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
- Print the answer
Below is the implementation of the above approach:
C++
// A recursive C++ program for partition problem#include <bits/stdc++.h>using namespace std;// A utility function that returns true if there is// a subset of arr[] with sum equal to given sumbool 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 falsebool 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 codeint 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 <stdbool.h>#include <stdio.h>// A utility function that returns true if there is// a subset of arr[] with sum equal to given sumbool 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 falsebool 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 codeint 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 problemimport 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 sumdef 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 falsedef 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 codeif __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 problemusing 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
<?php// A recursive PHP solution for partition problem// A utility function that returns true if there is // a subset of arr[] with sum equal to given sum function isSubsetSum ($arr, $n, $sum) { // Base Cases if ($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 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 callif (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
<script>// A recursive Javascript solution for partition problem // A utility function that returns true if there is a // subset of arr[] with sum equal to given sum function isSubsetSum(arr,n,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 function findPartition(arr,n) { // Calculate sum of the elements in array let sum = 0; for (let 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, Math.floor(sum / 2)); } // Driver code let arr=[3, 1, 5, 9, 12 ]; let n = arr.length; // Function call if (findPartition(arr, n) == true) document.write("Can be divided into two " + "subsets of equal sum"); else document.write( "Can not be divided into " + "two subsets of equal sum"); // This code is contributed by unknown2108</script> |
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
- Print the answer
Below is the implementation of the above approach:
C++
// A recursive C++ program for partition problem#include <bits/stdc++.h>using namespace std;// A utility function that returns true if there is// a subset of arr[] with sun equal to given sumbool isSubsetSum(int arr[], int n, int sum, vector<vector<int> >& 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 falsebool 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<vector<int> > 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 codeint 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 problemimport 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 sumdef 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 falsedef 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 codearr = [3, 1, 5, 9, 12]n = len(arr)# Function callif (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 problemusing 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
<script>// 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 sumfunction isSubsetSum(arr,n,sum,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 falsefunction findPartiion(arr, n){ // Calculate sum of the elements in array let sum = 0; for (let 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 let dp = new Array(n + 1).fill(new Array(sum+1).fill(-1)); // Find if there is subset with sum equal to // half of total sum return isSubsetSum(arr, n, sum / 2, dp);}// Driver codelet arr = [ 3, 1, 5, 9, 12 ];let n = arr.length; // Function callif (findPartiion(arr, n) == true) document.write("Can be divided into two subsets of equal sum");else document.write("Can not be divided into two subsets of equal sum");let arr2 = [ 3, 1, 5, 9, 14 ];let n2 = arr2.length;if (findPartiion(arr2, n2) == true) document.write("</br>","Can be divided into two subsets of equal sum");else document.write("</br>","Can not be divided into two subsets of equal sum"); // This code is contributed by shinjanpatra.</script> |
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
- Print the answer
Below is the implementation of the above approach:
C++
// A Dynamic Programming based// C++ program to partition problem#include <bits/stdc++.h>using namespace std;// Returns true if arr[] can be partitioned// in two subsets of equal sum, otherwise falsebool 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<<part[i][j]; cout<<endl; } */ return part[sum / 2][n];}// Driver Codeint main(){ int arr[] = { 3, 1, 1, 2, 2, 1 }; 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;} |
C
// A Dynamic Programming based C program to partition// problem#include <stdio.h>#include <stdbool.h>// Returns true if arr[] can be partitioned in two subsets// of equal sum, otherwise falsebool 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 codeint 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// problemimport 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 falsedef 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 Codearr = [3, 1, 1, 2, 2, 1]n = len(arr)# Function callif 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 problemusing 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
<script>// A dynamic programming based javascript // program for partition// problemclass Partition // Returns true if arr can be partitioned in two // subsets of equal sum, otherwise false function findPartition(arr , n) { var sum = 0; var i, j; // Calculate sum of all elements for (i = 0; i < n; i++) sum += arr[i]; if (sum % 2 != 0) return false; var part = Array(parseInt(sum / 2) + 1). fill().map(()=>Array(n + 1).fill(0)); // 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 <= parseInt(sum / 2); i++) part[i][0] = false; // Fill the partition table in bottom up manner for (i = 1; i <= parseInt(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[parseInt(sum / 2)][n]; } // Driver code var arr = [ 3, 1, 1, 2, 2, 1 ]; var n = arr.length; if (findPartition(arr, n) == true) document.write( "Can be divided into two subsets of equal sum" ); else document.write( "Can not be divided into two subsets of equal sum" );// This code contributed by Rajput-Ji</script> |
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 <bits/stdc++.h>using namespace std;// Returns true if arr[] can be partitioned// in two subsets of equal sum, otherwise falsebool 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 Codeint 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 problemimport 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 falsedef 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 codearr = [1, 3, 3, 2, 3, 2]n = len(arr)# Function callif (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 problemusing 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
<script>// A Dynamic Programming based Javascript// program to partition problem// Returns true if arr[] can be partitioned// in two subsets of equal sum, otherwise falsefunction findPartiion(arr, n){ let sum = 0; let i, j; // Calculate sum of all elements for(i = 0; i < n; i++) sum += arr[i]; if (sum % 2 != 0) return false; let part = new Array(parseInt(sum / 2 + 1, 10)); // Initialize the part array // as 0 for(i = 0; i <= parseInt(sum / 2, 10); 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 = parseInt(sum / 2, 10); 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[parseInt(sum / 2, 10)];} // Driver code let arr = [ 1, 3, 3, 2, 3, 2 ];let n = arr.length;// Function callif (findPartiion(arr, n) == true) document.write("Can be divided into two " + "subsets of equal sum");else document.write("Can not be divided into " + "two subsets of equal sum");// This code is contributed by suresh07 </script> |
Output
Can be divided into two subsets of equal sum
Time Complexity: O(sum * N)
Auxiliary Space: O(sum)
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