Python 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.

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

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). 

Python3

# Python Program for Maximum
# difference between groups
# of size two

# import the module
import sys

def CalculateMax(arr,n):
    first_min=sys.maxsize
    second_min=sys.maxsize

    for i in range(n):
        # If current element is smaller than first_min
        # then update both first_min and second_min
        if(arr[i]<first_min):
            second_min=first_min
            first_min = arr[i]

        # If arr[i] is in between first and second
        # then update second
        elif(arr[i]<second_min and arr[i]!=first_min):
            second_min=arr[i]

    first_max=-1*sys.maxsize
    second_max=-1*sys.maxsize

    for i in range(n):
        # If current element is greater than first_max
        # then update both first and second
        if(arr[i]>first_max):
            second_max=first_max
            first_max = arr[i]

        # If arr[i] is in between first_max and
        # second_max then update second
        elif(arr[i]>second_max and arr[i]!=first_max):
            second_max=arr[i]

    return abs(first_max+second_max-first_min-second_min)


# Driver code
arr = [ 6, 7, 1, 11 ];
n = len(arr)
print(CalculateMax(arr, n))

# This code is contributed by Pushpesh Raj

Output

11

Time Complexity: O(N)
Auxiliary Space: O(1)

Please refer complete article on Maximum difference between groups of size two for more details!