Given a set of integers S, the task is to divide the given set into two non-empty sets S1 and S2 such that the absolute difference between their sums is maximum, i.e., abs(sum(S1) – sum(S2)) is maximum.
Example:
Input: S[] = { 1, 2, 1 }
Output: 2
Explanation:
The subsets are {1} and {2, 1}. Their absolute difference is
abs(1 – (2+1)) = 2, which is maximum.Input: S[] = { -2, 3, -1, 5 }
Output: 11
Explanation:
The subsets are {-1, -2} and {3, 5}. Their absolute difference is
abs((-1, -2) – (3+5)) = 11, which is maximum.
Naive Approach: Generate and store all the subsets of the set of integers and find the maximum absolute difference between the sum of the subset and the difference between the total sum of the set and the sum of that subset, i.e, abs(sum(S1) – (totalSum – sum(S1)).
Time Complexity: O(2N)
Auxiliary Space: O(2N)
Efficient Approach: To optimize the naive approach, the idea is to use some mathematical observations. The problem can be divided into two cases:
- If the set contains only positive integers or only negative integers, then the maximum difference is obtained by splitting the set such that one subset contains only the minimum element of the set and the other subset contains all the remaining elements of the set, i.e.,
abs((totalSum – min(S)) – min(S)) or abs(totalSum – 2×min(S)), where S is the set of integers
- If the set contains both positive and negative integers, then the maximum difference is obtained by splitting the set such that one subset contains all the positive integers and the other subset contains all the negative integers, i.e.,
abs(sum(S1) – sum(S2)) or abs(sum(S)), where S1, S2 is the set of positive and negative integers respectively.
Below is the implementation of the above approach:
C++
// C++ Program for above approach #include <bits/stdc++.h> using namespace std; // Function to return the maximum // difference between the subset sums int maxDiffSubsets( int arr[], int N) { // Stores the total // sum of the array int totalSum = 0; // Checks for positive // and negative elements bool pos = false , neg = false ; // Stores the minimum element // from the given array int min = INT_MAX; // Traverse the array for ( int i = 0; i < N; i++) { // Calculate total sum totalSum += abs (arr[i]); // Mark positive element // present in the set if (arr[i] > 0) pos = true ; // Mark negative element // present in the set if (arr[i] < 0) neg = true ; // Find the minimum // element of the set if (arr[i] < min) min = arr[i]; } // If the array contains both // positive and negative elements if (pos && neg) return totalSum; // Otherwise else return totalSum - 2 * min; } // Driver Code int main() { // Given the array int S[] = {1, 2, 1}; // Length of the array int N = sizeof (S) / sizeof (S[0]); if (N < 2) cout << ( "Not Possible" ); else // Function Call cout << (maxDiffSubsets(S, N)); } // This code is contributed by Chitranayal |
Java
// Java Program for above approach import java.util.*; import java.lang.*; class GFG { // Function to return the maximum // difference between the subset sums static int maxDiffSubsets( int [] arr) { // Stores the total // sum of the array int totalSum = 0 ; // Checks for positive // and negative elements boolean pos = false , neg = false ; // Stores the minimum element // from the given array int min = Integer.MAX_VALUE; // Traverse the array for ( int i = 0 ; i < arr.length; i++) { // Calculate total sum totalSum += Math.abs(arr[i]); // Mark positive element // present in the set if (arr[i] > 0 ) pos = true ; // Mark negative element // present in the set if (arr[i] < 0 ) neg = true ; // Find the minimum // element of the set if (arr[i] < min) min = arr[i]; } // If the array contains both // positive and negative elements if (pos && neg) return totalSum; // Otherwise else return totalSum - 2 * min; } // Driver Code public static void main(String[] args) { // Given the array int [] S = { 1 , 2 , 1 }; // Length of the array int N = S.length; if (N < 2 ) System.out.println( "Not Possible" ); else // Function Call System.out.println(maxDiffSubsets(S)); } } |
Python3
# Python3 program for above approach import sys # Function to return the maximum # difference between the subset sums def maxDiffSubsets(arr): # Stores the total # sum of the array totalSum = 0 # Checks for positive # and negative elements pos = False neg = False # Stores the minimum element # from the given array min = sys.maxsize # Traverse the array for i in range ( len (arr)): # Calculate total sum totalSum + = abs (arr[i]) # Mark positive element # present in the set if (arr[i] > 0 ): pos = True # Mark negative element # present in the set if (arr[i] < 0 ): neg = True # Find the minimum # element of the set if (arr[i] < min ): min = arr[i] # If the array contains both # positive and negative elements if (pos and neg): return totalSum # Otherwise else : return totalSum - 2 * min # Driver Code if __name__ = = '__main__' : # Given the array S = [ 1 , 2 , 1 ] # Length of the array N = len (S) if (N < 2 ): print ( "Not Possible" ) else : # Function Call print (maxDiffSubsets(S)) # This code is contributed by mohit kumar 29 |
C#
// C# Program for above approach using System; class GFG{ // Function to return the maximum // difference between the subset sums static int maxDiffSubsets( int [] arr) { // Stores the total // sum of the array int totalSum = 0; // Checks for positive // and negative elements bool pos = false , neg = false ; // Stores the minimum element // from the given array int min = int .MaxValue; // Traverse the array for ( int i = 0; i < arr.Length; i++) { // Calculate total sum totalSum += Math.Abs(arr[i]); // Mark positive element // present in the set if (arr[i] > 0) pos = true ; // Mark negative element // present in the set if (arr[i] < 0) neg = true ; // Find the minimum // element of the set if (arr[i] < min) min = arr[i]; } // If the array contains both // positive and negative elements if (pos && neg) return totalSum; // Otherwise else return totalSum - 2 * min; } // Driver Code public static void Main(String[] args) { // Given the array int [] S = {1, 2, 1}; // Length of the array int N = S.Length; if (N < 2) Console.WriteLine( "Not Possible" ); else // Function Call Console.WriteLine(maxDiffSubsets(S)); } } // This code is contributed by Rajput-Ji |
2
Time Complexity: O(N)
Auxiliary Space: O(1)
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