Given a set of non-negative integers, and a value sum, determine if there is a subset of the given set with sum equal to given sum.
Example:
Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 9 Output: True There is a subset (4, 5) with sum 9. Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 30 Output: False There is no subset that add up to 30.
Method 1: Recursion.
Approach: For the recursive approach we will consider two cases.
- Consider the last element and now the required sum = target sum – value of ‘last’ element and number of elements = total elements – 1
- Leave the ‘last’ element and now the required sum = target sum and number of elements = total elements – 1
Following is the recursive formula for isSubsetSum() problem.
isSubsetSum(set, n, sum) = isSubsetSum(set, n-1, sum) || isSubsetSum(set, n-1, sum-set[n-1]) Base Cases: isSubsetSum(set, n, sum) = false, if sum > 0 and n == 0 isSubsetSum(set, n, sum) = true, if sum == 0
Let’s take a look at the simulation of above approach-:
set[]={3, 4, 5, 2} sum=9 (x, y)= 'x' is the left number of elements, 'y' is the required sum (4, 9) {True} / \ (3, 6) (3, 9) / \ / \ (2, 2) (2, 6) (2, 5) (2, 9) {True} / \ (1, -3) (1, 2) {False} {True} / \ (0, 0) (0, 2) {True} {False}
C++
// A recursive solution for subset sum problem #include <stdio.h> // Returns true if there is a subset // of set[] with sum equal to given sum bool isSubsetSum( int set[], int n, int sum) { // Base Cases if (sum == 0) return true ; if (n == 0) return false ; // If last element is greater than sum, // then ignore it if (set[n - 1] > sum) return isSubsetSum(set, 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(set, n - 1, sum) || isSubsetSum(set, n - 1, sum - set[n - 1]); } // Driver code int main() { int set[] = { 3, 34, 4, 12, 5, 2 }; int sum = 9; int n = sizeof (set) / sizeof (set[0]); if (isSubsetSum(set, n, sum) == true ) printf ( "Found a subset with given sum" ); else printf ( "No subset with given sum" ); return 0; } |
Java
// A recursive solution for subset sum // problem class GFG { // Returns true if there is a subset // of set[] with sum equal to given sum static boolean isSubsetSum( int set[], int n, int sum) { // Base Cases if (sum == 0 ) return true ; if (n == 0 ) return false ; // If last element is greater than // sum, then ignore it if (set[n - 1 ] > sum) return isSubsetSum(set, 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(set, n - 1 , sum) || isSubsetSum(set, n - 1 , sum - set[n - 1 ]); } /* Driver code */ public static void main(String args[]) { int set[] = { 3 , 34 , 4 , 12 , 5 , 2 }; int sum = 9 ; int n = set.length; if (isSubsetSum(set, n, sum) == true ) System.out.println( "Found a subset" + " with given sum" ); else System.out.println( "No subset with" + " given sum" ); } } /* This code is contributed by Rajat Mishra */ |
Python3
# A recursive solution for subset sum # problem # Returns true if there is a subset # of set[] with sun equal to given sum def isSubsetSum( set , n, sum ): # Base Cases if ( sum = = 0 ): return True if (n = = 0 ): return False # If last element is greater than # sum, then ignore it if ( set [n - 1 ] > sum ): return isSubsetSum( set , 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( set , n - 1 , sum ) or isSubsetSum( set , n - 1 , sum - set [n - 1 ]) # Driver code set = [ 3 , 34 , 4 , 12 , 5 , 2 ] sum = 9 n = len ( set ) if (isSubsetSum( set , n, sum ) = = True ): print ( "Found a subset with given sum" ) else : print ( "No subset with given sum" ) # This code is contributed by Nikita Tiwari. |
C#
// A recursive solution for subset sum problem using System; class GFG { // Returns true if there is a subset of set[] with sum // equal to given sum static bool isSubsetSum( int [] set , int n, int sum) { // Base Cases if (sum == 0) return true ; if (n == 0) return false ; // If last element is greater than sum, // then ignore it if ( set [n - 1] > sum) return isSubsetSum( set , 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( set , n - 1, sum) || isSubsetSum( set , n - 1, sum - set [n - 1]); } // Driver code public static void Main() { int [] set = { 3, 34, 4, 12, 5, 2 }; int sum = 9; int n = set .Length; if (isSubsetSum( set , n, sum) == true ) Console.WriteLine( "Found a subset with given sum" ); else Console.WriteLine( "No subset with given sum" ); } } // This code is contributed by Sam007 |
PHP
<?php // A recursive solution for subset sum problem // Returns true if there is a subset of set // with sun equal to given sum function isSubsetSum( $set , $n , $sum ) { // Base Cases if ( $sum == 0) return true; if ( $n == 0) return false; // If last element is greater // than sum, then ignore it if ( $set [ $n - 1] > $sum ) return isSubsetSum( $set , $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( $set , $n - 1, $sum ) || isSubsetSum( $set , $n - 1, $sum - $set [ $n - 1]); } // Driver Code $set = array (3, 34, 4, 12, 5, 2); $sum = 9; $n = 6; if (isSubsetSum( $set , $n , $sum ) == true) echo "Found a subset with given sum" ; else echo "No subset with given sum" ; // This code is contributed by Anuj_67 ?> |
Found a subset with given sum
Complexity Analysis: The above solution may try all subsets of given set in worst case. Therefore time complexity of the above solution is exponential. The problem is in-fact NP-Complete (There is no known polynomial time solution for this problem).
