Given the weights and values of n items, put these items in a knapsack of capacity W to get the maximum total value in the knapsack. In other words, given two integer arrays val[0..n-1] and wt[0..n-1] which represent values and weights associated with n items respectively. Also given an integer W which represents knapsack capacity, find out the maximum value subset of val[] such that sum of the weights of this subset is smaller than or equal to W. We cannot break an item, either pick the complete item or don’t pick it (0-1 property).
Here W <= 2000000 and n <= 500
Examples:
Input: W = 10, n = 3
val[] = {7, 8, 4}
wt[] = {3, 8, 6}
Output: 11
Explanation: We get maximum value by picking items of 3 KG and 6 KG.
We have discussed a Dynamic Programming based solution here. In the previous solution, we used a n * W matrix. We can reduce the used extra space. The idea behind the optimization is, to compute mat[i][j], we only need solution of previous row. In 0-1 Knapsack Problem if we are currently on mat[i][j] and we include ith element then we move j-wt[i] steps back in previous row and if we exclude the current element we move on jth column in the previous row. So here we can observe that at a time we are working only with 2 consecutive rows.
In the below solution, we create a matrix of size 2*W. If n is odd, then the final answer will be at mat[0][W] and if n is even then the final answer will be at mat[1][W] because index starts from 0.
C++
// C++ program of a space optimized DP solution for// 0-1 knapsack problem.#include<bits/stdc++.h>using namespace std;// val[] is for storing maximum profit for each weight// wt[] is for storing weights// n number of item// W maximum capacity of bag// mat[2][W+1] to store final resultint KnapSack(int val[], int wt[], int n, int W){ // matrix to store final result int mat[2][W+1]; memset(mat, 0, sizeof(mat)); // iterate through all items int i = 0; while (i < n) // one by one traverse each element { int j = 0; // traverse all weights j <= W // if i is odd that mean till now we have odd // number of elements so we store result in 1th // indexed row if (i%2!=0) { while (++j <= W) // check for each value { if (wt[i] <= j) // include element mat[1][j] = max(val[i] + mat[0][j-wt[i]], mat[0][j] ); else // exclude element mat[1][j] = mat[0][j]; } } // if i is even that mean till now we have even number // of elements so we store result in 0th indexed row else { while(++j <= W) { if (wt[i] <= j) mat[0][j] = max(val[i] + mat[1][j-wt[i]], mat[1][j]); else mat[0][j] = mat[1][j]; } } i++; } // Return mat[0][W] if n is odd, else mat[1][W] return (n%2 != 0)? mat[0][W] : mat[1][W];}// Driver program to test the casesint main(){ int val[] = {7, 8, 4}, wt[] = {3, 8, 6}, W = 10, n = 3; cout << KnapSack(val, wt, n, W) << endl; return 0;} |
Java
// Java program of a space optimized DP solution for // 0-1 knapsack problem. class GFG{ // val[] is for storing maximum // profit for each weight // wt[] is for storing weights // n number of item // W maximum capacity of bag // mat[2][W+1] to store final result static int KnapSack(int val[], int wt[], int n, int W) { // matrix to store final result int mat[][] = new int[2][W + 1]; // iterate through all items int i = 0; while (i < n) // one by one traverse each element { int j = 0; // traverse all weights j <= W // if i is odd that mean till now we have odd // number of elements so we store result in 1th // indexed row if (i % 2 != 0) { while (++j <= W) // check for each value { if (wt[i] <= j) // include element { mat[1][j] = Math.max(val[i] + mat[0][j - wt[i]], mat[0][j]); } else // exclude element { mat[1][j] = mat[0][j]; } } } // if i is even that means till now // we have even number of elements // so we store result in 0th indexed row else { while (++j <= W) { if (wt[i] <= j) { mat[0][j] = Math.max(val[i] + mat[1][j - wt[i]], mat[1][j]); } else { mat[0][j] = mat[1][j]; } } } i++; } // Return mat[0][W] if n is odd, else mat[1][W] return (n % 2 != 0) ? mat[0][W] : mat[1][W]; } // Driver Code 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)); }}// This code is contributed by PrinciRaj1992 |
Python3
