Given an array of even number of elements, form groups of 2 using these array elements such that the difference between the group with highest sum and the one with lowest sum is maximum.
Note: An element can be a part of one group only and it has to be a part of at least 1 group.
Examples:
Input : arr[] = {1, 4, 9, 6}
Output : 10
Groups formed will be (1, 4) and (6, 9),
the difference between highest sum group
(6, 9) i.e 15 and lowest sum group (1, 4)
i.e 5 is 10.
Input : arr[] = {6, 7, 1, 11}
Output : 11
Groups formed will be (1, 6) and (7, 11),
the difference between highest sum group
(7, 11) i.e 18 and lowest sum group (1, 6)
i.e 7 is 11.
Simple Approach: We can solve this problem by making all possible combinations and checking each set of combination difference between the group with highest sum and with the lowest sum. A total of n*(n-1)/2 such groups would be formed (nC2).
Time Complexity: O(n^3), because it will take O(n^2) to generate groups and to check against each group n iterations will be needed thus overall it takes O(n^3) time.
Efficient Approach: We can use the greedy approach. Sort the whole array and our result is sum of last two elements minus sum of first two elements.
C++
// CPP program to find minimum difference // between groups of highest and lowest // sums. #include <bits/stdc++.h> #define ll long long int using namespace std; ll CalculateMax(ll arr[], int n) { // Sorting the whole array. sort(arr, arr + n); int min_sum = arr[0] + arr[1]; int max_sum = arr[n-1] + arr[n-2]; return abs(max_sum - min_sum); } // Driver code int main() { ll arr[] = { 6, 7, 1, 11 }; int n = sizeof(arr) / sizeof(arr[0]); cout << CalculateMax(arr, n) << endl; return 0; } |
Java
// Java program to find minimum difference // between groups of highest and lowest // sums. import java.util.Arrays; import java.io.*; class GFG { static int CalculateMax(int arr[], int n) { // Sorting the whole array. Arrays.sort(arr); int min_sum = arr[0] + arr[1]; int max_sum = arr[n-1] + arr[n-2]; return (Math.abs(max_sum - min_sum)); } // Driver code public static void main (String[] args) { int arr[] = { 6, 7, 1, 11 }; int n = arr.length; System.out.println (CalculateMax(arr, n)); } } |
Python3
# Python 3 program to find minimum difference # between groups of highest and lowest def CalculateMax(arr, n): # Sorting the whole array. arr.sort() min_sum = arr[0] + arr[1] max_sum = arr[n - 1] + arr[n - 2] return abs(max_sum - min_sum) # Driver code arr = [6, 7, 1, 11] n = len(arr) print(CalculateMax(arr, n)) # This code is contributed # by Shrikant13 |
C#
// C# program to find minimum difference // between groups of highest and lowest // sums. using System; public class GFG{ static int CalculateMax(int []arr, int n) { // Sorting the whole array. Array.Sort(arr); int min_sum = arr[0] + arr[1]; int max_sum = arr[n-1] + arr[n-2]; return (Math.Abs(max_sum - min_sum)); } // Driver code static public void Main (){ int []arr = { 6, 7, 1, 11 }; int n = arr.Length; Console.WriteLine(CalculateMax(arr, n)); } //This code is contributed by Sachin. } |
PHP
<?php // PHP program to find minimum // difference between groups of // highest and lowest sums. function CalculateMax($arr, $n) { // Sorting the whole array. sort($arr); $min_sum = $arr[0] + $arr[1]; $max_sum = $arr[$n - 1] + $arr[$n - 2]; return abs($max_sum - $min_sum); } // Driver code $arr = array (6, 7, 1, 11 ); $n = sizeof($arr); echo CalculateMax($arr, $n), "\n" ; // This code is contributed by ajit ?> |
Output:
11
Time Complexity: O (n * log n)
Further Optimization :
Instead of sorting, we can find maximum two and minimum two in linear time and reduce time complexity to O(n).
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