text: Dynamic Programming | A recursive solution for subset sum problem ; Returns true if there is a subset of set [ ] with sum equal to given sum ; Base Cases ; If last element is greater than sum , then ignore it ; else , check if sum can be obtained by any of the following ( a ) including the last element ( b ) excluding the last element ; Driver code
code: class GFG { static boolean isSubsetSum ( int set [ ] , int n , int sum ) { if ( sum == 0 ) return true ; if ( n == 0 ) return false ; if ( set [ n - 1 ] > sum ) return isSubsetSum ( set , n - 1 , sum ) ; return isSubsetSum ( set , n - 1 , sum ) || isSubsetSum ( set , n - 1 , sum - set [ n - 1 ] ) ; } 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 " ) ; } }