum(arr, n, k));``    ``}``}` `// This code is contributed by Nitin Mittal.`

## PHP

 ` stores``// the index of current minimum``// element for j'th operation``function` `maximumSum(``\$arr``, ``\$n``, ``\$k``)``{``    ``\$INT_MAX` `= 0;``    ``// Modify array K``    ``// number of times``    ``for` `(``\$i` `= 1; ``\$i` `<= ``\$k``; ``\$i``++)``    ``{``        ``\$min` `= ``\$INT_MAX``;``        ``\$index` `= -1;` `        ``// Find minimum element in``        ``// array for current operation``        ``// and modify it i.e;``        ``// arr[j] --> -arr[j]``        ``for` `(``\$j` `= 0; ``\$j` `< ``\$n``; ``\$j``++)``        ``{``            ``if` `(``\$arr``[``\$j``] < ``\$min``)``            ``{``                ``\$min` `= ``\$arr``[``\$j``];``                ``\$index` `= ``\$j``;``            ``}``        ``}` `        ``// this the condition if we``        ``// find 0 as minimum element, so``        ``// it will useless to replace 0``        ``// by -(0) for remaining operations``        ``if` `(``\$min` `== 0)``            ``break``;` `        ``// Modify element of array``        ``\$arr``[``\$index``] = -``\$arr``[``\$index``];``    ``}` `    ``// Calculate sum of array``    ``\$sum` `= 0;``    ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++)``        ``\$sum` `+= ``\$arr``[``\$i``];``    ``return` `\$sum``;``}` `// Driver Code``\$arr` `= ``array``(-2, 0, 5, -1, 2);``\$k` `= 4;``\$n` `= sizeof(``\$arr``) / sizeof(``\$arr``[0]);``echo` `maximumSum(``\$arr``, ``\$n``, ``\$k``);``    ` `// This code is contributed``// by nitin mittal.``?>`

## Javascript

 ``
Output
`10`

Time Complexity: O(k*n)
Auxiliary Space: O(1)

1. Approach 2 (Using Sort):
This approach is somewhat better than the above-discussed method. In this method, we will first sort the given array using the java in-built sort function which has the worst running time complexity of O(nlogn).
2. Then for a given value of k we will continue to iterate through the array till k remains greater than 0, If the value of the array at any index is less than 0 we will change its sign and decrement k by 1.
3. If we find a 0 in the array we will immediately set k equal to 0 to maximize our result.
4. In some cases, if we have all the values in an array greater than 0 we will change the sign of positive values, as our array is already sorted we will be changing signs of lower values present in the array which will eventually maximize our sum.

Below is the implementation of the above approach:

## C++

 `// C++ program to find maximum array sum``// after at most k negations.``#include ` `using` `namespace` `std;` `int` `sol(``int` `arr[], ``int` `n, ``int` `k)``{``    ``int` `sum = 0;``    ``int` `i = 0;``  ` `    ``// Sorting given array using in-built``    ``// sort function``    ``sort(arr, arr + n);``    ``while` `(k > 0)``    ``{``        ``// If we find a 0 in our``        ``// sorted array, we stop``        ``if` `(arr[i] >= 0)``            ``k = 0;``        ``else``        ``{``            ``arr[i] = (-1) * arr[i];``            ``k = k - 1;``        ``}``        ``i++;``       ` `    ``}` `    ``// Calculating sum``    ``for``(``int` `j = 0; j < n; j++)``    ``{``        ``sum += arr[j];``    ``}``    ``return` `sum;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { -2, 0, 5, -1, 2 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ` `    ``cout << sol(arr, n, 4) << endl;` `    ``return` `0;``}` `// This code is contributed by pratham76`

## Java

 `// Java program to find maximum array sum``// after at most k negations.``import` `java.util.Arrays;` `public` `class` `GFG {` `    ``static` `int` `sol(``int` `arr[], ``int` `k)``    ``{``        ``// Sorting given array using in-built``        ``// java sort function``        ``Arrays.sort(arr);``        ``int` `sum = ``0``;` `        ``int` `i = ``0``;``        ``while` `(k > ``0``)``        ``{``            ``// If we find a 0 in our``            ``// sorted array, we stop``            ``if` `(arr[i] >= ``0``)``                ``k = ``0``;` `            ``else``            ``{``                ``arr[i] = (-``1``) * arr[i];``                ``k = k - ``1``;``            ``}` `            ``i++;``        ``}` `        ``// Calculating sum``        ``for` `(``int` `j = ``0``; j < arr.length; j++)``        ``{``            ``sum += arr[j];``        ``}``        ``return` `sum;``    ``}``  ` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { -``2``, ``0``, ``5``, -``1``, ``2` `};``        ``System.out.println(sol(arr, ``4``));``    ``}``}`

