Subset Sum Problem | DP-25
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 <iostream> using namespace std; // 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 ) cout << "Found a subset with given sum" ; else cout << "No subset with given sum" ; return 0; } // This code is contributed by shivanisinghss2110 |
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 // import java.io.*; 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 ?> |
Javascript
<script> // A recursive solution for subset sum problem // Returns true if there is a subset of set[] with sum // 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]); } let set = [ 3, 34, 4, 12, 5, 2 ]; let sum = 9; let n = set.length; if (isSubsetSum(set, n, sum) == true ) document.write( "Found a subset with given sum" ); else document.write( "No subset with given sum" ); // This code is contributed by mukesh07. </script> |
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-1] > j) DP[i][j] = DP[i-1][j] else DP[i][j] = DP[i-1][j] OR DP[i-1][j-A[i-1]]
- 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 <iostream> using namespace std; // Returns true if there is a subset of set[] // with sum 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 bottom 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]); cout <<"\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 ) cout << "Found a subset with given sum" ; else cout << "No subset with given sum" ; return 0; } // This code is contributed by shivanisinghss2110 |
C
// A Dynamic Programming 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) { // 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 bottom 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 import java.io.*; 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) { // 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 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 ]; } } /* // 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 sum 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 bottom 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 sum 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 sum 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 bottom // 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. ?> |
Javascript
<script> // A Dynamic Programming solution for subset sum problem // Returns true if there is a subset of // set[] with sum 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 let subset = new Array(sum + 1); for (let i = 0; i < sum + 1; i++) { subset[i] = new Array(sum + 1); for (let j = 0; j < n + 1; j++) { subset[i][j] = 0; } } // If sum is 0, then answer is true for (let i = 0; i <= n; i++) subset[0][i] = true ; // If sum is not 0 and set is empty, // then answer is false for (let i = 1; i <= sum; i++) subset[i][0] = false ; // Fill the subset table in bottom // up manner for (let i = 1; i <= sum; i++) { for (let 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]; } let set = [ 3, 34, 4, 12, 5, 2 ]; let sum = 9; let n = set.length; if (isSubsetSum(set, n, sum) == true ) document.write( "Found a subset" + " with given sum" ); else document.write( "No subset with" + " given sum" ); // This code is contributed by decode2207. </script> |
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. + O(n) for recursive stack space
Memoization Technique for finding Subset Sum:
Method:
- In this method, we also follow the recursive approach but In this method, we use another 2-D matrix in we first initialize with -1 or any negative value.
- In this method, we avoid the few of the recursive call which is repeated itself that’s why we use 2-D matrix. In this matrix we store the value of the previous call value.
Below is the implementation of the above approach:
C++
// CPP program for the above approach #include <bits/stdc++.h> using namespace std; // Taking the matrix as globally int tab[2000][2000]; // Check if possible subset with // given sum is possible or not int subsetSum( int a[], int n, int sum) { // If the sum is zero it means // we got our expected sum if (sum == 0) return 1; if (n <= 0) return 0; // If the value is not -1 it means it // already call the function // with the same value. // it will save our from the repetition. if (tab[n - 1][sum] != -1) return tab[n - 1][sum]; // if the value of a[n-1] is // greater than the sum. // we call for the next value if (a[n - 1] > sum) return tab[n - 1][sum] = subsetSum(a, n - 1, sum); else { // Here we do two calls because we // don't know which value is // full-fill our criteria // that's why we doing two calls return tab[n - 1][sum] = subsetSum(a, n - 1, sum) || subsetSum(a, n - 1, sum - a[n - 1]); } } // Driver Code int main() { // Storing the value -1 to the matrix memset (tab, -1, sizeof (tab)); int n = 5; int a[] = {1, 5, 3, 7, 4}; int sum = 12; if (subsetSum(a, n, sum)) { cout << "YES" << endl; } else cout << "NO" << endl; /* This Code is Contributed by Ankit Kumar*/ } |
Java