# Python 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 # mat[2][W+1] to store final result def KnapSack(val, wt, n, W): # matrix to store final result mat = [[0 for i in range(W + 1)] for i in range(2)] # iterate through all items i = 0 while i < n: # one by one traverse # each element j = 0 # traverse all weights j <= W # if i is odd that mean till # now we have odd number of # elements so we store result # in 1th indexed row if i % 2 == 0: while j < W: # check for each value j += 1 if wt[i] <= j: # include element mat[1][j] = max(val[i] + mat[0][j - wt[i]], mat[0][j]) else: # exclude element mat[1][j] = mat[0][j] # if i is even that mean till # now we have even number # of elements so we store # result in 0th indexed row else: while j < W: j += 1 if wt[i] <= j: mat[0][j] = max(val[i] + mat[1][j - wt[i]], mat[1][j]) else: mat[0][j] = mat[1][j] i += 1 # Return mat[0][W] if n is # odd, else mat[1][W] if n % 2 == 0: return mat[0][W] else: return mat[1][W]# Driver codeval = [7, 8, 4]wt = [3, 8, 6] W = 10n = 3print(KnapSack(val, wt, n, W))# This code is contributed# by sahilshelangia |
C#
// C# program of a space optimized DP solution for // 0-1 knapsack problem. using System; class GFG{ // val[] is for storing maximum // profit for each weight // wt[] is for storing weights // n number of item // W maximum capacity of bag // mat[2,W+1] to store final result static int KnapSack(int []val, int []wt, int n, int W) { // matrix to store final result int [,]mat = new int[2, W + 1]; // iterate through all items int i = 0; while (i < n) // one by one traverse each element { int j = 0; // traverse all weights j <= W // if i is odd that mean till now we have odd // number of elements so we store result in 1th // indexed row if (i % 2 != 0) { while (++j <= W) // check for each value { if (wt[i] <= j) // include element { mat[1, j] = Math.Max(val[i] + mat[0, j - wt[i]], mat[0, j]); } else // exclude element { mat[1,j] = mat[0,j]; } } } // if i is even that means till now // we have even number of elements // so we store result in 0th indexed row else { while (++j <= W) { if (wt[i] <= j) { mat[0, j] = Math.Max(val[i] + mat[1, j - wt[i]], mat[1, j]); } else { mat[0, j] = mat[1, j]; } } } i++; } // Return mat[0,W] if n is odd, else mat[1,W] return (n % 2 != 0) ? mat[0, W] : mat[1, W]; } // Driver Code public static void Main(String[] args) { int []val = {7, 8, 4}; int []wt = {3, 8, 6}; int W = 10, n = 3; Console.WriteLine(KnapSack(val, wt, n, W)); }}// This code is contributed by 29AjayKumar |
PHP
<?php // PHP 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// mat[2][W+1] to store final resultfunction KnapSack(&$val, &$wt, $n, $W){ // matrix to store final result $mat = array_fill(0, 2, array_fill(0, $W + 1, NULL)); // iterate through all items $i = 0; while ($i < $n) // one by one traverse // each element { $j = 0; // traverse all weights j <= W // if i is odd that mean till now we // have odd number of elements so we // store result in 1th indexed row if ($i % 2 != 0) { while (++$j <= $W) // check for each value { if ($wt[$i] <= $j) // include element $mat[1][$j] = max($val[$i] + $mat[0][$j - $wt[$i]], $mat[0][$j]); else // exclude element $mat[1][$j] = $mat[0][$j]; } } // if i is even that mean till now we have // even number of elements so we store result // in 0th indexed row else { while(++$j <= $W) { if ($wt[$i] <= $j) $mat[0][$j] = max($val[$i] + $mat[1][$j - $wt[$i]], $mat[1][$j]); else $mat[0][$j] = $mat[1][$j]; } } $i++; } // Return mat[0][W] if n is odd, // else mat[1][W] if ($n % 2 != 0) return $mat[0][$W]; else return $mat[1][$W];}// Driver Code$val = array(7, 8, 4);$wt = array(3, 8, 6);$W = 10;$n = 3;echo KnapSack($val, $wt, $n, $W) . "\n";// This code is contributed // by Akanksha Rai?> |
Javascript
<script>// Javascript 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 // mat[2][W+1] to store final result function KnapSack(val, wt, n, W) { // matrix to store final result let mat = new Array(2); for(let i = 0; i < 2; i++) { mat[i] = new Array(W + 1); } for(let i = 0; i < 2; i++) { for(let j = 0; j < W + 1; j++) { mat[i][j] = 0; } } // iterate through all items let i = 0; while (i < n) // one by one traverse each element { let j = 0; // traverse all weights j <= W // if i is odd that mean till now we have odd // number of elements so we store result in 1th // indexed row if (i % 2 != 0) { while (++j <= W) // check for each value { if (wt[i] <= j) // include element { mat[1][j] = Math.max(val[i] + mat[0][j - wt[i]], mat[0][j]); } else // exclude element { mat[1][j] = mat[0][j]; } } } // if i is even that means till now // we have even number of elements // so we store result in 0th indexed row else { while (++j <= W) { if (wt[i] <= j) { mat[0][j] = Math.max(val[i] + mat[1][j - wt[i]], mat[1][j]); } else { mat[0][j] = mat[1][j]; } } } i++; } // Return mat[0][W] if n is odd, else mat[1][W] return (n % 2 != 0) ? mat[0][W] : mat[1][W]; } let val=[7, 8, 4]; let wt=[3, 8, 6]; let W = 10, n = 3; document.write(KnapSack(val, wt, n, W)); // This code is contributed by rag2127</script> |
