Partition problem | DP-18

The partition problem is to determine whether a given set can be partitioned into two subsets such that the sum of elements in both subsets is the same.

Examples:

Input: arr[] = {1, 5, 11, 5}
Output: true
The array can be partitioned as {1, 5, 5} and {11}

Input: arr[] = {1, 5, 3}
Output: false
The array cannot be partitioned into equal sum sets.

We strongly recommend that you click here and practice it, before moving on to the solution.

The following are the two main steps to solve this problem:

• Calculate the sum of the array. If the sum is odd, there can not be two subsets with an equal sum, so return false.
• If the sum of the array elements is even, calculate sum/2 and find a subset of the array with a sum equal to sum/2.
The first step is simple. The second step is crucial, it can be solved either using recursion or Dynamic Programming.

Partition problem using recursion:

To solve the problem follow the below idea:

Let isSubsetSum(arr, n, sum/2) be the function that returns true if there is a subset of arr[0..n-1] with sum equal to sum/2
The isSubsetSum problem can be divided into two subproblems

•  isSubsetSum() without considering last element (reducing n to n-1)
•  isSubsetSum considering the last element (reducing sum/2 by arr[n-1] and n to n-1)

If any of the above subproblems return true, then return true.
isSubsetSum (arr, n, sum/2) = isSubsetSum (arr, n-1, sum/2) || isSubsetSum (arr, n-1, sum/2 – arr[n-1])

Follow the below steps to solve the problem:

• First, check if the sum of the elements is even or not
• After checking, call the recursive function isSubsetSum with parameters as input array, array size, and sum/2
• If the sum is equal to zero then return true (Base case)
• If n is equal to 0 and sum is not equal to zero then return false (Base case)
• Check if the value of the last element is greater than the remaining sum then call this function again by removing the last element
• else call this function again for both the cases stated above and return true, if anyone of them returns true

Below is the implementation of the above approach:

C++

 `// A recursive C++ program for partition problem` `#include ` `using` `namespace` `std;`   `// A utility function that returns true if there is` `// a subset of arr[] with sum equal to given sum` `bool` `isSubsetSum(``int` `arr[], ``int` `n, ``int` `sum)` `{` `    ``// Base Cases` `    ``if` `(sum == 0)` `        ``return` `true``;` `    ``if` `(n == 0 && sum != 0)` `        ``return` `false``;`   `    ``// If last element is greater than sum, then` `    ``// ignore it` `    ``if` `(arr[n - 1] > sum)` `        ``return` `isSubsetSum(arr, n - 1, sum);`   `    ``/* else, check if sum can be obtained by any of` `        ``the following` `        ``(a) including the last element` `        ``(b) excluding the last element` `    ``*/` `    ``return` `isSubsetSum(arr, n - 1, sum)` `           ``|| isSubsetSum(arr, n - 1, sum - arr[n - 1]);` `}`   `// Returns true if arr[] can be partitioned in two` `// subsets of equal sum, otherwise false` `bool` `findPartiion(``int` `arr[], ``int` `n)` `{` `    ``// Calculate sum of the elements in array` `    ``int` `sum = 0;` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``sum += arr[i];`   `    ``// If sum is odd, there cannot be two subsets` `    ``// with equal sum` `    ``if` `(sum % 2 != 0)` `        ``return` `false``;`   `    ``// Find if there is subset with sum equal to` `    ``// half of total sum` `    ``return` `isSubsetSum(arr, n, sum / 2);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 3, 1, 5, 9, 12 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``// Function call` `    ``if` `(findPartiion(arr, n) == ``true``)` `        ``cout << ``"Can be divided into two subsets "` `                ``"of equal sum"``;` `    ``else` `        ``cout << ``"Can not be divided into two subsets"` `                ``" of equal sum"``;` `    ``return` `0;` `}`   `// This code is contributed by rathbhupendra`

C

 `// A recursive C program for partition problem` `#include ` `#include `   `// A utility function that returns true if there is` `// a subset of arr[] with sum equal to given sum` `bool` `isSubsetSum(``int` `arr[], ``int` `n, ``int` `sum)` `{` `    ``// Base Cases` `    ``if` `(sum == 0)` `        ``return` `true``;` `    ``if` `(n == 0 && sum != 0)` `        ``return` `false``;`   `    ``// If last element is greater than sum, then` `    ``// ignore it` `    ``if` `(arr[n - 1] > sum)` `        ``return` `isSubsetSum(arr, n - 1, sum);`   `    ``/* else, check if sum can be obtained by any of` `       ``the following` `       ``(a) including the last element` `       ``(b) excluding the last element` `    ``*/` `    ``return` `isSubsetSum(arr, n - 1, sum)` `           ``|| isSubsetSum(arr, n - 1, sum - arr[n - 1]);` `}`   `// Returns true if arr[] can be partitioned in two` `//  subsets of equal sum, otherwise false` `bool` `findPartiion(``int` `arr[], ``int` `n)` `{` `    ``// Calculate sum of the elements in array` `    ``int` `sum = 0;` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``sum += arr[i];`   `    ``// If sum is odd, there cannot be two subsets` `    ``// with equal sum` `    ``if` `(sum % 2 != 0)` `        ``return` `false``;`   `    ``// Find if there is subset with sum equal to` `    ``// half of total sum` `    ``return` `isSubsetSum(arr, n, sum / 2);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 3, 1, 5, 9, 12 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``// Function call` `    ``if` `(findPartiion(arr, n) == ``true``)` `        ``printf``(``"Can be divided into two subsets "` `               ``"of equal sum"``);` `    ``else` `        ``printf``(``"Can not be divided into two subsets"` `               ``" of equal sum"``);` `    ``return` `0;` `}`

