Partition problem | DP-18
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:
arr[] = {1, 5, 11, 5} Output: true The array can be partitioned as {1, 5, 5} and {11} 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.
Following are the two main steps to solve this problem:
1) Calculate sum of the array. If sum is odd, there can not be two subsets with equal sum, so return false.
2) If sum of array elements is even, calculate sum/2 and find a subset of array with 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.
Recursive Solution
Following is the recursive property of the second step mentioned above.
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 a) isSubsetSum() without considering last element (reducing n to n-1) b) 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])
Below is the implementation of the above code:
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 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 <stdbool.h> #include <stdio.h> // 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 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
<?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 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
<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> |
Can be divided into two subsets of equal sum
Time Complexity: O(2^n) In the worst case, this solution tries two possibilities (whether to include or exclude) for every element.
Dynamic Programming Solution
1. Top-Down: Memoization
We can avoid the repeated work done in method 1 by storing the result calculated so far.
We just need to store all the values in a matrix.
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 sum bool 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 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<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 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; } |
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 )] * (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. |
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 sum function 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 false function 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 code let arr = [ 3, 1, 5, 9, 12 ]; let n = arr.length; // Function call if (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> |
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)
2. Bottom-Up: Tabulation
The problem can be solved using dynamic programming when the sum of the elements is not too big. We can create a 2D array part[][] of size (sum/2 + 1)*(n+1). And we can construct the solution in a bottom-up manner such that every filled entry has the following property
part[i][j] = true if a subset of {arr[0], arr[1], ..arr[j-1]} has sum equal to i, otherwise false
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 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<<part[i][j]; cout<<endl; } */ 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 ) 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> // 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
<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> |
Can be divided into two subsets of equal sum
Following diagram shows the values in the partition table.
Time Complexity: O(sum*n)
Auxiliary Space: O(sum*n)
Please note that this solution will not be feasible for arrays with big sum.
Dynamic Programming Solution (Space Complexity Optimized)
Instead of creating a 2-D array of size (sum/2 + 1)*(n + 1), we can solve this problem using an array of size (sum/2 + 1 ) only.
part[j] = true if there is a subset with sum equal to j, otherwise false.
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 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 ] # Drive 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
<script> // A Dynamic Programming based Javascript // program to partition problem // Returns true if arr[] can be partitioned // in two subsets of equal sum, otherwise false function 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 call if (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> |
Can be divided into two subsets of equal sum
Time Complexity: O(sum * n)
Auxiliary Space: O(sum)
Please note that this solution will not be feasible for arrays with big sum.
References:
http://en.wikipedia.org/wiki/Partition_problem
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.