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Find maximum subset sum formed by partitioning any subset of array into 2 partitions with equal sum

  • Difficulty Level : Expert
  • Last Updated : 08 Jun, 2021

Given an array A containing N elements. Partition any subset of this array into two disjoint subsets such that both the subsets have an identical sum. Obtain the maximum sum that can be obtained after partitioning. 

Note: It is not necessary to partition the entire array, that is any element might not contribute to any of the partition.

Examples: 

Input: A = [1, 2, 3, 6] 
Output:
Explanation: We have two disjoint subsets {1, 2, 3} and {6}, which have the same sum = 6

Input: A = [1, 2, 3, 4, 5, 6] 
Output: 10 
Explanation: We have two disjoint subsets {2, 3, 5} and {4, 6}, which have the same sum = 10.



Input: A = [1, 2] 
Output:
Explanation: No subset can be partitioned into 2 disjoint subsets with identical sum 
 

Naive Approach: 
The above problem can be solved by brute force method using recursion. All the elements have three possibilities. Either it will contribute to partition 1 or partition 2 or will not be included in any of the partitions. We will perform these three operations on each of the elements and proceed to the next element in each recursive step.

Below is the implementation of the above approach:  

C++




// CPP implementation for the
// above mentioned recursive approach
 
#include <bits/stdc++.h>
 
using namespace std;
 
// Function to find the maximum subset sum
int maxSum(int p0, int p1, int a[], int pos, int n)
{
    if (pos == n) {
        if (p0 == p1)
            return p0;
        else
            return 0;
    }
    // Ignore the current element
    int ans = maxSum(p0, p1, a, pos + 1, n);
 
    // including element in partition 1
    ans = max(ans, maxSum(p0 + a[pos], p1, a, pos + 1, n));
 
    // including element in partition 2
    ans = max(ans, maxSum(p0, p1 + a[pos], a, pos + 1, n));
    return ans;
}
 
// Driver code
int main()
{
    // size of the array
    int n = 4;
    int a[n] = { 1, 2, 3, 6 };
    cout << maxSum(0, 0, a, 0, n);
    return 0;
}

Java




// Java implementation for the
// above mentioned recursive approach
class GFG {
     
    // Function to find the maximum subset sum
    static int maxSum(int p0, int p1, int a[], int pos, int n)
    {
        if (pos == n) {
            if (p0 == p1)
                return p0;
            else
                return 0;
        }
 
        // Ignore the current element
        int ans = maxSum(p0, p1, a, pos + 1, n);
     
        // including element in partition 1
        ans = Math.max(ans, maxSum(p0 + a[pos], p1, a, pos + 1, n));
     
        // including element in partition 2
        ans = Math.max(ans, maxSum(p0, p1 + a[pos], a, pos + 1, n));
        return ans;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        // size of the array
        int n = 4;
        int a[] = { 1, 2, 3, 6 };
        System.out.println(maxSum(0, 0, a, 0, n));
         
    }
}
 
// This code is contributed by AnkitRai01

Python3




# Python3 implementation for the
# above mentioned recursive approach
 
# Function to find the maximum subset sum
def maxSum(p0, p1, a, pos, n) :
 
    if (pos == n) :
        if (p0 == p1) :
            return p0;
        else :
            return 0;
     
    # Ignore the current element
    ans = maxSum(p0, p1, a, pos + 1, n);
 
    # including element in partition 1
    ans = max(ans, maxSum(p0 + a[pos], p1, a, pos + 1, n));
 
    # including element in partition 2
    ans = max(ans, maxSum(p0, p1 + a[pos], a, pos + 1, n));
     
    return ans;
 
# Driver code
if __name__ == "__main__" :
 
    # size of the array
    n = 4;
    a = [ 1, 2, 3, 6 ];
     
    print(maxSum(0, 0, a, 0, n));
 
# This code is contributed by AnkitRai01

C#




// C# implementation for the
// above mentioned recursive approach
 
using System;
 
public class GFG {
     
    // Function to find the maximum subset sum
    static int maxSum(int p0, int p1, int []a, int pos, int n)
    {
        if (pos == n) {
            if (p0 == p1)
                return p0;
            else
                return 0;
        }
 
        // Ignore the current element
        int ans = maxSum(p0, p1, a, pos + 1, n);
     
        // including element in partition 1
        ans = Math.Max(ans, maxSum(p0 + a[pos], p1, a, pos + 1, n));
     
        // including element in partition 2
        ans = Math.Max(ans, maxSum(p0, p1 + a[pos], a, pos + 1, n));
        return ans;
    }
     
    // Driver code
    public static void Main (string[] args)
    {
        // size of the array
        int n = 4;
        int []a = { 1, 2, 3, 6 };
        Console.WriteLine(maxSum(0, 0, a, 0, n));
         
    }
}
 
// This code is contributed by AnkitRai01

Javascript




<script>
 
// Javascript implementation for the
// above mentioned recursive approach
 
// Function to find the maximum subset sum
function maxSum(p0, p1, a, pos, n)
{
    if (pos == n)
    {
        if (p0 == p1)
            return p0;
        else
            return 0;
    }
     