Java

 `// A recursive Java solution for partition problem` `import` `java.io.*;`   `class` `Partition {` `    ``// A utility function that returns true if there is a` `    ``// subset of arr[] with sum equal to given sum` `    ``static` `boolean` `isSubsetSum(``int` `arr[], ``int` `n, ``int` `sum)` `    ``{` `        ``// Base Cases` `        ``if` `(sum == ``0``)` `            ``return` `true``;` `        ``if` `(n == ``0` `&& sum != ``0``)` `            ``return` `false``;`   `        ``// If last element is greater than sum, then ignore` `        ``// it` `        ``if` `(arr[n - ``1``] > sum)` `            ``return` `isSubsetSum(arr, n - ``1``, sum);`   `        ``/* else, check if sum can be obtained by any of` `           ``the following` `        ``(a) including the last element` `        ``(b) excluding the last element` `        ``*/` `        ``return` `isSubsetSum(arr, n - ``1``, sum)` `            ``|| isSubsetSum(arr, n - ``1``, sum - arr[n - ``1``]);` `    ``}`   `    ``// Returns true if arr[] can be partitioned in two` `    ``// subsets of equal sum, otherwise false` `    ``static` `boolean` `findPartition(``int` `arr[], ``int` `n)` `    ``{` `        ``// Calculate sum of the elements in array` `        ``int` `sum = ``0``;` `        ``for` `(``int` `i = ``0``; i < n; i++)` `            ``sum += arr[i];`   `        ``// If sum is odd, there cannot be two subsets` `        ``// with equal sum` `        ``if` `(sum % ``2` `!= ``0``)` `            ``return` `false``;`   `        ``// Find if there is subset with sum equal to half` `        ``// of total sum` `        ``return` `isSubsetSum(arr, n, sum / ``2``);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``int` `arr[] = { ``3``, ``1``, ``5``, ``9``, ``12` `};` `        ``int` `n = arr.length;`   `        ``// Function call` `        ``if` `(findPartition(arr, n) == ``true``)` `            ``System.out.println(``"Can be divided into two "` `                               ``+ ``"subsets of equal sum"``);` `        ``else` `            ``System.out.println(` `                ``"Can not be divided into "` `                ``+ ``"two subsets of equal sum"``);` `    ``}` `}` `/* This code is contributed by Devesh Agrawal */`

Python3

 `# A recursive Python3 program for` `# partition problem`   `# A utility function that returns` `# true if there is a subset of` `# arr[] with sum equal to given sum`     `def` `isSubsetSum(arr, n, ``sum``):` `    ``# Base Cases` `    ``if` `sum` `=``=` `0``:` `        ``return` `True` `    ``if` `n ``=``=` `0` `and` `sum` `!``=` `0``:` `        ``return` `False`   `    ``# If last element is greater than sum, then` `    ``# ignore it` `    ``if` `arr[n``-``1``] > ``sum``:` `        ``return` `isSubsetSum(arr, n``-``1``, ``sum``)`   `    ``''' else, check if sum can be obtained by any of ` `    ``the following` `    ``(a) including the last element` `    ``(b) excluding the last element'''`   `    ``return` `isSubsetSum(arr, n``-``1``, ``sum``) ``or` `isSubsetSum(arr, n``-``1``, ``sum``-``arr[n``-``1``])`   `# Returns true if arr[] can be partitioned in two` `# subsets of equal sum, otherwise false`     `def` `findPartion(arr, n):` `    ``# Calculate sum of the elements in array` `    ``sum` `=` `0` `    ``for` `i ``in` `range``(``0``, n):` `        ``sum` `+``=` `arr[i]` `    ``# If sum is odd, there cannot be two subsets` `    ``# with equal sum` `    ``if` `sum` `%` `2` `!``=` `0``:` `        ``return` `false`   `    ``# Find if there is subset with sum equal to` `    ``# half of total sum` `    ``return` `isSubsetSum(arr, n, ``sum` `/``/` `2``)`     `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `  ``arr ``=` `[``3``, ``1``, ``5``, ``9``, ``12``]` `  ``n ``=` `len``(arr)`   `  ``# Function call` `  ``if` `findPartion(arr, n) ``=``=` `True``:` `      ``print``(``"Can be divided into two subsets of equal sum"``)` `  ``else``:` `      ``print``(``"Can not be divided into two subsets of equal sum"``)`   `# This code is contributed by shreyanshi_arun.`

C#

 `// A recursive C# solution for partition problem` `using` `System;`   `class` `GFG {`   `    ``// A utility function that returns true if there is a` `    ``// subset of arr[] with sum equal to given sum` `    ``static` `bool` `isSubsetSum(``int``[] arr, ``int` `n, ``int` `sum)` `    ``{` `        ``// Base Cases` `        ``if` `(sum == 0)` `            ``return` `true``;` `        ``if` `(n == 0 && sum != 0)` `            ``return` `false``;`   `        ``// If last element is greater than sum, then ignore` `        ``// it` `        ``if` `(arr[n - 1] > sum)` `            ``return` `isSubsetSum(arr, n - 1, sum);`   `        ``/* else, check if sum can be obtained by any of` `        ``the following` `        ``(a) including the last element` `        ``(b) excluding the last element` `        ``*/` `        ``return` `isSubsetSum(arr, n - 1, sum)` `            ``|| isSubsetSum(arr, n - 1, sum - arr[n - 1]);` `    ``}`   `    ``// Returns true if arr[] can be partitioned in two` `    ``// subsets of equal sum, otherwise false` `    ``static` `bool` `findPartition(``int``[] arr, ``int` `n)` `    ``{` `        ``// Calculate sum of the elements in array` `        ``int` `sum = 0;` `        ``for` `(``int` `i = 0; i < n; i++)` `            ``sum += arr[i];`   `        ``// If sum is odd, there cannot be two subsets` `        ``// with equal sum` `        ``if` `(sum % 2 != 0)` `            ``return` `false``;`   `        ``// Find if there is subset with sum equal to half` `        ``// of total sum` `        ``return` `isSubsetSum(arr, n, sum / 2);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{`   `        ``int``[] arr = { 3, 1, 5, 9, 12 };` `        ``int` `n = arr.Length;`   `        ``// Function call` `        ``if` `(findPartition(arr, n) == ``true``)` `            ``Console.Write(``"Can be divided into two "` `                          ``+ ``"subsets of equal sum"``);` `        ``else` `            ``Console.Write(``"Can not be divided into "` `                          ``+ ``"two subsets of equal sum"``);` `    ``}` `}`   `// This code is contributed by Sam007`

