Maximum Weight Difference

You are given an array W, W, …, W[N]. Choose K numbers among them such that the absolute difference between the sum of chosen numbers and the sum of remaining numbers is as large as possible.

Examples :

Input : arr[] = [8, 4, 5, 2, 10]
k = 2
Output: 17

Input : arr[] = [1, 1, 1, 1, 1, 1, 1, 1]
k = 3
Output: 2

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

There are two possibilities to get the desired answer. These two are:Choose k largest numbers or Choose k smallest numbers. Choose the best-suited option which fits according to the given values. This is because there are some cases in which the sum of smallest k numbers can be greater than rest of the array and there are some cases in which the sum of largest k numbers can be greater than rest of the sum of the numbers.

Approach :

• Sort the given array.
• Get the sum of all the numbers of the array and store it in sum
• Get the sum of first k numbers of the array and store it in sum1
• Get the sum of last k numbers of the array and store it in sum2
• Output the result which is : max(abs(S1-(S-S1)), abs(S2-(S-S2)))

C++

 // C++ Program to find maximum weight  // difference #include #include using namespace std;    // return the max value of two numbers int max(int a, int b) {     return a > b ? a : b; }    int solve(int array[], int n, int k) {     // sort the given array     sort(array, array + n);        // Initializing the value to 0     int sum = 0, sum1 = 0, sum2 = 0;        // Getting the sum of the array     for (int i = 0; i < n; i++) {         sum += array[i];     }        // Getting the sum of first k elements     for (int i = 0; i < k; i++) {         sum1 += array[i];     }        // Getting the sum of (n-k) elements     for (int i = k; i < n; i++) {         sum2 += array[i];     }        // Returning the maximum possible difference.     return max(abs(sum1 - (sum - sum1)), abs(sum2 -                                    (sum - sum2))); }    // Driver function int main() {     int k = 2;     int array[] = { 8, 4, 5, 2, 10 };        // calculate the numbers of elements in the array     int n = sizeof(array) / sizeof(array);        // call the solve function     cout << solve(array, n, k);        return 0; }

Java

 // JAVA Code for Maximum Weight Difference import java.util.*;    class GFG {            public static int solve(int array[], int n,                                         int k)     {         // sort the given array         Arrays.sort(array);                 // Initializing the value to 0         int sum = 0, sum1 = 0, sum2 = 0;                 // Getting the sum of the array         for (int i = 0; i < n; i++) {             sum += array[i];         }                 // Getting the sum of first k elements         for (int i = 0; i < k; i++) {             sum1 += array[i];         }                 // Getting the sum of (n-k) elements         for (int i = k; i < n; i++) {             sum2 += array[i];         }                 // Returning the maximum possible difference.         return Math.max(Math.abs(sum1 - (sum - sum1)),                        Math.abs(sum2 - (sum - sum2)));     }            /* Driver program to test above function */     public static void main(String[] args)      {         int k = 2;         int array[] = { 8, 4, 5, 2, 10 };                 // calculate the numbers of elements         // in the array         int n = array.length;                 // call the solve function         System.out.print(solve(array, n, k));                    } } // This code is contributed by Arnav Kr. Mandal.

Python

 def solve(array, k):        # Sorting array   array.sort()      # Getting the sum of all the elements   sum = sum(array)      # Getting the sum of first k elements   sum1 = sum(array[:k])      # Getting the sum last (n-k) elements   sum2 = sum(array[k:])      # Returning the maximum possible difference   return max(abs(s1-(s-s1)), abs(s2-(s-s2)))      # Driver function k = 2 array =[8, 4, 5, 2, 10] print(solve(array, k))

C#

 // C# Code for Maximum Weight Difference using System;    class GFG {            public static int solve(int []array, int n,                                         int k)     {                    // sort the given array         Array.Sort(array);                // Initializing the value to 0         int sum = 0, sum1 = 0, sum2 = 0;                // Getting the sum of the array         for (int i = 0; i < n; i++) {             sum += array[i];         }                // Getting the sum of first k elements         for (int i = 0; i < k; i++) {             sum1 += array[i];         }                // Getting the sum of (n-k) elements         for (int i = k; i < n; i++) {             sum2 += array[i];         }                // Returning the maximum possible difference.         return Math.Max(Math.Abs(sum1 - (sum - sum1)),                         Math.Abs(sum2 - (sum - sum2)));     }            /* Driver program to test above function */     public static void Main()      {         int k = 2;         int []array = { 8, 4, 5, 2, 10 };                // calculate the numbers of elements         // in the array         int n = array.Length;                // call the solve function         Console.WriteLine(solve(array, n, k));                    } }    // This code is contributed by vt_m.

PHP

 \$b)      {         return \$a;     }      else     {         return \$b;     } }    function solve(&\$arr, \$n, \$k) {     // sort the given array     sort(\$arr);        // Initializing the value to 0     \$sum = 0;     \$sum1 = 0;     \$sum2 = 0;        // Getting the sum of the array     for (\$i = 0; \$i < \$n; \$i++)      {         \$sum += \$arr[\$i];     }        // Getting the sum of first k elements     for (\$i = 0; \$i < \$k; \$i++)     {         \$sum1 += \$arr[\$i];     }        // Getting the sum of (n-k) elements     for (\$i = \$k; \$i < \$n; \$i++)      {         \$sum2 += \$arr[\$i];     }        // Returning the maximum possible difference.     return maxi(abs(\$sum1 - (\$sum - \$sum1)),                  abs(\$sum2 - (\$sum - \$sum2))); }    // DriverCode \$k = 2; \$arr = array(8, 4, 5, 2, 10 );    // calculate the numbers of  // elements in the array \$n = sizeof(\$arr);    // call the solve function echo (solve(\$arr, \$n, \$k));    // This code is contributed  // by Shivi_Aggarwal  ?>

Output:

17

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Improved By : vt_m, Shivi_Aggarwal

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