    // Ignore the current element
    var ans = maxSum(p0, p1, a, pos + 1, n);
 
    // Including element in partition 1
    ans = Math.max(ans, maxSum(
        p0 + a[pos], p1, a, pos + 1, n));
 
    // Including element in partition 2
    ans = Math.max(ans, maxSum(
        p0, p1 + a[pos], a, pos + 1, n));
    return ans;
}
 
// Driver code
 
// Size of the array
var n = 4;
var a = [ 1, 2, 3, 6 ];
 
document.write(maxSum(0, 0, a, 0, n));
 
// This code is contributed by importantly
 
</script>
Output: 
6

 

Time Complexity: O(3^n)

Efficient Approach: 
The above method can be optimized using Dynamic Programming method. 
We will define our DP state as follows : 

dp[i][j] = Max sum of group g0 considering the first i elements such that,
the difference between the sum of g0 and g1 is (sum of all elements – j), where j is the difference.
So, the answer would be dp[n][sum]

Now we might encounter, the difference between the sums is negative, lying in the range [-sum, +sum], where the sum is a summation of all elements. The minimum and maximum ranges occurring when one of the subsets is empty and the other one has all the elements. Due to this, in the DP state, we have defined j as (sum – diff). Thus, j will range from [0, 2*sum].

Below is the implementation of the above approach: 

C++




// CPP implementation for the above mentioned
// Dynamic Programming  approach
 
#include <bits/stdc++.h>
 
using namespace std;
 
// Function to find the maximum subset sum
int maxSum(int a[], int n)
{
    // sum of all elements
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += a[i];
 
    int limit = 2 * sum + 1;
 
    // bottom up lookup table;
    int dp[n + 1][limit];
 
    // initialising dp table with INT_MIN
    // where, INT_MIN means no solution
    for (int i = 0; i < n + 1; i++) {
        for (int j = 0; j < limit; j++)
            dp[i][j] = INT_MIN;
    }
 
    // Case when diff is 0
    dp[0][sum] = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < limit; j++) {
 
            // Putting ith element in g0
            if ((j - a[i - 1]) >= 0 && dp[i - 1][j - a[i - 1]] != INT_MIN)
 
                dp[i][j] = max(dp[i][j], dp[i - 1][j - a[i - 1]]
                                             + a[i - 1]);
 
            // Putting ith element in g1
            if ((j + a[i - 1]) < limit && dp[i - 1][j + a[i - 1]] != INT_MIN)
 
                dp[i][j] = max(dp[i][j], dp[i - 1][j + a[i - 1]]);
 
            // Ignoring ith element
            if (dp[i - 1][j] != INT_MIN)
 
                dp[i][j] = max(dp[i][j], dp[i - 1][j]);
        }
    }
 
    return dp[n][sum];
}
 
// Driver code
 
int main()
{
    int n = 4;
    int a[n] = { 1, 2, 3, 6 };
    cout << maxSum(a, n);
    return 0;
}

Java




// Java implementation for the above mentioned
// Dynamic Programming approach
class GFG {
     
    final static int INT_MIN = Integer.MIN_VALUE;
     
    // Function to find the maximum subset sum
    static int maxSum(int a[], int n)
    {
        // sum of all elements
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += a[i];
     
        int limit = 2 * sum + 1;
     
        // bottom up lookup table;
        int dp[][] = new int[n + 1][limit];
     
        // initialising dp table with INT_MIN
        // where, INT_MIN means no solution
        for (int i = 0; i < n + 1; i++) {
            for (int j = 0; j < limit; j++)
                dp[i][j] = INT_MIN;
        }
     
        // Case when diff is 0
        dp[0][sum] = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < limit; j++) {
     
                // Putting ith element in g0
                if ((j - a[i - 1]) >= 0 && dp[i - 1][j - a[i - 1]] != INT_MIN)
     
                    dp[i][j] = Math.max(dp[i][j], dp[i - 1][j - a[i - 1]]
                                                + a[i - 1]);
     
                // Putting ith element in g1
                if ((j + a[i - 1]) < limit && dp[i - 1][j + a[i - 1]] != INT_MIN)
     
                    dp[i][j] = Math.max(dp[i][j], dp[i - 1][j + a[i - 1]]);
     
                // Ignoring ith element
                if (dp[i - 1][j] != INT_MIN)
     
                    dp[i][j] = Math.max(dp[i][j], dp[i - 1][j]);
            }
        }
     
        return dp[n][sum];
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 4;
        int []a = { 1, 2, 3, 6 };
        System.out.println(maxSum(a, n));
    }
}
 
// This code is contributed by AnkitRai01

Python3




# Python3 implementation for the above mentioned
# Dynamic Programming approach
import numpy as np
import sys
 
INT_MIN = -(sys.maxsize - 1)
 
# Function to find the maximum subset sum
def maxSum(a, n) :
 
    # sum of all elements
    sum = 0;
    for i in range(n) :
        sum += a[i];
 
    limit = 2 * sum + 1;
 