PHP

 ` ``\$sum``) ` `        ``return` `isSubsetSum (``\$arr``, ``\$n` `- 1, ``\$sum``); ` `    `  `    ``/* else, check if sum can be obtained ` `       ``by any of the following ` `        ``(a) including the last element ` `        ``(b) excluding the last element ` `    ``*/` `    ``return` `isSubsetSum (``\$arr``, ``\$n` `- 1, ``\$sum``) || ` `           ``isSubsetSum (``\$arr``, ``\$n` `- 1, ` `                        ``\$sum` `- ``\$arr``[``\$n` `- 1]); ` `}  `   `// Returns true if arr[] can be partitioned ` `// in two subsets of equal sum, otherwise false ` `function` `findPartiion (``\$arr``, ``\$n``) ` `{ ` `    ``// Calculate sum of the elements` `    ``// in array ` `    ``\$sum` `= 0; ` `    ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++) ` `    ``\$sum` `+= ``\$arr``[``\$i``]; `   `    ``// If sum is odd, there cannot be ` `    ``// two subsets with equal sum ` `    ``if` `(``\$sum` `% 2 != 0) ` `    ``return` `false; `   `    ``// Find if there is subset with sum` `    ``// equal to half of total sum ` `    ``return` `isSubsetSum (``\$arr``, ``\$n``, ``\$sum` `/ 2); ` `} `   `// Driver Code` `\$arr` `= ``array``(3, 1, 5, 9, 12); ` `\$n` `= ``count``(``\$arr``); `   `// Function call` `if` `(findPartiion(``\$arr``, ``\$n``) == true) ` `    ``echo` `"Can be divided into two subsets of equal sum"``; ` `else` `    ``echo` `"Can not be divided into two subsets of equal sum"``; `   `// This code is contributed by rathbhupendra` `?>`

Javascript

 ``

Output

`Can be divided into two subsets of equal sum`

Time Complexity: O(2N) In the worst case, this solution tries two possibilities (whether to include or exclude) for every element.
Auxiliary Space: O(N). Recursion stack space

Partition problem using memoization:

To solve the problem follow the below idea:

As the above recursive solution has overlapping subproblems so we can declare a 2-D array to save the values for different states of the recursive function instead of solving them more than once

Follow the below steps to solve the problem:

• Declare a 2-D array of size N+1 X sum+1
• Call the recursive function with parameters as input array, size, sum, and dp array
• In this recursive function
• If the sum is equal to zero then return true (Base case)
• If n is equal to 0 and sum is not equal to zero then return false (Base case)
• If the value of this subproblem is already calculated then return the answer from dp array
• Else calculate the answer for this subproblem using the recursive formula in the above approach and save the answer in the dp array
• Return the answer as true or false

Below is the implementation of the above approach:

C++

 `// A recursive C++ program for partition problem` `#include ` `using` `namespace` `std;`   `// A utility function that returns true if there is` `// a subset of arr[] with sun equal to given sum` `bool` `isSubsetSum(``int` `arr[], ``int` `n, ``int` `sum,` `                 ``vector >& dp)` `{` `    ``// Base Cases` `    ``if` `(sum == 0)` `        ``return` `true``;` `    ``if` `(n == 0 && sum != 0)` `        ``return` `false``;`   `    ``// return solved subproblem` `    ``if` `(dp[n][sum] != -1) {` `        ``return` `dp[n][sum];` `    ``}`   `    ``// If last element is greater than sum, then` `    ``// ignore it` `    ``if` `(arr[n - 1] > sum)` `        ``return` `isSubsetSum(arr, n - 1, sum, dp);`   `    ``/* else, check if sum can be obtained by any of` `        ``the following` `        ``(a) including the last element` `        ``(b) excluding the last element` `    ``*/` `    ``// also store the subproblem in dp matrix` `    ``return` `dp[n][sum]` `           ``= isSubsetSum(arr, n - 1, sum, dp)` `             ``|| isSubsetSum(arr, n - 1, sum - arr[n - 1],` `                            ``dp);` `}`   `// Returns true if arr[] can be partitioned in two` `// subsets of equal sum, otherwise false` `bool` `findPartiion(``int` `arr[], ``int` `n)` `{` `    ``// Calculate sum of the elements in array` `    ``int` `sum = 0;` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``sum += arr[i];`   `    ``// If sum is odd, there cannot be two subsets` `    ``// with equal sum` `    ``if` `(sum % 2 != 0)` `        ``return` `false``;`   `    ``// To store overlapping subproblems` `    ``vector > dp(n + 1,` `                            ``vector<``int``>(sum + 1, -1));`   `    ``// Find if there is subset with sum equal to` `    ``// half of total sum` `    ``return` `isSubsetSum(arr, n, sum / 2, dp);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 3, 1, 5, 9, 12 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``// Function call` `    ``if` `(findPartiion(arr, n) == ``true``)` `        ``cout << ``"Can be divided into two subsets "` `                ``"of equal sum"``;` `    ``else` `        ``cout << ``"Can not be divided into two subsets"` `                ``" of equal sum"``;`   `    ``int` `arr2[] = { 3, 1, 5, 9, 14 };` `    ``int` `n2 = ``sizeof``(arr2) / ``sizeof``(arr2[0]);`   `    ``if` `(findPartiion(arr2, n2) == ``true``)` `        ``cout << endl` `             ``<< ``"Can be divided into two subsets "` `                ``"of equal sum"``;` `    ``else` `        ``cout << endl` `             ``<< ``"Can not be divided into two subsets"` `                ``" of equal sum"``;` `    ``return` `0;` `}`

