0-1 knapsack queries

Given an integer array W[] consisting of weights of the items and some queries consisting of capacity C of knapsack, for each query find maximum weight we can put in the knapsack. Value of C doesn’t exceed a certain integer C_MAX.

Examples:

Input: W[] = {3, 8, 9} q = {11, 10, 4} 
Output:
11 
9 
3 
If C = 11: select 3 + 8 = 11 
If C = 10: select 9 
If C = 4: select 3

Input: W[] = {1, 5, 10} q = {6, 14} 
Output:
6 
11 

Its recommended that you go through this article on 0-1 knapsack before attempting this problem.

Naive approach: For each query, we can generate all possible subsets of weight and choose the one that has maximum weight among all those subsets that fits in the knapsack. Thus, answering each query will take exponential time.

Efficient approach: We will optimize answering each query using dynamic programming. 
0-1 knapsack is solved using 2-D DP, one state ‘i’ for current index(i.e select or reject) and one for remaining capacity ‘R’. 
Recurrence relation is 

dp[i][R] = max(arr[i] + dp[i + 1][R – arr[i]], dp[i + 1][R]) 

We will pre-compute the 2-d array dp[i][C] for every possible value of ‘C’ between 1 to C_MAX in O(C_MAX*i). 
Using this, pre-computation we can answer each queries in O(1) time.

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
#define C_MAX 30
#define max_arr_len 10
using namespace std;

// To store states of DP
int dp[max_arr_len][C_MAX + 1];

// To check if a state has
// been solved
bool v[max_arr_len][C_MAX + 1];

// Function to compute the states
int findMax(int i, int r, int w[], int n)
{

    // Base case
    if (r < 0)
        return INT_MIN;
    if (i == n)
        return 0;

    // Check if a state has
    // been solved
    if (v[i][r])
        return dp[i][r];

    // Setting a state as solved
    v[i][r] = 1;
    dp[i][r] = max(w[i] + findMax(i + 1, r - w[i], w, n),
                   findMax(i + 1, r, w, n));

    // Returning the solved state
    return dp[i][r];
}

// Function to pre-compute the states
// dp[0][0], dp[0][1], .., dp[0][C_MAX]
void preCompute(int w[], int n)
{
    for (int i = C_MAX; i >= 0; i--)
        findMax(0, i, w, n);
}

// Function to answer a query in O(1)
int ansQuery(int w)
{
    return dp[0][w];
}

// Driver code
int main()
{
    int w[] = { 3, 8, 9 };
    int n = sizeof(w) / sizeof(int);

    // Performing required
    // pre-computation
    preCompute(w, n);

    int queries[] = { 11, 10, 4 };
    int q = sizeof(queries) / sizeof(queries[0]);

    // Perform queries
    for (int i = 0; i < q; i++)
        cout << ansQuery(queries[i]) << endl;

    return 0;
}

Java

// Java implementation of the approach
class GFG
{
    static int C_MAX = 30;
    static int max_arr_len = 10;

    // To store states of DP
    static int dp [][] = new int [max_arr_len][C_MAX + 1];

    // To check if a state has
    // been solved
    static boolean v[][]= new boolean [max_arr_len][C_MAX + 1];

    // Function to compute the states
    static int findMax(int i, int r, int w[], int n)
    {

        // Base case
        if (r < 0)
            return Integer.MIN_VALUE;
        if (i == n)
            return 0;

        // Check if a state has
        // been solved
        if (v[i][r])
            return dp[i][r];

        // Setting a state as solved
        v[i][r] = true;
        dp[i][r] = Math.max(w[i] + findMax(i + 1, r - w[i], w, n),
                                        findMax(i + 1, r, w, n));

        // Returning the solved state
        return dp[i][r];
    }

    // Function to pre-compute the states
    // dp[0][0], dp[0][1], .., dp[0][C_MAX]
    static void preCompute(int w[], int n)
    {
        for (int i = C_MAX; i >= 0; i--)
            findMax(0, i, w, n);
    }

