Given an array of n-integers. Array may contain repetitive elements but the highest frequency of any elements must not exceed two. You have to make two subsets such that difference of their elements sum is maximum and both of them jointly contains all of elements of given array along with the most important condition, no subset should contain repetitive elements.
Examples:
Input : arr[] = {5, 8, -1, 4} Output : Maximum Difference = 18 Explanation : Let Subset A = {5, 8, 4} & Subset B = {-1} Sum of elements of subset A = 17, of subset B = -1 Difference of Sum of Both subsets = 17 - (-1) = 18 Input : arr[] = {5, 8, 5, 4} Output : Maximum Difference = 12 Explanation : Let Subset A = {5, 8, 4} & Subset B = {5} Sum of elements of subset A = 17, of subset B = 5 Difference of Sum of Both subsets = 17 - 5 = 12
Before solving this question we have to take care of some given conditions and they are listed as:
- While building up the subsets, take care that no subset should contain repetitive elements. And for this we can conclude that all such elements whose frequency are 2, going to be part of both subsets and hence overall they don’t have any impact on difference of subset sum. So, we can easily ignore them.
- For making the difference of sum of elements of both subset maximum we have to make subset in such a way that all positive elements belongs to one subset and negative ones to other subset.
Algorithm with time complexity O(n2):
for i=0 to n-1 isSingleOccurance = true; for j= i+1 to n-1 // if frequency of any element is two // make both equal to zero if arr[i] equals arr[j] arr[i] = arr[j] = 0 isSingleOccurance = false; break; if isSingleOccurance == true if (arr[i] > 0) SubsetSum_1 += arr[i]; else SubsetSum_2 += arr[i]; return abs(SubsetSum_1 - SubsetSum2)
leave a comment
0 Comments