Java

 `// Java program for partition problem` `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG {`   `    ``// A utility function that returns true if there is` `    ``// a subset of arr[] with sum equal to given sum` `    ``static` `int` `isSubsetSum(``int` `arr[], ``int` `n, ``int` `sum,` `                           ``int``[][] dp)` `    ``{` `        ``// Base Cases` `        ``if` `(sum == ``0``)` `            ``return` `1``;` `        ``if` `(n == ``0` `&& sum != ``0``)` `            ``return` `0``;`   `        ``// return solved subproblem` `        ``if` `(dp[n][sum] != -``1``) {` `            ``return` `dp[n][sum];` `        ``}`   `        ``// If last element is greater than sum, then` `        ``// ignore it` `        ``if` `(arr[n - ``1``] > sum)` `            ``return` `isSubsetSum(arr, n - ``1``, sum, dp);`   `        ``/* else, check if sum can be obtained by any of` `                ``the following` `                ``(a) including the last element` `                ``(b) excluding the last element` `        ``*/` `        ``// also store the subproblem in dp matrix` `        ``if` `(isSubsetSum(arr, n - ``1``, sum, dp) != ``0` `            ``|| isSubsetSum(arr, n - ``1``, sum - arr[n - ``1``], dp)` `                   ``!= ``0``)` `            ``return` `dp[n][sum] = ``1``;` `        ``return` `dp[n][sum] = ``0``;` `        ``// return dp[n][sum] = isSubsetSum(arr, n - 1, sum,` `        ``// dp) || isSubsetSum(arr, n - 1, sum - arr[n - 1],` `        ``// dp);` `    ``}`   `    ``// Returns true if arr[] can be partitioned in two` `    ``// subsets of equal sum, otherwise false` `    ``static` `int` `findPartiion(``int` `arr[], ``int` `n)` `    ``{` `        ``// Calculate sum of the elements in array` `        ``int` `sum = ``0``;` `        ``for` `(``int` `i = ``0``; i < n; i++)` `            ``sum += arr[i];`   `        ``// If sum is odd, there cannot be two subsets` `        ``// with equal sum` `        ``if` `(sum % ``2` `!= ``0``)` `            ``return` `0``;`   `        ``// To store overlapping subproblems` `        ``int` `dp[][] = ``new` `int``[n + ``1``][sum + ``1``];` `        ``for` `(``int` `row[] : dp)` `            ``Arrays.fill(row, -``1``);`   `        ``// Find if there is subset with sum equal to` `        ``// half of total sum` `        ``return` `isSubsetSum(arr, n, sum / ``2``, dp);` `    ``}`   `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `arr[] = { ``3``, ``1``, ``5``, ``9``, ``12` `};` `        ``int` `n = arr.length;`   `        ``// Function call` `        ``if` `(findPartiion(arr, n) == ``1``)` `            ``System.out.println(` `                ``"Can be divided into two subsets of equal sum"``);` `        ``else` `            ``System.out.println(` `                ``"Can not be divided into two subsets of equal sum"``);`   `        ``int` `arr2[] = { ``3``, ``1``, ``5``, ``9``, ``14` `};` `        ``int` `n2 = arr2.length;`   `        ``if` `(findPartiion(arr2, n2) == ``1``)` `            ``System.out.println(` `                ``"Can be divided into two subsets of equal sum"``);` `        ``else` `            ``System.out.println(` `                ``"Can not be divided into two subsets of equal sum"``);` `    ``}` `}`

Python3

 `# A recursive JavaScript program for partition problem`   `# A utility function that returns true if there is` `# a subset of arr[] with sun equal to given sum`     `def` `isSubsetSum(arr, n, ``sum``, dp):`   `    ``# Base Cases` `    ``if` `(``sum` `=``=` `0``):` `        ``return` `True` `    ``if` `(n ``=``=` `0` `and` `sum` `!``=` `0``):` `        ``return` `False`   `    ``# return solved subproblem` `    ``if` `(dp[n][``sum``] !``=` `-``1``):` `        ``return` `dp[n][``sum``]`   `    ``# If last element is greater than sum, then` `    ``# ignore it` `    ``if` `(arr[n ``-` `1``] > ``sum``):` `        ``return` `isSubsetSum(arr, n ``-` `1``, ``sum``, dp)`   `        ``# else, check if sum can be obtained by any of` `        ``# the following` `        ``# (a) including the last element` `        ``# (b) excluding the last element`   `    ``# also store the subproblem in dp matrix` `    ``dp[n][``sum``] ``=` `isSubsetSum(` `        ``arr, n ``-` `1``, ``sum``, dp) ``or` `isSubsetSum(arr, n ``-` `1``, ``sum` `-` `arr[n ``-` `1``], dp)`   `    ``return` `dp[n][``sum``]`   `# Returns true if arr[] can be partitioned in two` `# subsets of equal sum, otherwise false`     `def` `findPartiion(arr, n):`   `    ``# Calculate sum of the elements in array` `    ``sum` `=` `0` `    ``for` `i ``in` `range``(n):` `        ``sum` `+``=` `arr[i]`   `    ``# If sum is odd, there cannot be two subsets` `    ``# with equal sum` `    ``if` `(``sum` `%` `2` `!``=` `0``):` `        ``return` `False`   `    ``# To store overlapping subproblems` `    ``dp ``=` `[[``-``1``]``*``(``sum``+``1``) ``for` `i ``in` `range``(n``+``1``)]`   `    ``# Find if there is subset with sum equal to` `    ``# half of total sum` `    ``return` `isSubsetSum(arr, n, ``sum` `/``/` `2``, dp)`   `# Driver code`     `arr ``=` `[``3``, ``1``, ``5``, ``9``, ``12``]` `n ``=` `len``(arr)`   `# Function call` `if` `(findPartiion(arr, n) ``=``=` `True``):` `    ``print``(``"Can be divided into two subsets of equal sum"``)` `else``:` `    ``print``(``"Can not be divided into two subsets of equal sum"``)`   `arr2 ``=` `[``3``, ``1``, ``5``, ``9``, ``14``]` `n2 ``=` `len``(arr2)`   `if` `(findPartiion(arr2, n2) ``=``=` `True``):` `    ``print``(``"Can be divided into two subsets of equal sum"``)` `else``:` `    ``print``(``"Can not be divided into two subsets of equal sum"``)`   `# This code is contributed by shinjanpatra.`