## Python3

 `# Python3 program to find maximum array``# sum after at most k negations`  `def` `sol(arr, k):` `    ``# Sorting given array using``    ``# in-built java sort function``    ``arr.sort()` `    ``Sum` `=` `0``    ``i ``=` `0` `    ``while` `(k > ``0``):` `        ``# If we find a 0 in our``        ``# sorted array, we stop``        ``if` `(arr[i] >``=` `0``):``            ``k ``=` `0``        ``else``:``            ``arr[i] ``=` `(``-``1``) ``*` `arr[i]``            ``k ``=` `k ``-` `1` `        ``i ``+``=` `1` `    ``# Calculating sum``    ``for` `j ``in` `range``(``len``(arr)):``        ``Sum` `+``=` `arr[j]` `    ``return` `Sum`  `# Driver code``arr ``=` `[``-``2``, ``0``, ``5``, ``-``1``, ``2``]` `print``(sol(arr, ``4``))` `# This code is contributed by avanitrachhadiya2155`

## C#

 `// C# program to find maximum array sum``// after at most k negations.``using` `System;`` ` `class` `GFG{`` ` `static` `int` `sol(``int` `[]arr, ``int` `k)``{``    ` `    ``// Sorting given array using``    ``// in-built java sort function``    ``Array.Sort(arr);``    ` `    ``int` `sum = 0;``    ``int` `i = 0;``    ` `    ``while` `(k > 0)``    ``{``        ` `        ``// If we find a 0 in our``        ``// sorted array, we stop``        ``if` `(arr[i] >= 0)``            ``k = 0;` `        ``else``        ``{``            ``arr[i] = (-1) * arr[i];``            ``k = k - 1;``        ``}``        ``i++;``    ``}``    ` `    ``// Calculating sum``    ``for``(``int` `j = 0; j < arr.Length; j++)``    ``{``        ``sum += arr[j];``    ``}``    ``return` `sum;``}` `// Driver code``public` `static` `void` `Main(``string``[] args)``{``    ``int` `[]arr = { -2, 0, 5, -1, 2 };``    ` `    ``Console.Write(sol(arr, 4));``}``}` `// This code is contributed by rutvik_56`

## Javascript

 ``
Output
`10`

Time Complexity: O(n*logn)
Auxiliary Space: O(1)

#### Approach 3(Using Sort):

The above approach 2 is optimal when there is a need to negate at most k elements. To solve when there are exactly k negations the algorithm is given below.

1. Sort the array in ascending order. Initialize i = 0.
2. Increment i and multiply all negative elements by -1 till k becomes or a positive element is reached.
3. Check if the end of the array has occurred. If true then go to (n-1)th element.
4. If k ==0 or k is even, return the sum of all elements. Else multiply the absolute of minimum of ith or (i-1) th element by -1.
5. Return sum of the array.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``#include ``using` `namespace` `std;` `// Function to calculate sum of the array``long` `long` `int` `sumArray(``long` `long` `int``* arr, ``int` `n)``{``    ``long` `long` `int` `sum = 0;` `    ``// Iterate from 0 to n - 1``    ``for` `(``int` `i = 0; i < n; i++) {``        ``sum += arr[i];``    ``}``    ``return` `sum;``}` `// Function to maximize sum``long` `long` `int` `maximizeSum(``long` `long` `int` `arr[], ``int` `n, ``int` `k)``{``    ``sort(arr, arr + n);``    ``int` `i = 0;` `    ``// Iterate from 0 to n - 1``    ``for` `(i = 0; i < n; i++) {``        ``if` `(k && arr[i] < 0) {``            ``arr[i] *= -1;``            ``k--;``            ``continue``;``        ``}``        ``break``;``    ``}``    ``if` `(i == n)``        ``i--;` `    ``if` `(k == 0 || k % 2 == 0) {``        ``return` `sumArray(arr, n);``    ``}` `    ``if` `(i != 0 && ``abs``(arr[i]) >= ``abs``(arr[i - 1])) {``        ``i--;``    ``}` `    ``arr[i] *= -1;``    ``return` `sumArray(arr, n);``}` `// Driver Code``int` `main()``{``    ``int` `n = 5;``    ``int` `k = 4;``    ``long` `long` `int` `arr[5] = { -3, -2, -1, 5, 6 };` `    ``// Function Call``    ``cout << maximizeSum(arr, n, k) << endl;``    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG{` `// Function to calculate sum of the array``static` `int` `sumArray( ``int``[] arr, ``int` `n)``{``    ``int` `sum = ``0``;``   ` `    ``// Iterate from 0 to n - 1``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``        ``sum += arr[i];``    ``}``    ``return` `sum;``}`` ` `// Function to maximize sum``static` `int` `maximizeSum(``int` `arr[], ``int` `n, ``int` `k)``{``    ``Arrays.sort(arr);``    ``int` `i = ``0``;``   ` `    ``// Iterate from 0 to n - 1``    ``for``(i = ``0``; i < n; i++)``    ``{``        ``if` `(k != ``0` `&& arr[i] < ``0``)``        ``{``            ``arr[i] *= -``1``;``            ``k--;``            ``continue``;``        ``}``        ``break``;``    ``}``    ``if` `(i == n)``        ``i--;`` ` `    ``if` `(k == ``0` `|| k % ``2` `== ``0``)``    ``{``        ``return` `sumArray(arr, n);``    ``}`` ` `    ``if` `(i != ``0` `&& Math.abs(arr[i]) >=``        ``Math.abs(arr[i - ``1``]))``    ``{``        ``i--;``    ``}`` ` `    ``arr[i] *= -``1``;``    ``return` `sumArray(arr, n);``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``int` `n = ``5``;``    ``int` `k = ``4``;``    ``int` `arr[] = { -``3``, -``2``, -``1``, ``5``, ``6` `};``   ` `    ``// Function Call``    ``System.out.print(maximizeSum(arr, n, k));``}``}` `// This code is contributed by sanjoy_62`

