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.
Javascript
<script> // Javascript program to // find minimum difference // between groups of highest and lowest // sums. function CalculateMax(arr, n) { // Sorting the whole array. arr.sort(function(a, b){return a - b}); let min_sum = arr[0] + arr[1]; let max_sum = arr[n-1] + arr[n-2]; return (Math.abs(max_sum - min_sum)); } let arr = [ 6, 7, 1, 11 ]; let n = arr.length; document.write(CalculateMax(arr, n)); </script> |
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.
Javascript
<script> // Javascript program to // find minimum difference // between groups of highest and lowest // sums. function CalculateMax(arr, n) { let first_min = Math.min.apply(Math,arr);; let second_min = Number.MAX_VALUE; for(let 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); } let first_max = Math.max.apply(Math,arr);; let second_max = Number.MIN_VALUE; for (let 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); } let arr = [ 6, 7, 1, 11 ]; let n = arr.length; document.write(CalculateMax(arr, n)); </script> |
Time Complexity: O(N)
Auxiliary Space: O(1)
Please refer complete article on Maximum difference between groups of size two for more details!
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