C#

 `// C# program for partition problem` `using` `System;` `public` `class` `GFG {`   `    ``// A utility function that returns true if there is` `    ``// a subset of arr[] with sum equal to given sum` `    ``static` `int` `isSubsetSum(``int``[] arr, ``int` `n, ``int` `sum,` `                           ``int``[, ] dp)` `    ``{`   `        ``// Base Cases` `        ``if` `(sum == 0)` `            ``return` `1;` `        ``if` `(n == 0 && sum != 0)` `            ``return` `0;`   `        ``// return solved subproblem` `        ``if` `(dp[n, sum] != -1) {` `            ``return` `dp[n, sum];` `        ``}`   `        ``// If last element is greater than sum, then` `        ``// ignore it` `        ``if` `(arr[n - 1] > sum)` `            ``return` `isSubsetSum(arr, n - 1, sum, dp);`   `        ``/* else, check if sum can be obtained by any of` `                    ``the following` `                    ``(a) including the last element` `                    ``(b) excluding the last element` `            ``*/` `        ``// also store the subproblem in dp matrix` `        ``if` `(isSubsetSum(arr, n - 1, sum, dp) != 0` `            ``|| isSubsetSum(arr, n - 1, sum - arr[n - 1], dp)` `                   ``!= 0)` `            ``return` `dp[n, sum] = 1;` `        ``return` `dp[n, sum] = 0;` `        ``// return dp[n][sum] = isSubsetSum(arr, n - 1, sum,` `        ``// dp) || isSubsetSum(arr, n - 1, sum - arr[n - 1],` `        ``// dp);` `    ``}`   `    ``// Returns true if arr[] can be partitioned in two` `    ``// subsets of equal sum, otherwise false` `    ``static` `int` `findPartiion(``int``[] arr, ``int` `n)` `    ``{` `        ``// Calculate sum of the elements in array` `        ``int` `sum = 0;` `        ``for` `(``int` `i = 0; i < n; i++)` `            ``sum += arr[i];`   `        ``// If sum is odd, there cannot be two subsets` `        ``// with equal sum` `        ``if` `(sum % 2 != 0)` `            ``return` `0;`   `        ``// To store overlapping subproblems` `        ``int``[, ] dp = ``new` `int``[n + 1, sum + 1];`   `        ``for` `(``int` `i = 0; i <= n; i++) {` `            ``for` `(``int` `j = 0; j <= sum; ++j) {` `                ``dp[i, j] = -1;` `            ``}` `        ``}`   `        ``// Find if there is subset with sum equal to` `        ``// half of total sum` `        ``return` `isSubsetSum(arr, n, sum / 2, dp);` `    ``}`   `    ``static` `public` `void` `Main()` `    ``{`   `        ``// Code` `        ``int``[] arr = { 3, 1, 5, 9, 12 };` `        ``int` `n = arr.Length;`   `        ``// Function call` `        ``if` `(findPartiion(arr, n) == 1)` `            ``Console.WriteLine(` `                ``"Can be divided into two subsets of equal sum"``);` `        ``else` `            ``Console.WriteLine(` `                ``"Can not be divided into two subsets of equal sum"``);`   `        ``int``[] arr2 = { 3, 1, 5, 9, 14 };` `        ``int` `n2 = arr2.Length;`   `        ``if` `(findPartiion(arr2, n2) == 1)` `            ``Console.WriteLine(` `                ``"Can be divided into two subsets of equal sum"``);` `        ``else` `            ``Console.WriteLine(` `                ``"Can not be divided into two subsets of equal sum"``);` `    ``}` `}`   `// This code is contributed by lokeshmvs21.`

