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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 : 09 Aug, 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 ` `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

 ``
Output:
`6`

Time Complexity: Auxiliary Space: O(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 ` `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[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

 ``
Output:
`6`

Time Complexity: , where Sum represents sum of all array elements.
Auxiliary Space: My Personal Notes arrow_drop_up