0 1 Knapsack

Easy

0/40

Average time to solve is 15m

Contributed by

379 upvotes

Asked in companies

Problem statement

A thief is robbing a store and can carry a maximal weight of W into his knapsack. There are N items and the ith item weighs wi and is of value vi. Considering the constraints of the maximum weight that a knapsack can carry, you have to find and return the maximum value that a thief can generate by stealing items.

Detailed explanation ( Input/output format, Notes, Images )

Input Format:

The first line contains a single integer T representing the number of test cases.      
The T-test cases are as follows:

Line 1:The first line contains an integer, that denotes the value of N. 
Line 2:The following line contains N space-separated integers, that denote the values of the weight of items. 
Line 3:The following line contains N space-separated integers, that denote the values associated with the items. 
Line 4:The following line contains an integer that denotes the value of W. W denotes the maximum weight that a thief can carry.

Output Format :

The first and only line of output contains the maximum value that a thief can generate, as described in the task. 
The output of every test case is printed in a separate line.

Constraints:

1 <= T <= 10
1 <= N <= 10^2
1<= wi <= 50
1 <= vi <= 10^2
1 <= W <= 10^3

Time Limit: 1 second

Sample Input:

Sample Output:

Hint 1

Can you proceed if already some items are filled?

Approaches (1)

Approach 1

Dynamic Programming

Approach: In the Dynamic programming we will work considering the same cases as mentioned in the recursive approach. In a DP[][] table let’s consider all the possible weights from ‘1’ to ‘W’ as the columns and weights that can be kept as the rows.
The state DP[i][j] will denote maximum value of ‘j-weight’ considering all values from ‘1 to ith’. So if we consider ‘wi’ (weight in ‘ith’ row) we can fill it in all columns which have ‘weight values > wi’. Now two possibilities can take place:

Fill ‘wi’ in the given column.
Do not fill ‘wi’ in the given column.

Now we have to take a maximum of these two possibilities, formally if we do not fill ‘ith’ weight in ‘jth’ column then DP[i][j] state will be same as DP[i-1][j] but if we fill the weight, DP[i][j] will be equal to the value of ‘wi’+ value of the column weighing ‘j-wi’ in the previous row. So we take the maximum of these two possibilities to fill the current state. This visualization will make the concept clear:

Time Complexity

O(N * W), where ‘N’ is the number of items and ‘W’ is the weight of bag.

Space Complexity

O(W), where ‘W’ is the weight of bag.

(75% EXP penalty)

(100% EXP penalty)

0 1 Knapsack

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