Javascript

 ``

Output

```Can be divided into two subsets of equal sum
Can not be divided into two subsets of equal sum```

Time Complexity: O(sum * N)
Auxiliary Space: O(sum * N)

Partition problem using dynamic programming:

To solve the problem follow the below idea:

The problem can be solved using dynamic programming when the sum of the elements is not too big. As the recomputations of the same subproblems can be avoided by constructing a temporary array part[][] in a bottom-up manner using the above recursive formula and it should satisfy the following formula:

part[i][j] = true if a subset of {arr[0], arr[1], ..arr[j-1]} has sum equal to i, otherwise false

Follow the below steps to solve the problem:

• First, check if the sum of the elements is even or not
• Declare a 2-D array part[][] of size (sum/2)+1 * (N + 1)
• Run a for loop for 0 <= i <= n and set part[0][i] equal to true as zero-sum is always possible
• Run a for loop for 1 <= i <= sum/2 and set part[i][0] equal to zero as any sum with zero elements is never possible
• Run a nested for loop for 1 <= i <= sum/2 and 1 <= j <= N
• Set part[i][j] equal to part[i][j-1]
• If i is greater than or equal to arr[j-1], if part[i – arr[j-1]][j-1] is true then set part[i][j] as true

Below is the implementation of the above approach:

C++

 `// A Dynamic Programming based` `// C++ program to partition problem` `#include ` `using` `namespace` `std;`   `// Returns true if arr[] can be partitioned` `// in two subsets of equal sum, otherwise false` `bool` `findPartiion(``int` `arr[], ``int` `n)` `{` `    ``int` `sum = 0;` `    ``int` `i, j;`   `    ``// Calculate sum of all elements` `    ``for` `(i = 0; i < n; i++)` `        ``sum += arr[i];`   `    ``if` `(sum % 2 != 0)` `        ``return` `false``;`   `    ``bool` `part[sum / 2 + 1][n + 1];`   `    ``// initialize top row as true` `    ``for` `(i = 0; i <= n; i++)` `        ``part[0][i] = ``true``;`   `    ``// initialize leftmost column,` `    ``// except part[0][0], as 0` `    ``for` `(i = 1; i <= sum / 2; i++)` `        ``part[i][0] = ``false``;`   `    ``// Fill the partition table in bottom up manner` `    ``for` `(i = 1; i <= sum / 2; i++) {` `        ``for` `(j = 1; j <= n; j++) {` `            ``part[i][j] = part[i][j - 1];` `            ``if` `(i >= arr[j - 1])` `                ``part[i][j] = part[i][j]` `                             ``|| part[i - arr[j - 1]][j - 1];` `        ``}` `    ``}`   `    ``/* // uncomment this part to print table` `    ``for (i = 0; i <= sum/2; i++)` `    ``{` `    ``for (j = 0; j <= n; j++)` `        ``cout<

C

 `// A Dynamic Programming based C program to partition` `// problem` `#include ` `#include `   `// Returns true if arr[] can be partitioned in two subsets` `// of equal sum, otherwise false` `bool` `findPartiion(``int` `arr[], ``int` `n)` `{` `    ``int` `sum = 0;` `    ``int` `i, j;`   `    ``// Calculate sum of all elements` `    ``for` `(i = 0; i < n; i++)` `        ``sum += arr[i];`   `    ``if` `(sum % 2 != 0)` `        ``return` `false``;`   `    ``bool` `part[sum / 2 + 1][n + 1];`   `    ``// initialize top row as true` `    ``for` `(i = 0; i <= n; i++)` `        ``part[0][i] = ``true``;`   `    ``// initialize leftmost column, except part[0][0], as 0` `    ``for` `(i = 1; i <= sum / 2; i++)` `        ``part[i][0] = ``false``;`   `    ``// Fill the partition table in bottom up manner` `    ``for` `(i = 1; i <= sum / 2; i++) {` `        ``for` `(j = 1; j <= n; j++) {` `            ``part[i][j] = part[i][j - 1];` `            ``if` `(i >= arr[j - 1])` `                ``part[i][j] = part[i][j]` `                             ``|| part[i - arr[j - 1]][j - 1];` `        ``}` `    ``}`   `    ``/* // uncomment this part to print table` `     ``for (i = 0; i <= sum/2; i++)` `     ``{` `       ``for (j = 0; j <= n; j++)` `          ``printf ("%4d", part[i][j]);` `       ``printf("\n");` `     ``} */`   `    ``return` `part[sum / 2][n];` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 3, 1, 1, 2, 2, 1 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``// Function call` `    ``if` `(findPartiion(arr, n) == ``true``)` `        ``printf``(` `            ``"Can be divided into two subsets of equal sum"``);` `    ``else` `        ``printf``(``"Can not be divided into two subsets of "` `               ``"equal sum"``);` `    ``getchar``();` `    ``return` `0;` `}`

Java

 `// A dynamic programming based Java program for partition` `// problem` `import` `java.io.*;`   `class` `Partition {`   `    ``// Returns true if arr[] can be partitioned in two` `    ``// subsets of equal sum, otherwise false` `    ``static` `boolean` `findPartition(``int` `arr[], ``int` `n)` `    ``{` `        ``int` `sum = ``0``;` `        ``int` `i, j;`   `        ``// Calculate sum of all elements` `        ``for` `(i = ``0``; i < n; i++)` `            ``sum += arr[i];`   `        ``if` `(sum % ``2` `!= ``0``)` `            ``return` `false``;`   `        ``boolean` `part[][] = ``new` `boolean``[sum / ``2` `+ ``1``][n + ``1``];`   `        ``// initialize top row as true` `        ``for` `(i = ``0``; i <= n; i++)` `            ``part[``0``][i] = ``true``;`   `        ``// initialize leftmost column, except part[0][0], as` `        ``// 0` `        ``for` `(i = ``1``; i <= sum / ``2``; i++)` `            ``part[i][``0``] = ``false``;`   `        ``// Fill the partition table in bottom up manner` `        ``for` `(i = ``1``; i <= sum / ``2``; i++) {` `            ``for` `(j = ``1``; j <= n; j++) {` `                ``part[i][j] = part[i][j - ``1``];` `                ``if` `(i >= arr[j - ``1``])` `                    ``part[i][j]` `                        ``= part[i][j]` `                          ``|| part[i - arr[j - ``1``]][j - ``1``];` `            ``}` `        ``}`   `        ``/* // uncomment this part to print table` `        ``for (i = 0; i <= sum/2; i++)` `        ``{` `            ``for (j = 0; j <= n; j++)` `                ``printf ("%4d", part[i][j]);` `            ``printf("\n");` `        ``} */`   `        ``return` `part[sum / ``2``][n];` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `arr[] = { ``3``, ``1``, ``1``, ``2``, ``2``, ``1` `};` `        ``int` `n = arr.length;` `        ``if` `(findPartition(arr, n) == ``true``)` `            ``System.out.println(` `                ``"Can be divided into two subsets of equal sum"``);` `        ``else` `            ``System.out.println(` `                ``"Can not be divided into two subsets of equal sum"``);` `    ``}` `}` `/* This code is contributed by Devesh Agrawal */`

