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.
// 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));
}
} |
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.
// 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)
{ int first_min = Arrays.stream(arr).min().getAsInt();
int second_min = Integer.MAX_VALUE;
for ( int i = 0 ; i < n ; i ++)
{
// If arr[i] is not equal to first min
if (arr[i] != first_min)
second_min = Math.min(arr[i],second_min);
}
int first_max = Arrays.stream(arr).max().getAsInt();
int second_max = Integer.MIN_VALUE;
for ( int i = 0 ; i < n ; i ++)
{
// If arr[i] is not equal to first max
if (arr[i] != first_max)
second_max = Math.max(arr[i],second_max);
}
return Math.abs(first_max+second_max-first_min-second_min);
} // Driver code public static void main (String[] args) {
int arr[] = { 6 , 7 , 1 , 11 };
int n = arr.length;
System.out.println (CalculateMax(arr, n));
}
} // This code is contributed by Aman Kumar |
11
Time Complexity: O(N)
Auxiliary Space: O(1)
Please refer complete article on Maximum difference between groups of size two for more details!