Minimum difference between groups of size two
Given an array of even number of elements, form groups of 2 using these array elements such that the difference between the group with the highest sum and the one with the lowest sum is minimum.
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[] = {2, 6, 4, 3} Output : 1 Groups formed will be (2, 6) and (4, 3), the difference between highest sum group (2, 6) i.e 8 and lowest sum group (3, 4) i.e 7 is 1. Input : arr[] = {11, 4, 3, 5, 7, 1} Output : 3 Groups formed will be (1, 11), (4, 5) and (3, 7), the difference between highest sum group (1, 11) i.e 12 and lowest sum group (4, 5) i.e 9 is 3.
Simple Approach: A simple approach would be to try against all combinations of array elements and check against each set of combination difference between the group with the highest sum and the one with the lowest sum. A total of n*(n-1)/2 such groups would be formed (nC2).
Time Complexity: O(n^3) To generate groups n^2 iterations will be needed and to check against each group n iterations will be needed and hence n^3 iterations will be needed in the worst case.
Efficient Approach: Efficient approach would be to use the greedy approach. Sort the whole array and generate groups by selecting one element from the start of the array and one from the end.
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 calculate(ll a[], ll n) { // Sorting the whole array. sort(a, a + n); // Generating sum groups. vector<ll> s; for ( int i = 0, j = n - 1; i < j; i++, j--) s.push_back(a[i] + a[j]); ll mini = *min_element(s.begin(), s.end()); ll maxi = *max_element(s.begin(), s.end()); return abs (maxi - mini); } int main() { ll a[] = { 2, 6, 4, 3 }; int n = sizeof (a) / ( sizeof (a[0])); cout << calculate(a, n) << endl; return 0; } |
Java
// Java program to find minimum // difference between groups of // highest and lowest sums. import java.util.Arrays; import java.util.Collections; import java.util.Vector; class GFG { static long calculate( long a[], int n) { // Sorting the whole array. Arrays.sort(a); int i,j; // Generating sum groups. Vector<Long> s = new Vector<>(); for (i = 0 , j = n - 1 ; i < j; i++, j--) s.add((a[i] + a[j])); long mini = Collections.min(s); long maxi = Collections.max(s); return Math.abs(maxi - mini); } // Driver code public static void main(String[] args) { long a[] = { 2 , 6 , 4 , 3 }; int n = a.length; System.out.println(calculate(a, n)); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 program to find minimum # difference between groups of # highest and lowest sums. def calculate(a, n): # Sorting the whole array. a.sort(); # Generating sum groups. s = []; i = 0 ; j = n - 1 ; while (i < j): s.append((a[i] + a[j])); i + = 1 ; j - = 1 ; mini = min (s); maxi = max (s); return abs (maxi - mini); # Driver Code a = [ 2 , 6 , 4 , 3 ]; n = len (a); print (calculate(a, n)); # This is contributed by mits |
C#
// C# program to find minimum // difference between groups of // highest and lowest sums. using System; using System.Linq; using System.Collections.Generic; class GFG { static long calculate( long []a, int n) { // Sorting the whole array. Array.Sort(a); int i, j; // Generating sum groups. List< long > s = new List< long >(); for (i = 0, j = n - 1; i < j; i++, j--) s.Add((a[i] + a[j])); long mini = s.Min(); long maxi = s.Max(); return Math.Abs(maxi - mini); } // Driver code public static void Main() { long []a = { 2, 6, 4, 3 }; int n = a.Length; Console.WriteLine(calculate(a, n)); } } //This code is contributed by Rajput-Ji |
PHP
<?php // PHP program to find minimum // difference between groups of // highest and lowest sums. function calculate( $a , $n ) { // Sorting the whole array. sort( $a ); // Generating sum groups. $s = array (); for ( $i = 0, $j = $n - 1; $i < $j ; $i ++, $j --) array_push ( $s , ( $a [ $i ] + $a [ $j ])); $mini = min( $s ); $maxi = max( $s ); return abs ( $maxi - $mini ); } // Driver Code $a = array ( 2, 6, 4, 3 ); $n = sizeof( $a ); echo calculate( $a , $n ); // This is contributed by mits ?> |
1
Time Complexity: O (n * log n)
Asked in: Inmobi
Reference: https://www.hackerearth.com/problem/algorithm/project-team/
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