Python3

 `# Dynamic Programming based python` `# program to partition problem`   `# Returns true if arr[] can be` `# partitioned in two subsets of` `# equal sum, otherwise false`     `def` `findPartition(arr, n):` `    ``sum` `=` `0` `    ``i, j ``=` `0``, ``0`   `    ``# calculate sum of all elements` `    ``for` `i ``in` `range``(n):` `        ``sum` `+``=` `arr[i]`   `    ``if` `sum` `%` `2` `!``=` `0``:` `        ``return` `false`   `    ``part ``=` `[[``True` `for` `i ``in` `range``(n ``+` `1``)]` `            ``for` `j ``in` `range``(``sum` `/``/` `2` `+` `1``)]`   `    ``# initialize top row as true` `    ``for` `i ``in` `range``(``0``, n ``+` `1``):` `        ``part[``0``][i] ``=` `True`   `    ``# initialize leftmost column,` `    ``# except part[0][0], as 0` `    ``for` `i ``in` `range``(``1``, ``sum` `/``/` `2` `+` `1``):` `        ``part[i][``0``] ``=` `False`   `    ``# fill the partition table in` `    ``# bottom up manner` `    ``for` `i ``in` `range``(``1``, ``sum` `/``/` `2` `+` `1``):`   `        ``for` `j ``in` `range``(``1``, n ``+` `1``):` `            ``part[i][j] ``=` `part[i][j ``-` `1``]`   `            ``if` `i >``=` `arr[j ``-` `1``]:` `                ``part[i][j] ``=` `(part[i][j] ``or` `                              ``part[i ``-` `arr[j ``-` `1``]][j ``-` `1``])`   `    ``return` `part[``sum` `/``/` `2``][n]`     `# Driver Code` `arr ``=` `[``3``, ``1``, ``1``, ``2``, ``2``, ``1``]` `n ``=` `len``(arr)`   `# Function call` `if` `findPartition(arr, n) ``=``=` `True``:` `    ``print``(``"Can be divided into two"``,` `          ``"subsets of equal sum"``)` `else``:` `    ``print``(``"Can not be divided into "``,` `          ``"two subsets of equal sum"``)`   `# This code is contributed` `# by mohit kumar 29`

C#

 `// A dynamic programming based C# program` `// for partition problem` `using` `System;`   `class` `GFG {`   `    ``// Returns true if arr[] can be partitioned` `    ``// in two subsets of equal sum, otherwise` `    ``// false` `    ``static` `bool` `findPartition(``int``[] arr, ``int` `n)` `    ``{`   `        ``int` `sum = 0;` `        ``int` `i, j;`   `        ``// Calculate sum of all elements` `        ``for` `(i = 0; i < n; i++)` `            ``sum += arr[i];`   `        ``if` `(sum % 2 != 0)` `            ``return` `false``;`   `        ``bool``[, ] part = ``new` `bool``[sum / 2 + 1, n + 1];`   `        ``// initialize top row as true` `        ``for` `(i = 0; i <= n; i++)` `            ``part[0, i] = ``true``;`   `        ``// initialize leftmost column, except` `        ``// part[0][0], as 0` `        ``for` `(i = 1; i <= sum / 2; i++)` `            ``part[i, 0] = ``false``;`   `        ``// Fill the partition table in bottom` `        ``// up manner` `        ``for` `(i = 1; i <= sum / 2; i++) {` `            ``for` `(j = 1; j <= n; j++) {` `                ``part[i, j] = part[i, j - 1];` `                ``if` `(i >= arr[j - 1])` `                    ``part[i, j]` `                        ``= part[i, j - 1]` `                          ``|| part[i - arr[j - 1], j - 1];` `            ``}` `        ``}`   `        ``/* // uncomment this part to print table` `        ``for (i = 0; i <= sum/2; i++)` `        ``{` `            ``for (j = 0; j <= n; j++)` `                ``printf ("%4d", part[i][j]);` `            ``printf("\n");` `        ``} */`   `        ``return` `part[sum / 2, n];` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int``[] arr = { 3, 1, 1, 2, 2, 1 };` `        ``int` `n = arr.Length;`   `        ``// Function call` `        ``if` `(findPartition(arr, n) == ``true``)` `            ``Console.Write(``"Can be divided"` `                          ``+ ``" into two subsets of"` `                          ``+ ``" equal sum"``);` `        ``else` `            ``Console.Write(``"Can not be "` `                          ``+ ``"divided into two subsets"` `                          ``+ ``" of equal sum"``);` `    ``}` `}`   `// This code is contributed by Sam007.`

Javascript

 ``

Output

`Can be divided into two subsets of equal sum`

The following diagram shows the values in the partition table:

Time Complexity: O(sum * N)
Auxiliary Space: O(sum * N)

Note: this solution will not be feasible for arrays with a big sum

Space-optimized approach for the above solution:

To solve the problem follow the below idea:

We can space optimize the above dp approach as for calculating the values of the current row we require only previous row

Below is the implementation of the above approach:

C++

 `// A Dynamic Programming based` `// C++ program to partition problem` `#include ` `using` `namespace` `std;`   `// Returns true if arr[] can be partitioned` `// in two subsets of equal sum, otherwise false` `bool` `findPartiion(``int` `arr[], ``int` `n)` `{` `    ``int` `sum = 0;` `    ``int` `i, j;`   `    ``// Calculate sum of all elements` `    ``for` `(i = 0; i < n; i++)` `        ``sum += arr[i];`   `    ``if` `(sum % 2 != 0)` `        ``return` `false``;`   `    ``bool` `part[sum / 2 + 1];`   `    ``// Initialize the part array` `    ``// as 0` `    ``for` `(i = 0; i <= sum / 2; i++) {` `        ``part[i] = 0;` `    ``}`   `    ``// Fill the partition table in bottom up manner`   `    ``for` `(i = 0; i < n; i++) {` `        ``// the element to be included` `        ``// in the sum cannot be` `        ``// greater than the sum` `        ``for` `(j = sum / 2; j >= arr[i];` `             ``j--) { ``// check if sum - arr[i]` `            ``// could be formed` `            ``// from a subset` `            ``// using elements` `            ``// before index i` `            ``if` `(part[j - arr[i]] == 1 || j == arr[i])` `                ``part[j] = 1;` `        ``}` `    ``}`   `    ``return` `part[sum / 2];` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `arr[] = { 1, 3, 3, 2, 3, 2 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``// Function call` `    ``if` `(findPartiion(arr, n) == ``true``)` `        ``cout << ``"Can be divided into two subsets of equal "` `                ``"sum"``;` `    ``else` `        ``cout << ``"Can not be divided into"` `             ``<< ``" two subsets of equal sum"``;` `    ``return` `0;` `}`

