C# Program for Subset Sum Problem | DP-25
Write a C# program for a given set of non-negative integers and a value sum, the task is to check if there is a subset of the given set whose sum is equal to the given sum.
Examples:
Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 9
Output: True
Explanation: There is a subset (4, 5) with sum 9.Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 30
Output: False
Explanation: There is no subset that adds up to 30.
C# Program for Subset Sum Problem using Recursion:
For the recursive approach, there will be two cases.
- Consider the ‘last’ element to be a part of the subset. Now the new required sum = required sum – value of ‘last’ element.
- Don’t include the ‘last’ element in the subset. Then the new required sum = old required sum.
In both cases, the number of available elements decreases by 1.
Step-by-step approach:
- Build a recursive function and pass the index to be considered (here gradually moving from the last end) and the remaining sum amount.
- For each index check the base cases and utilize the above recursive call.
- If the answer is true for any recursion call, then there exists such a subset. Otherwise, no such subset exists.
Below is the implementation of the above approach.
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 |
Found a subset with given sum
Time Complexity: O(2n)
Auxiliary space: O(n)
C# Program for Subset Sum Problem using Memoization:
As seen in the previous recursion method, each state of the solution can be uniquely identified using two variables – the index and the remaining sum. So create a 2D array to store the value of each state to avoid recalculation of the same state.
Below is the implementation of the above approach:
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 |
YES
Time Complexity: O(sum*n)
Auxiliary space: O(n)
C# Program for Subset Sum Problem using Dynamic Programming:
We can solve the problem in Pseudo-polynomial time we can use the Dynamic programming approach.
So we will create a 2D array of size (n + 1) * (sum + 1) of type boolean. The state dp[i][j] will be true if there exists a subset of elements from set[0 . . . i] with sum value = ‘j’.
The dynamic programming relation is as follows:
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-set[i-1]]
Below is the implementation of the above approach:
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 |
Found a subset with given sum
Time Complexity: O(sum * n), where n is the size of the array.
Auxiliary Space: O(sum*n), as the size of the 2-D array is sum*n.
C# Program for Subset Sum Problem using Dynamic Programming with space optimization to linear:
In previous approach of dynamic programming we have derive the relation between states as given below:
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-set[i-1]]
If we observe that for calculating current dp[i][j] state we only need previous row dp[i-1][j] or dp[i-1][j-set[i-1]].
There is no need to store all the previous states just one previous state is used to compute result.
Step-by-step approach:
- Define two arrays prev and curr of size Sum+1 to store the just previous row result and current row result respectively.
- Once curr array is calculated then curr becomes our prev for the next row.
- When all rows are processed the answer is stored in prev array.
Below is the implementation of the above approach:
C#
using System; class Program { // Returns true if there is a subset of set[] // with sum equal to given sum static bool IsSubsetSum( int [] set , int n, int sum) { // Create a 2D array 'dp' with dimensions (n+1) x (sum+1) bool [,] dp = new bool [n + 1, sum + 1]; // If sum is 0, then answer is true for ( int i = 0; i <= n; i++) dp[i, 0] = true ; // If sum is not 0 and set is empty, // then answer is false for ( int i = 1; i <= sum; i++) dp[0, i] = false ; for ( int i = 1; i <= n; i++) { for ( int j = 1; j <= sum; j++) { if (j < set [i - 1]) dp[i, j] = dp[i - 1, j]; if (j >= set [i - 1]) dp[i, j] = dp[i - 1, j] || dp[i - 1, j - set [i - 1]]; } } return dp[n, sum]; } static void Main() { int [] set = { 3, 34, 4, 12, 5, 2 }; int sum = 9; int n = set .Length; if (IsSubsetSum( set , n, sum)) Console.WriteLine( "Found a subset with given sum" ); else Console.WriteLine( "No subset with given sum" ); } } |
Found a subset with given sum
Time Complexity: O(sum * n), where n is the size of the array.
Auxiliary Space: O(sum), as the size of the 1-D array is sum+1.
Please refer complete article on Subset Sum Problem | DP-25 for more details!