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Minimum difference between groups of size two

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  • Difficulty Level : Medium
  • Last Updated : 17 Aug, 2022
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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.  

Implementation:

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
?>

Javascript




<script>
 
// Javascript program to find minimum
// difference between groups of
// highest and lowest sums
   
    function calculate(a, n)
{
    // Sorting the whole array.
    a.sort();
    let i,j;
       
    // Generating sum groups.
    let s = [];
    for (i = 0, j = n - 1; i < j; i++, j--)
        s.push((a[i] + a[j]));
           
    let mini = Math.min(...s);
    let maxi = Math.max(...s);
    return Math.abs(maxi - mini);
}
 
// driver code
 
    let a = [ 2, 6, 4, 3 ];
    let n = a.length;
    document.write(calculate(a, n));
     
</script>

Output

1

Complexity Analysis:

  • Time Complexity: O (n * log n) 
  • Auxiliary Space: O(n)

Optimizing the above approach: 

We can modify the above efficient approach by reducing the auxiliary space from O(n) to O(1). Instead of pushing all the groups in a vector we can directly traverse through the array and keep the track of maximum and minimum element groups.

Below is the code for the above approach:

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);
     
    ll mini = a[0]+a[n-1];
    ll maxi = a[0]+a[n-1];
     
    for (int i = 1, j = n - 2; i < j; i++, j--)
    {
        if(a[i]+a[j]>maxi)
        {
            maxi=a[i]+a[j];
        }
        if(a[i]+a[j]<mini)
        {
            mini=a[i]+a[j];
        }
    }
 
    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;
}
 
// This code is contributed by Pushpesh Raj

Java




// Java program for the above approach
 
import java.io.*;
import java.util.*;
 
class GFG {
 
    static int calculate(int[] a, int n)
    {
 
        // sort the array.
        Arrays.sort(a);
 
        int mini = a[0] + a[n - 1];
        int maxi = a[0] + a[n - 1];
 
        for (int i = 1, j = n - 1; i < j; i++, j--) {
            if (a[i] + a[j] > maxi) {
                maxi = a[i] + a[j];
            }
            if (a[i] + a[j] < mini) {
                mini = a[i] + a[j];
            }
        }
 
        return Math.abs(maxi - mini);
    }
 
    public static void main(String[] args)
    {
        int[] a = { 2, 6, 4, 3 };
        int n = a.length;
 
        System.out.print(calculate(a, n));
    }
}
 
// This code is contributed by lokesh (lokeshmvs21).

Output

1

Complexity Analysis:

  • Time Complexity: O (n*log(n))
  • Auxiliary Space: O(1)

Asked in: Inmobi


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