Java

 `// A Dynamic Programming based` `// Java program to partition problem` `import` `java.io.*;`   `class` `GFG {`   `    ``// Returns true if arr[] can be partitioned` `    ``// in two subsets of equal sum, otherwise false` `    ``public` `static` `boolean` `findPartiion(``int` `arr[], ``int` `n)` `    ``{` `        ``int` `sum = ``0``;` `        ``int` `i, j;`   `        ``// Calculate sum of all elements` `        ``for` `(i = ``0``; i < n; i++)` `            ``sum += arr[i];`   `        ``if` `(sum % ``2` `!= ``0``)` `            ``return` `false``;`   `        ``boolean``[] part = ``new` `boolean``[sum / ``2` `+ ``1``];`   `        ``// Initialize the part array` `        ``// as 0` `        ``for` `(i = ``0``; i <= sum / ``2``; i++) {` `            ``part[i] = ``false``;` `        ``}`   `        ``// Fill the partition table in` `        ``// bottom up manner` `        ``for` `(i = ``0``; i < n; i++) {`   `            ``// The element to be included` `            ``// in the sum cannot be` `            ``// greater than the sum` `            ``for` `(j = sum / ``2``; j >= arr[i]; j--) {`   `                ``// Check if sum - arr[i] could be` `                ``// formed from a subset using elements` `                ``// before index i` `                ``if` `(part[j - arr[i]] == ``true` `|| j == arr[i])` `                    ``part[j] = ``true``;` `            ``}` `        ``}` `        ``return` `part[sum / ``2``];` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `arr[] = { ``1``, ``3``, ``3``, ``2``, ``3``, ``2` `};` `        ``int` `n = ``6``;`   `        ``// Function call` `        ``if` `(findPartiion(arr, n) == ``true``)` `            ``System.out.println(``"Can be divided into two "` `                               ``+ ``"subsets of equal sum"``);` `        ``else` `            ``System.out.println(` `                ``"Can not be divided into "` `                ``+ ``"two subsets of equal sum"``);` `    ``}` `}`   `// This code is contributed by RohitOberoi`

Python3

 `# A Dynamic Programming based` `# Python3 program to partition problem`   `# Returns true if arr[] can be partitioned` `# in two subsets of equal sum, otherwise false`     `def` `findPartiion(arr, n):` `    ``Sum` `=` `0`   `    ``# Calculate sum of all elements` `    ``for` `i ``in` `range``(n):` `        ``Sum` `+``=` `arr[i]` `    ``if` `(``Sum` `%` `2` `!``=` `0``):` `        ``return` `0` `    ``part ``=` `[``0``] ``*` `((``Sum` `/``/` `2``) ``+` `1``)`   `    ``# Initialize the part array as 0` `    ``for` `i ``in` `range``((``Sum` `/``/` `2``) ``+` `1``):` `        ``part[i] ``=` `0`   `    ``# Fill the partition table in bottom up manner` `    ``for` `i ``in` `range``(n):`   `        ``# the element to be included` `        ``# in the sum cannot be` `        ``# greater than the sum` `        ``for` `j ``in` `range``(``Sum` `/``/` `2``, arr[i] ``-` `1``, ``-``1``):`   `            ``# check if sum - arr[i]` `            ``# could be formed` `            ``# from a subset` `            ``# using elements` `            ``# before index i` `            ``if` `(part[j ``-` `arr[i]] ``=``=` `1` `or` `j ``=``=` `arr[i]):` `                ``part[j] ``=` `1`   `    ``return` `part[``Sum` `/``/` `2``]`     `# Driver code` `arr ``=` `[``1``, ``3``, ``3``, ``2``, ``3``, ``2``]` `n ``=` `len``(arr)`   `# Function call` `if` `(findPartiion(arr, n) ``=``=` `1``):` `    ``print``(``"Can be divided into two subsets of equal sum"``)` `else``:` `    ``print``(``"Can not be divided into two subsets of equal sum"``)`   `    ``# This code is contributed by divyeshrabadiya07`

C#

 `// A Dynamic Programming based` `// C# program to partition problem` `using` `System;` `class` `GFG {`   `    ``// Returns true if arr[] can be partitioned` `    ``// in two subsets of equal sum, otherwise false` `    ``static` `bool` `findPartiion(``int``[] arr, ``int` `n)` `    ``{` `        ``int` `sum = 0;` `        ``int` `i, j;`   `        ``// Calculate sum of all elements` `        ``for` `(i = 0; i < n; i++)` `            ``sum += arr[i];` `        ``if` `(sum % 2 != 0)` `            ``return` `false``;` `        ``bool``[] part = ``new` `bool``[sum / 2 + 1];`   `        ``// Initialize the part array` `        ``// as 0` `        ``for` `(i = 0; i <= sum / 2; i++) {` `            ``part[i] = ``false``;` `        ``}`   `        ``// Fill the partition table in` `        ``// bottom up manner` `        ``for` `(i = 0; i < n; i++) {`   `            ``// The element to be included` `            ``// in the sum cannot be` `            ``// greater than the sum` `            ``for` `(j = sum / 2; j >= arr[i]; j--) {`   `                ``// Check if sum - arr[i] could be` `                ``// formed from a subset using elements` `                ``// before index i` `                ``if` `(part[j - arr[i]] == ``true` `|| j == arr[i])` `                    ``part[j] = ``true``;` `            ``}` `        ``}` `        ``return` `part[sum / 2];` `    ``}`   `    ``// Driver code` `    ``static` `void` `Main()` `    ``{` `        ``int``[] arr = { 1, 3, 3, 2, 3, 2 };` `        ``int` `n = 6;`   `        ``// Function call` `        ``if` `(findPartiion(arr, n) == ``true``)` `            ``Console.WriteLine(``"Can be divided into two "` `                              ``+ ``"subsets of equal sum"``);` `        ``else` `            ``Console.WriteLine(``"Can not be divided into "` `                              ``+ ``"two subsets of equal sum"``);` `    ``}` `}`   `// This code is contributed by divyesh072019`

Javascript

 ``

Output

`Can be divided into two subsets of equal sum`

Time Complexity: O(sum * N)
Auxiliary Space: O(sum)

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