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 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);    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 codearr = [-2, 0, 5, -1, 2]k = 4n = 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 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);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 codeint main(){    int arr[] = { -2, 0, 5, -1, 2 };     int n = sizeof(arr) / sizeof(arr);         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 codearr = [-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 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

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 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 = { -3, -2, -1, 5, 6 };     // Function Call    cout << maximizeSum(arr, n, k) << endl;    return 0;}

Java

 // Java program for the above approachimport java.util.*; 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){    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 Codepublic 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 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 = [ -3, -2, -1, 5, 6 ]   # Function Callprint(maximizeSum(arr, n, k)) # This code is contributed by rohitsingh07052

C#

 // 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 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.