    # bottom up lookup table;
    dp = np.zeros((n + 1,limit));
 
    # initialising dp table with INT_MIN
    # where, INT_MIN means no solution
    for i in range(n + 1) :
        for j in range(limit) :
            dp[i][j] = INT_MIN;
 
    # Case when diff is 0
    dp[0][sum] = 0;
    for i in range(1, n + 1) :
        for j in range(limit) :
 
            # Putting ith element in g0
            if ((j - a[i - 1]) >= 0 and dp[i - 1][j - a[i - 1]] != INT_MIN) :
 
                dp[i][j] = max(dp[i][j], dp[i - 1][j - a[i - 1]]
                                            + a[i - 1]);
 
            # Putting ith element in g1
            if ((j + a[i - 1]) < limit and dp[i - 1][j + a[i - 1]] != INT_MIN) :
 
                dp[i][j] = max(dp[i][j], dp[i - 1][j + a[i - 1]]);
 
            # Ignoring ith element
            if (dp[i - 1][j] != INT_MIN) :
 
                dp[i][j] = max(dp[i][j], dp[i - 1][j]);
                 
    return dp[n][sum];
 
# Driver code
 
if __name__ == "__main__" :
 
    n = 4;
    a = [ 1, 2, 3, 6 ];
    print(maxSum(a, n));
 
# This code is contributed by Yash_R

C#




// C# implementation for the above mentioned
// Dynamic Programming approach
using System;
 
class GFG {
     
    static int INT_MIN = int.MinValue;
     
    // Function to find the maximum subset sum
    static int maxSum(int []a, int n)
    {
        // sum of all elements
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += a[i];
     
        int limit = 2 * sum + 1;
     
        // bottom up lookup table;
        int [,]dp = new int[n + 1,limit];
     
        // initialising dp table with INT_MIN
        // where, INT_MIN means no solution
        for (int i = 0; i < n + 1; i++) {
            for (int j = 0; j < limit; j++)
                dp[i,j] = INT_MIN;
        }
     
        // Case when diff is 0
        dp[0,sum] = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < limit; j++) {
     
                // Putting ith element in g0
                if ((j - a[i - 1]) >= 0 && dp[i - 1,j - a[i - 1]] != INT_MIN)
     
                    dp[i,j] = Math.Max(dp[i,j], dp[i - 1,j - a[i - 1]]
                                                + a[i - 1]);
     
                // Putting ith element in g1
                if ((j + a[i - 1]) < limit && dp[i - 1,j + a[i - 1]] != INT_MIN)
     
                    dp[i,j] = Math.Max(dp[i,j], dp[i - 1,j + a[i - 1]]);
     
                // Ignoring ith element
                if (dp[i - 1,j] != INT_MIN)
     
                    dp[i,j] = Math.Max(dp[i,j], dp[i - 1,j]);
            }
        }
     
        return dp[n,sum];
    }
     
    // Driver code
    public static void Main()
    {
        int n = 4;
        int []a = { 1, 2, 3, 6 };
        Console.WriteLine(maxSum(a, n));
    }
}
 
// This code is contributed by Yash_R

Javascript




<script>
 
 
// JavaScript implementation for the above mentioned
// Dynamic Programming  approach
 
// Function to find the maximum subset sum
function maxSum(a, n)
{
    // sum of all elements
    var sum = 0;
    for (var i = 0; i < n; i++)
        sum += a[i];
 
    var limit = 2 * sum + 1;
 
    // bottom up lookup table;
    var dp = Array.from(Array(n+1), ()=>Array(limit));
 
    // initialising dp table with -1000000000
    // where, -1000000000 means no solution
    for (var i = 0; i < n + 1; i++) {
        for (var j = 0; j < limit; j++)
            dp[i][j] = -1000000000;
    }
 
    // Case when diff is 0
    dp[0][sum] = 0;
    for (var i = 1; i <= n; i++) {
        for (var j = 0; j < limit; j++) {
 
            // Putting ith element in g0
            if ((j - a[i - 1]) >= 0 &&
            dp[i - 1][j - a[i - 1]] != -1000000000)
 
                dp[i][j] = Math.max(dp[i][j],
                dp[i - 1][j - a[i - 1]] + a[i - 1]);
 
            // Putting ith element in g1
            if ((j + a[i - 1]) < limit &&
            dp[i - 1][j + a[i - 1]] != -1000000000)
 
                dp[i][j] = Math.max(dp[i][j],
                dp[i - 1][j + a[i - 1]]);
 
            // Ignoring ith element
            if (dp[i - 1][j] != -1000000000)
 
                dp[i][j] = Math.max(dp[i][j], dp[i - 1][j]);
        }
    }
 
    return dp[n][sum];
}
 
// Driver code
var n = 4;
var a = [1, 2, 3, 6];
document.write( maxSum(a, n));
 
</script>
Output: 
6

 

Time Complexity: O(N*Sum)    , where Sum represents sum of all array elements. 
Auxiliary Space: O(N*Sum)
 




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