You are given an array W[1], W[2], …, 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: 17Input : arr[] = [1, 1, 1, 1, 1, 1, 1, 1]
k = 3
Output: 2
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)))
Implementation:
C++
// C++ Program to find maximum weight // difference #include <iostream> #include <algorithm> using namespace std; // return the max value of two numbers 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 smallest k elements for ( int i = 0; i < k; i++) { sum1 += array[i]; } sort(array, array+n, greater< int >()); // Getting the sum of k largest elements for ( int i = 0; i < k; 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[0]); // 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 s = sum (array) # Getting the sum of smallest k elements s1 = sum (array[:k]) # Getting the sum greatest k elements array.sort(reverse = True ) s2 = 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
<?php // PHP Program to find maximum weight // difference // return the max value of two numbers function maxi( $a , $b ) { if ( $a > $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 ?> |
Javascript
<script> // JavaScript Code for Maximum Weight Difference function solve(array, n, k) { // sort the given array array.sort( function (a, b){ return a - b}); // Initializing the value to 0 let sum = 0, sum1 = 0, sum2 = 0; // Getting the sum of the array for (let i = 0; i < n; i++) { sum += array[i]; } // Getting the sum of first k elements for (let i = 0; i < k; i++) { sum1 += array[i]; } // Getting the sum of (n-k) elements for (let 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))); } let k = 2; let array = [ 8, 4, 5, 2, 10 ]; // calculate the numbers of elements // in the array let n = array.length; // call the solve function document.write(solve(array, n, k)); </script> |
17
Time Complexity: O(n log n), where n is the length of the array.
Auxiliary Space: O(1)
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