Output
11
Time Complexity: O(n * W)
Auxiliary Space: O(W)
Here is an optimized code contributed by Gaurav Mamgain
C++14
// C++ program of a space optimized DP solution for// 0-1 knapsack problem.#include <bits/stdc++.h>using namespace std;// 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 resultint KnapSack(int val[], int wt[], int n, int W){ // array to store final result // dp[i] stores the profit with KnapSack capacity "i" int dp[W + 1]; // initially profit with 0 to W KnapSack capacity is 0 memset(dp, 0, sizeof(dp)); // iterate through all items for (int i = 0; i < n; i++) // traverse dp array from right to left for (int j = W; j >= wt[i]; j--) dp[j] = max(dp[j], val[i] + dp[j - wt[i]]); /*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") */ return dp[W];}// Driver program to test the casesint main(){ int val[] = { 7, 8, 4 }, wt[] = { 3, 8, 6 }, W = 10, n = 3; cout << KnapSack(val, wt, n, W) << endl; return 0;}// This code is contributed by Gaurav Mamgain |
Java
// Java program of a space optimized DP solution for// 0-1 knapsack problem.import java.util.Arrays;class GFG {// 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 resultstatic int KnapSack(int val[], int wt[], int n, int W){ // array to store final result //dp[i] stores the profit with KnapSack capacity "i" int []dp = new int[W+1]; //initially profit with 0 to W KnapSack capacity is 0 Arrays.fill(dp, 0); // iterate through all items for(int i=0; i < n; i++) //traverse dp array from right to left for(int j = W; j >= wt[i]; j--) dp[j] = Math.max(dp[j] , val[i] + dp[j - wt[i]]); /*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") */ return dp[W];}// Driver codepublic 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));}}// This code is contributed by Princi Singh |
Python3
# Python 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 resultdef KnapSack(val, wt, n, W): # array to store final result # dp[i] stores the profit with KnapSack capacity "i" dp = [0]*(W+1); # iterate through all items for i in range(n): #traverse dp array from right to left for j in range(W,wt[i]-1,-1): dp[j] = max(dp[j] , val[i] + dp[j-wt[i]]); '''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") *''' return dp[W];# Driver program to test the casesval = [7, 8, 4];wt = [3, 8, 6];W = 10; n = 3;print(KnapSack(val, wt, n, W));# This code is contributed by Princi Singh |
C#
// C# program of a space optimized DP solution for // 0-1 knapsack problem. using System;class GFG { // 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 static int KnapSack(int []val, int []wt, int n, int W) { // array to store final result //dp[i] stores the profit with KnapSack capacity "i" int []dp = new int[W + 1]; //initially profit with 0 to W KnapSack capacity is 0 for(int i = 0; i < W + 1; i++) dp[i] = 0; // iterate through all items for(int i = 0; i < n; i++) //traverse dp array from right to left for(int j = W; j >= wt[i]; j--) dp[j] = Math.Max(dp[j] , val[i] + dp[j - wt[i]]); /*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") */ return dp[W]; } // Driver code public static void Main(String[] args) { int []val = {7, 8, 4}; int []wt = {3, 8, 6}; int W = 10, n = 3; Console.WriteLine(KnapSack(val, wt, n, W)); } } // This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript 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 function KnapSack(val,wt,n,W) { // array to store final result // dp[i] stores the profit with KnapSack capacity "i" let dp = new Array(W+1); // initially profit with 0 to W KnapSack capacity is 0 for(let i=0;i<W+1;i++) { dp[i]=0; } // iterate through all items for(let i=0; i < n; i++) // traverse dp array from right to left for(let j = W; j >= wt[i]; j--) dp[j] = Math.max(dp[j] , val[i] + dp[j - wt[i]]); /*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") */ return dp[W]; } // Driver code let val=[7, 8, 4]; let wt=[3, 8, 6]; let W = 10, n = 3; document.write(KnapSack(val, wt, n, W)); // This code is contributed by avanitrachhadiya2155 </script> |
Output
11
Time complexity: O(n * W)
Auxiliary space: O(W)
This article is contributed by Shashank Mishra ( Gullu ). This article is reviewed by team neveropen.
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