Method 2: To solve the problem in Pseudo-polynomial time use the Dynamic programming.
So we will create a 2D array of size (arr.size() + 1) * (target + 1) of type boolean. The state DP[i][j] will be true if there exists a subset of elements from A[0….i] with sum value = ‘j’. The approach for the problem is:
if (A[i] > j) DP[i][j] = DP[i-1][j] else DP[i][j] = DP[i-1][j] OR DP[i-1][j-A[i]]
- This means that if current element has value greater than ‘current sum value’ we will copy the answer for previous cases
- And if the current sum value is greater than the ‘ith’ element we will see if any of previous states have already experienced the sum=’j’ OR any previous states experienced a value ‘j – A[i]’ which will solve our purpose.
The below simulation will clarify the above approach:
set[]={3, 4, 5, 2} target=6 0 1 2 3 4 5 6 0 T F F F F F F 3 T F F T F F F 4 T F F T T F F 5 T F F T T T F 2 T F T T T T T
Below is the implementation of the above approach:
C++
// A Dynamic Programming solution // for subset sum problem #include <stdio.h> // Returns true if there is a subset of set[] // with sun equal to given sum bool isSubsetSum( int set[], int n, int sum) { // The value of subset[i][j] will be true if // there is a subset of set[0..j-1] with sum // equal to i bool subset[n + 1][sum + 1]; // If sum is 0, then answer is true for ( int i = 0; i <= n; i++) subset[i][0] = true ; // If sum is not 0 and set is empty, // then answer is false for ( int i = 1; i <= sum; i++) subset[0][i] = false ; // Fill the subset table in botton up manner for ( int i = 1; i <= n; i++) { for ( int j = 1; j <= sum; j++) { if (j < set[i - 1]) subset[i][j] = subset[i - 1][j]; if (j >= set[i - 1]) subset[i][j] = subset[i - 1][j] || subset[i - 1][j - set[i - 1]]; } } /* // uncomment this code to print table for (int i = 0; i <= n; i++) { for (int j = 0; j <= sum; j++) printf ("%4d", subset[i][j]); printf("\n"); }*/ return subset[n][sum]; } // Driver code int main() { int set[] = { 3, 34, 4, 12, 5, 2 }; int sum = 9; int n = sizeof (set) / sizeof (set[0]); if (isSubsetSum(set, n, sum) == true ) printf ( "Found a subset with given sum" ); else printf ( "No subset with given sum" ); return 0; } // This code is contributed by Arjun Tyagi. |
Java
// A Dynamic Programming solution for subset // sum problem class GFG { // Returns true if there is a subset of // set[] with sun equal to given sum static boolean isSubsetSum( int set[], int n, int sum) { // The value of subset[i][j] will be // true if there is a subset of // set[0..j-1] with sum equal to i boolean subset[][] = new boolean [sum + 1 ][n + 1 ]; // If sum is 0, then answer is true for ( int i = 0 ; i <= n; i++) subset[ 0 ][i] = true ; // If sum is not 0 and set is empty, // then answer is false for ( int i = 1 ; i <= sum; i++) subset[i][ 0 ] = false ; // Fill the subset table in botton // up manner for ( int i = 1 ; i <= sum; i++) { for ( int j = 1 ; j <= n; j++) { subset[i][j] = subset[i][j - 1 ]; if (i >= set[j - 1 ]) subset[i][j] = subset[i][j] || subset[i - set[j - 1 ]][j - 1 ]; } } /* // uncomment this code to print table for (int i = 0; i <= sum; i++) { for (int j = 0; j <= n; j++) System.out.println (subset[i][j]); } */ return subset[sum][n]; } /* Driver code*/ public static void main(String args[]) { int set[] = { 3 , 34 , 4 , 12 , 5 , 2 }; int sum = 9 ; int n = set.length; if (isSubsetSum(set, n, sum) == true ) System.out.println( "Found a subset" + " with given sum" ); else System.out.println( "No subset with" + " given sum" ); } } /* This code is contributed by Rajat Mishra */ |
Python3
