Maximize array sum after K negations | Set 1

05 Jul, 2022
Given an array of size n and a number k. We must modify array K a 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, the sum of the array must be maximum?
Examples : 
Input : arr[] = {-2, 0, 5, -1, 2}, K = 4Output: 10Explanation: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 = 3Output: 20
Naive approach: This problem has a very simple solution, we just have to replace the minimum element arr[i] in the array by -arr[i] for the current operation. In this way, we can make sum of the array maximum after K operations. One interesting case is, that once the minimum element becomes 0, we don’t need to make any more changes. 
Implementation:
C++
C
Java
Python3
C#
PHP
Javascript
// C++ program to maximize array sum after// k operations.#include <bits/stdc++.h>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 operationint 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 codeint 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;} // This code is contributed by Aditya Kumar (adityakumar129)
// C program to maximize array sum after// k operations.#include<stdio.h>#include<stdlib.h>#include <limits.h> // 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 operationint 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 codeint main(){    int arr[] = { -2, 0, 5, -1, 2 };    int k = 4;    int n = sizeof(arr) / sizeof(arr[0]);    printf("%d",maximumSum(arr, n, k));    return 0;} // This code is contributed by Aditya Kumar (adityakumar129)
// 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 Aditya Kumar (adityakumar129)
# 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 codearr = [-2, 0, 5, -1, 2]k = 4n = len(arr)print(maximumSum(arr, n, k)) # This code is contributed by Anant Agarwal.
// 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// PHP 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 operationfunction 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.?>
<script>     // Javascript 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 operationfunction maximumSum(arr, n, k){         // Modify array K number of times    for(let i = 1; i <= k; i++)    {        let min = +2147483647;        let index = -1;         // Find minimum element in array for        // current operation and modify it        // i.e; arr[j] --> -arr[j]        for(let 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    let sum = 0;    for(let i = 0; i < n; i++)        sum += arr[i];             return sum;}     // Driver codelet arr = [ -2, 0, 5, -1, 2 ];let k = 4;let n = arr.length; document.write(maximumSum(arr, n, k)); // This code is contributed by code_hunt </script>
10
Time Complexity: O(k*n) Auxiliary Space: O(1)
Approach 2 (Using Sort):  When there is a need to negate at most k elements.
Follow the steps below to solve this problem:
Sort the given array arr.Then for a given value of k, 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.If we find a 0 in the array we will immediately set k equal to 0 to maximize our result.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.
Sort the given array arr.
Then for a given value of k, 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.
If we find a 0 in the array we will immediately set k equal to 0 to maximize our result.
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
Java
Python3
C#
Javascript
// C++ program to find maximum array sum after at most k// negations.#include <bits/stdc++.h> 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 codeint 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 Aditya Kumar (adityakumar129)
// C++ program to find maximum array sum after at most k// negations.#include <stdio.h>#include <stdlib.h> int cmpfunc(const void* a, const void* b){    return (*(int*)a - *(int*)b);} int sol(int arr[], int n, int k){    int sum = 0;    int i = 0;     // Sorting given array using in-built sort function    qsort(arr, n, sizeof(int), cmpfunc);    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 codeint main(){    int arr[] = { -2, 0, 5, -1, 2 };    int n = sizeof(arr) / sizeof(arr[0]);    printf("%d", sol(arr, n, 4));    return 0;} // This code is contributed by Aditya Kumar (adityakumar129)
// 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));    }} // This code is contributed by Aditya Kumar (adityakumar129)
# 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 codearr = [-2, 0, 5, -1, 2] print(sol(arr, 4)) # This code is contributed by avanitrachhadiya2155
// 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 codepublic static void Main(string[] args){    int []arr = { -2, 0, 5, -1, 2 };         Console.Write(sol(arr, 4));}} // This code is contributed by rutvik_56
<script> // JavaScript program to find maximum array sum// after at most k negations.     function sol(arr, k)    {             // Sorting given array using in-built        // java sort function        arr.sort();        let sum = 0;          let 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 (let j = 0; j < arr.length; j++)        {            sum += arr[j];        }        return sum;    } // Driver code    let arr = [ -2, 0, 5, -1, 2 ];    document.write(sol(arr, 4)); // This code is contributed by souravghosh0416.                            </script>
10
Time Complexity: O(n*logn) Auxiliary Space: O(1)
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.
Sort the array in ascending order. Initialize i = 0.Increment i and multiply all negative elements by -1 till k becomes or a positive element is reached.Check if the end of the array has occurred. If true then go to (n-1)th element.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.Return sum of the array.
