A Space Optimized DP solution for 0 | Java program of a space optimized DP solution for 0 - 1 knapsack problem . ; val [ ] is for storing maximum profit for each weight wt [ ] is for storing weights n number of item W maximum capacity of bag dp [ W + 1 ] to store final result ; array to store final result dp [ i ] stores the profit with KnapSack capacity " i " ; initially profit with 0 to W KnapSack capacity is 0 ; iterate through all items ; traverse dp array from right to left ; above line finds out maximum of dp [ j ] ( excluding ith element value ) and val [ i ] + dp [ j - wt [ i ] ] ( including ith element value and the profit with " KnapSack ▁ capacity ▁ - ▁ ith ▁ element ▁ weight " ) ; Driver code
import java . util . Arrays ; class GFG { static int KnapSack ( int val [ ] , int wt [ ] , int n , int W ) { int [ ] dp = new int [ W + 1 ] ; Arrays . fill ( dp , 0 ) ; for ( int i = 0 ; i < n ; i ++ ) for ( int j = W ; j >= wt [ i ] ; j -- ) dp [ j ] = Math . max ( dp [ j ] , val [ i ] + dp [ j - wt [ i ] ] ) ; return dp [ W ] ; } public static void main ( String [ ] args ) { int val [ ] = { 7 , 8 , 4 } , wt [ ] = { 3 , 8 , 6 } , W = 10 , n = 3 ; System . out . println ( KnapSack ( val , wt , n , W ) ) ; } }