// Java program for the above approach import java.io.*; class GFG { // Check if possible subset with // given sum is possible or not static int subsetSum( int a[], int n, int sum) { // Storing the value -1 to the matrix int tab[][] = new int [n + 1 ][sum + 1 ]; for ( int i = 1 ; i <= n; i++) { for ( int j = 1 ; j <= sum; j++) { tab[i][j] = - 1 ; } } // If the sum is zero it means // we got our expected sum if (sum == 0 ) return 1 ; if (n <= 0 ) return 0 ; // If the value is not -1 it means it // already call the function // with the same value. // it will save our from the repetition. if (tab[n - 1 ][sum] != - 1 ) return tab[n - 1 ][sum]; // if the value of a[n-1] is // greater than the sum. // we call for the next value if (a[n - 1 ] > sum) return tab[n - 1 ][sum] = subsetSum(a, n - 1 , sum); else { // Here we do two calls because we // don't know which value is // full-fill our criteria // that's why we doing two calls if (subsetSum(a, n - 1 , sum) != 0 || subsetSum(a, n - 1 , sum - a[n - 1 ]) != 0 ) { return tab[n - 1 ][sum] = 1 ; } else return tab[n - 1 ][sum] = 0 ; } } // Driver Code public static void main(String[] args) { int n = 5 ; int a[] = { 1 , 5 , 3 , 7 , 4 }; int sum = 12 ; if (subsetSum(a, n, sum) != 0 ) { System.out.println( "YES\n" ); } else System.out.println( "NO\n" ); } } // This code is contributed by rajsanghavi9. |
Python3
# Python program for the above approach # Taking the matrix as globally tab = [[ - 1 for i in range ( 2000 )] for j in range ( 2000 )] # Check if possible subset with # given sum is possible or not def subsetSum(a, n, sum ): # If the sum is zero it means # we got our expected sum if ( sum = = 0 ): return 1 if (n < = 0 ): return 0 # If the value is not -1 it means it # already call the function # with the same value. # it will save our from the repetition. if (tab[n - 1 ][ sum ] ! = - 1 ): return tab[n - 1 ][ sum ] # if the value of a[n-1] is # greater than the sum. # we call for the next value if (a[n - 1 ] > sum ): tab[n - 1 ][ sum ] = subsetSum(a, n - 1 , sum ) return tab[n - 1 ][ sum ] else : # Here we do two calls because we # don't know which value is # full-fill our criteria # that's why we doing two calls tab[n - 1 ][ sum ] = subsetSum(a, n - 1 , sum ) return tab[n - 1 ][ sum ] or subsetSum(a, n - 1 , sum - a[n - 1 ]) # Driver Code n = 5 a = [ 1 , 5 , 3 , 7 , 4 ] sum = 12 if (subsetSum(a, n, sum )): print ( "YES" ) else : print ( "NO" ) # This code is contributed by shivani. |
C#
// C# program for the above approach using System; class GFG { // Check if possible subset with // given sum is possible or not static int subsetSum( int []a, int n, int sum) { // Storing the value -1 to the matrix int [,]tab = new int [n + 1,sum + 1]; for ( int i = 1; i <= n; i++) { for ( int j = 1; j <= sum; j++) { tab[i,j] = -1; } } // If the sum is zero it means // we got our expected sum if (sum == 0) return 1; if (n <= 0) return 0; // If the value is not -1 it means it // already call the function // with the same value. // it will save our from the repetition. if (tab[n - 1,sum] != -1) return tab[n - 1,sum]; // if the value of a[n-1] is // greater than the sum. // we call for the next value if (a[n - 1] > sum) return tab[n - 1,sum] = subsetSum(a, n - 1, sum); else { // Here we do two calls because we // don't know which value is // full-fill our criteria // that's why we doing two calls if (subsetSum(a, n - 1, sum) != 0 || subsetSum(a, n - 1, sum - a[n - 1]) != 0) { return tab[n - 1,sum] = 1; } else return tab[n - 1,sum] = 0; } } // Driver Code public static void Main(String[] args) { int n = 5; int []a = { 1, 5, 3, 7, 4 }; int sum = 12; if (subsetSum(a, n, sum) != 0) { Console.Write( "YES\n" ); } else Console.Write( "NO\n" ); } } // This code is contributed by shivanisinghss2110 |
Javascript
<script> // JavaScript Program for the above approach // Check if possible subset with // given sum is possible or not function subsetSum(a, n, sum) { // If the sum is zero it means // we got our expected sum if (sum == 0) return 1; if (n <= 0) return 0; // If the value is not -1 it means it // already call the function // with the same value. // it will save our from the repetition. if (tab[n - 1][sum] != -1) return tab[n - 1][sum]; // if the value of a[n-1] is // greater than the sum. // we call for the next value if (a[n - 1] > sum) return tab[n - 1][sum] = subsetSum(a, n - 1, sum); else { // Here we do two calls because we // don't know which value is // full-fill our criteria // that's why we doing two calls return tab[n - 1][sum] = subsetSum(a, n - 1, sum) || subsetSum(a, n - 1, sum - a[n - 1]); } } // Driver Code // Storing the value -1 to the matrix let tab = Array(2000).fill().map(() => Array(2000).fill(-1)); let n = 5; let a = [1, 5, 3, 7, 4]; let sum = 12; if (subsetSum(a, n, sum)) { document.write( "YES" + "<br>" ); } else { document.write( "NO" + "<br>" ); } // This code is contributed by Potta Lokesh </script> |
YES
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) + O(n) -> O(sum*n) = the size of 2-D array is sum*n and O(n)=auxiliary stack space.
Subset Sum Problem in O(sum) space
Perfect Sum Problem (Print all subsets with given sum)
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