Sort the array in ascending order. Initialize i = 0.
Increment i and multiply all negative elements by -1 till k becomes or a positive element is reached.
Check if the end of the array has occurred. If true then go to (n-1)th element.
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.
Return sum of the array.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
C
// C++ program for the above approach#include <bits/stdc++.h>using namespace std; // Function to calculate sum of the arraylong 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 sumlong 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 Codeint main(){    int n = 5;    int k = 4;    long long int arr[5] = { -2, 0, 5, -1, 2 };    // Function Call    cout << maximizeSum(arr, n, k) << endl;    return 0;} // This code is contributed by Aditya Kumar (adityakumar129)
// Java program for the above approachimport 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 arr[] = { -2, 0, 5, -1, 2 };        int k = 4;        // Function Call        System.out.print(maximizeSum(arr, n, k));    }} // This code is contributed by Aditya Kumar (adityakumar129)
# Python3 program for the above approach # Function to calculate sum of the arraydef sumArray(arr, n):    sum = 0         # Iterate from 0 to n - 1    for i in range(n):        sum += arr[i]             return sum # Function to maximize sumdef 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 Coden = 5k = 4arr = [ -2, 0, 5, -1, 2 ]   # Function Callprint(maximizeSum(arr, n, k)) # This code is contributed by rohitsingh07052
// C# program for the above approachusing System; class GFG{     // Function to calculate sum of the arraystatic 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 sumstatic 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 Codestatic public void Main(){    int n = 5;    int[] arr = { -2, 0, 5, -1, 2 };    int k = 4;        // Function Call    Console.Write(maximizeSum(arr, n, k));}} // This code is contributed by shubhamsingh10
<script>// Javascript program for the above approach // Function to calculate sum of the arrayfunction  sumArray(arr,n){    let sum = 0;         // Iterate from 0 to n - 1    for(let i = 0; i < n; i++)    {        sum += arr[i];    }    return sum;} // Function to maximize sumfunction maximizeSum(arr,n,k){    (arr).sort(function(a,b){return a-b;});    let 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 Codelet n = 5;let k = 4;   let arr=[ -2, 0, 5, -1, 2 ]; // Function Calldocument.write(maximizeSum(arr, n, k)); // This code is contributed by ab2127</script>
// C program for the above approach#include <stdio.h>#include<stdlib.h> // Function to calculate sum of the arraylong 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;} void merge(long long int arr[], int l, int m, int r){    int i, j, k;    int n1 = m - l + 1;    int n2 = r - m;      /* create temp arrays */    int L[n1], R[n2];      /* Copy data to temp arrays L[] and R[] */    for (i = 0; i < n1; i++)        L[i] = arr[l + i];    for (j = 0; j < n2; j++)        R[j] = arr[m + 1 + j];      /* Merge the temp arrays back into arr[l..r]*/    i = 0; // Initial index of first subarray    j = 0; // Initial index of second subarray    k = l; // Initial index of merged subarray    while (i < n1 && j < n2) {        if (L[i] <= R[j]) {            arr[k] = L[i];            i++;        }        else {            arr[k] = R[j];            j++;        }        k++;    }      /* Copy the remaining elements of L[], if there    are any */    while (i < n1) {        arr[k] = L[i];        i++;        k++;    }      /* Copy the remaining elements of R[], if there    are any */    while (j < n2) {        arr[k] = R[j];        j++;        k++;    }}  /* l is for left index and r is right index of thesub-array of arr to be sorted */void mergeSort(long long int arr[], int l, int r){    if (l < r) {        // Same as (l+r)/2, but avoids overflow for        // large l and h        int m = l + (r - l) / 2;          // Sort first and second halves        mergeSort(arr, l, m);        mergeSort(arr, m + 1, r);          merge(arr, l, m, r);    }} // Function to maximize sumlong long int maximizeSum(long long int arr[], int n, int k){    mergeSort(arr,0,n-1);    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 Codeint main(){    int n = 5;    int k = 4;    long long int arr[] = { -2, 0, 5, -1, 2 };    // Function Call    printf("%lld",maximizeSum(arr,n,k));    return 0;}
10
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.
nitin mittal
nidhi_biet
RitikGarg2
rutvik_56
avanitrachhadiya2155
pratham76
yashupadhyayece19
rjrachit
code_hunt
sanjoy_62
rohitsingh07052
souravghosh0416
ab2127
akashkumarsmet18
SHUBHAMSINGH10
harendrakumar123
adityakumar129
rexomkar
hardikkoriintern
Arrays
Greedy
Arrays
Greedy
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.