# A Dynamic Programming solution for subset # sum problem Returns true if there is a subset of # set[] with sun equal to given sum # Returns true if there is a subset of set[] # with sun equal to given sum def isSubsetSum( set , n, sum ): # The value of subset[i][j] will be # true if there is a # subset of set[0..j-1] with sum equal to i subset = ([[ False for i in range ( sum + 1 )] for i in range (n + 1 )]) # If sum is 0, then answer is true for i in range (n + 1 ): subset[i][ 0 ] = True # If sum is not 0 and set is empty, # then answer is false for i in range ( 1 , sum + 1 ): subset[ 0 ][i] = False # Fill the subset table in botton up manner for i in range ( 1 , n + 1 ): for j in range ( 1 , sum + 1 ): if j< set [i - 1 ]: subset[i][j] = subset[i - 1 ][j] if j> = set [i - 1 ]: subset[i][j] = (subset[i - 1 ][j] or subset[i - 1 ][j - set [i - 1 ]]) # uncomment this code to print table # for i in range(n + 1): # for j in range(sum + 1): # print (subset[i][j], end =" ") # print() return subset[n][ sum ] # Driver code if __name__ = = '__main__' : set = [ 3 , 34 , 4 , 12 , 5 , 2 ] sum = 9 n = len ( set ) if (isSubsetSum( set , n, sum ) = = True ): print ( "Found a subset with given sum" ) else : print ( "No subset with given sum" ) # This code is contributed by # sahil shelangia. |
C#
// A Dynamic Programming solution for // subset sum problem using System; class GFG { // Returns true if there is a subset // of set[] with sun equal to given sum static bool isSubsetSum( int [] set , int n, int sum) { // The value of subset[i][j] will be true if there // is a subset of set[0..j-1] with sum equal to i bool [, ] subset = new bool [sum + 1, n + 1]; // If sum is 0, then answer is true for ( int i = 0; i <= n; i++) subset[0, i] = true ; // If sum is not 0 and set is empty, // then answer is false for ( int i = 1; i <= sum; i++) subset[i, 0] = false ; // Fill the subset table in bottom up manner for ( int i = 1; i <= sum; i++) { for ( int j = 1; j <= n; j++) { subset[i, j] = subset[i, j - 1]; if (i >= set [j - 1]) subset[i, j] = subset[i, j] || subset[i - set [j - 1], j - 1]; } } return subset[sum, n]; } // Driver code public static void Main() { int [] set = { 3, 34, 4, 12, 5, 2 }; int sum = 9; int n = set .Length; if (isSubsetSum( set , n, sum) == true ) Console.WriteLine( "Found a subset with given sum" ); else Console.WriteLine( "No subset with given sum" ); } } // This code is contributed by Sam007 |
PHP
<?php // A Dynamic Programming solution for // subset sum problem // Returns true if there is a subset of // set[] with sun equal to given sum function isSubsetSum( $set , $n , $sum ) { // The value of subset[i][j] will // be true if there is a subset of // set[0..j-1] with sum equal to i $subset = array ( array ()); // If sum is 0, then answer is true for ( $i = 0; $i <= $n ; $i ++) $subset [ $i ][0] = true; // If sum is not 0 and set is empty, // then answer is false for ( $i = 1; $i <= $sum ; $i ++) $subset [0][ $i ] = false; // Fill the subset table in botton // up manner for ( $i = 1; $i <= $n ; $i ++) { for ( $j = 1; $j <= $sum ; $j ++) { if ( $j < $set [ $i -1]) $subset [ $i ][ $j ] = $subset [ $i -1][ $j ]; if ( $j >= $set [ $i -1]) $subset [ $i ][ $j ] = $subset [ $i -1][ $j ] || $subset [ $i - 1][ $j - $set [ $i -1]]; } } /* // uncomment this code to print table for (int i = 0; i <= n; i++) { for (int j = 0; j <= sum; j++) printf ("%4d", subset[i][j]); printf("n"); }*/ return $subset [ $n ][ $sum ]; } // Driver code $set = array (3, 34, 4, 12, 5, 2); $sum = 9; $n = count ( $set ); if (isSubsetSum( $set , $n , $sum ) == true) echo "Found a subset with given sum" ; else echo "No subset with given sum" ; // This code is contributed by anuj_67. ?> |
Found a subset with given sum
Complexity Analysis:
- Time Complexity: O(sum*n), where sum is the ‘target sum’ and ‘n’ is the size of array.
- Auxiliary Space: O(sum*n), as the size of 2-D array is sum*n.
Subset Sum Problem in O(sum) space
Perfect Sum Problem (Print all subsets with given sum)
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