C++ Program for Maximum 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 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 differences between the group with the 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++
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
ll CalculateMax(ll arr[], int n)
{
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);
}
int main()
{
ll arr[] = { 6, 7, 1, 11 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << CalculateMax(arr, n) << endl;
return 0;
}
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Output:
11
Time Complexity: O (n * log n)
Space Complexity: O(1) as no extra space has been taken.
Further Optimization :
Instead of sorting, we can find a maximum two and minimum of two in linear time and reduce the time complexity to O(n).
Below is the code for the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
int CalculateMax( int arr[], int n)
{
int first_min = *min_element(arr, arr + n);;
int second_min = INT_MAX;
for ( int i = 0; i < n ; i ++)
{
if (arr[i] != first_min)
second_min = min(arr[i],second_min);
}
int first_max = *max_element(arr, arr + n);
int second_max = INT_MIN;
for ( int i = 0; i < n ; i ++)
{
if (arr[i] != first_max)
second_max = max(arr[i],second_max);
}
return abs (first_max+second_max-first_min-second_min);
}
int main()
{
int arr[] = { 6, 7, 1, 11 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << CalculateMax(arr, n) << endl;
return 0;
}
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Time Complexity: O(N)
Auxiliary Space: O(1)
Please refer complete article on Maximum difference between groups of size two for more details!
Last Updated :
17 Aug, 2023
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