A Space Optimized DP solution for 0-1 Knapsack Problem

4.2

Given 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
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 previous row. So here we can observe that at a time we are working only with 2 consecutive rows.
In below solution, we create a matrix of size 2*W. If n is odd, then final answer will be at mat[0][W] and if n is even then final answer will be at mat[1][W] because index starts from 0.

// 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 result
int 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 wieghts 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 cases
int main()
{
    int val[] = {7, 8, 4}, wt[] = {3, 8, 6}, W = 10, n = 3;
    cout << KnapSack(val, wt, n, W) << endl;
    return 0;
}

Output:

 11

Time Complexity : O(n * W)
Auxiliary Space : O(W)

Here is an optimized code contributed by Gaurav Mamgain

// 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 result
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[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 cases
int 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

Output:

 11

This article is contributed by Shashank Mishra ( Gullu ). This article is reviwed by team geeksforgeeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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