## Python3

 `# Python3 program for the above approach` `# Function to calculate sum of the array``def` `sumArray(arr, n):``    ``sum` `=` `0``    ` `    ``# Iterate from 0 to n - 1``    ``for` `i ``in` `range``(n):``        ``sum` `+``=` `arr[i]``        ` `    ``return` `sum` `# Function to maximize sum``def` `maximizeSum(arr, n, k):``    ` `    ``arr.sort()``    ``i ``=` `0``  ` `    ``# Iterate from 0 to n - 1``    ``for` `i ``in` `range``(n):``        ``if` `(k ``and` `arr[i] < ``0``):``            ``arr[i] ``*``=` `-``1``            ``k ``-``=` `1``            ``continue``        ` `        ``break``    ` `    ``if` `(i ``=``=` `n):``        ``i ``-``=` `1` `    ``if` `(k ``=``=` `0` `or` `k ``%` `2` `=``=` `0``):``        ``return` `sumArray(arr, n)` `    ``if` `(i !``=` `0` `and` `abs``(arr[i]) >``=` `abs``(arr[i ``-` `1``])):``        ``i ``-``=` `1` `    ``arr[i] ``*``=` `-``1``    ``return` `sumArray(arr, n)` `# Driver Code``n ``=` `5``k ``=` `4``arr ``=` `[ ``-``3``, ``-``2``, ``-``1``, ``5``, ``6` `]``  ` `# Function Call``print``(maximizeSum(arr, n, k))` `# This code is contributed by rohitsingh07052`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{``    ` `// Function to calculate sum of the array``static` `int` `sumArray( ``int``[] arr, ``int` `n)``{``    ``int` `sum = 0;``   ` `    ``// Iterate from 0 to n - 1``    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ``sum += arr[i];``    ``}``    ``return` `sum;``}`` ` `// Function to maximize sum``static` `int` `maximizeSum(``int``[] arr, ``int` `n, ``int` `k)``{``    ``Array.Sort(arr);``    ``int` `i = 0;``   ` `    ``// Iterate from 0 to n - 1``    ``for``(i = 0; i < n; i++)``    ``{``        ``if` `(k != 0 && arr[i] < 0)``        ``{``            ``arr[i] *= -1;``            ``k--;``            ``continue``;``        ``}``        ``break``;``    ``}``    ``if` `(i == n)``        ``i--;`` ` `    ``if` `(k == 0 || k % 2 == 0)``    ``{``        ``return` `sumArray(arr, n);``    ``}`` ` `    ``if` `(i != 0 && Math.Abs(arr[i]) >=``                  ``Math.Abs(arr[i - 1]))``    ``{``        ``i--;``    ``}`` ` `    ``arr[i] *= -1;``    ``return` `sumArray(arr, n);``}` `// Driver Code``static` `public` `void` `Main()``{``    ``int` `n = 5;``    ``int` `k = 4;``    ``int``[] arr = { -3, -2, -1, 5, 6 };``   ` `    ``// Function Call``    ``Console.Write(maximizeSum(arr, n, k));``}``}` `// This code is contributed by shubhamsingh10`

## Javascript

 ``
Output
`15`

Time Complexity: O(n*logn)

Auxiliary Space: O(1)

Maximize array sum after K negations | Set 2

This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.