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 program to test above function 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; } |
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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 program to test above function */ 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 */ |
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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 program to test above function set = [3, 34, 4, 12, 5, 2] sum = 9n = 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. |
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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 program 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 |
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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 ?> |
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Output:
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][sum-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 program to test above function 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. |
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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 program to test above function */ 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 */ |
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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 program to test above function 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. |
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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 program 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 |
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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 program to test above function $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. ?> |
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Output:
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)
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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