    // Function to answer a query in O(1)
    static int ansQuery(int w)
    {
        return dp[0][w];
    }

    // Driver code
    public static void main (String[] args)
    {

        int w[] = new int []{ 3, 8, 9 };
        int n = w.length;

        // Performing required
        // pre-computation
        preCompute(w, n);

        int queries[] = new int [] { 11, 10, 4 };
        int q = queries.length;

        // Perform queries
        for (int i = 0; i < q; i++)
            System.out.println(ansQuery(queries[i]));
    }
}

// This code is contributed by ihritik

Python3

# Python3 implementation of the approach

import numpy as np
import sys

C_MAX = 30
max_arr_len = 10

# To store states of DP
dp = np.zeros((max_arr_len,C_MAX + 1));

# To check if a state has
# been solved
v = np.zeros((max_arr_len,C_MAX + 1));

INT_MIN = -(sys.maxsize) + 1

# Function to compute the states
def findMax(i, r, w, n) :

    # Base case
    if (r < 0) :
        return INT_MIN;

    if (i == n) :
        return 0;

    # Check if a state has
    # been solved
    if (v[i][r]) :
        return dp[i][r];

    # Setting a state as solved
    v[i][r] = 1;
    dp[i][r] = max(w[i] + findMax(i + 1, r - w[i], w, n),
                findMax(i + 1, r, w, n));

    # Returning the solved state
    return dp[i][r];


# Function to pre-compute the states
# dp[0][0], dp[0][1], .., dp[0][C_MAX]
def preCompute(w, n) :

    for i in range(C_MAX, -1, -1) :
        findMax(0, i, w, n);


# Function to answer a query in O(1)
def ansQuery(w) :

    return dp[0][w];


# Driver code
if __name__ == "__main__" :

    w = [ 3, 8, 9 ];
    n = len(w)

    # Performing required
    # pre-computation
    preCompute(w, n);

    queries = [ 11, 10, 4 ];
    q = len(queries)

    # Perform queries
    for i in range(q) :
        print(ansQuery(queries[i]));


# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{
static int C_MAX = 30;
    static int max_arr_len = 10;

    // To store states of DP
    static int[,] dp  = new int [max_arr_len, C_MAX + 1];

    // To check if a state has
    // been solved
    static bool[,] v = new bool [max_arr_len, C_MAX + 1];

    // Function to compute the states
    static int findMax(int i, int r, int[] w, int n)
    {

        // Base case
        if (r < 0)
            return Int32.MaxValue;
        if (i == n)
            return 0;

        // Check if a state has
        // been solved
        if (v[i, r])
            return dp[i, r];

        // Setting a state as solved
        v[i, r] = true;
        dp[i, r] = Math.Max(w[i] + findMax(i + 1, r - w[i], w, n),
                                        findMax(i + 1, r, w, n));

        // Returning the solved state
        return dp[i, r];
    }

    // Function to pre-compute the states
    // dp[0][0], dp[0][1], .., dp[0][C_MAX]
    static void preCompute(int[] w, int n)
    {
        for (int i = C_MAX; i >= 0; i--)
            findMax(0, i, w, n);
    }

    // Function to answer a query in O(1)
    static int ansQuery(int w)
    {
        return dp[0, w];
    }


    // Driver code
    public static void Main()
    {
        int[] w= { 3, 8, 9 };
        int n = w.Length;

        // Performing required
        // pre-computation
        preCompute(w, n);

        int[] queries = { 11, 10, 4 };
        int q = queries.Length;

        // Perform queries
        for (int i = 0; i < q; i++)
            Console.WriteLine(ansQuery(queries[i]));
    }
}

// This code is contributed by sanjoy_62.

Javascript

<script>

// Javascript implementation of the approach
var C_MAX = 30
var max_arr_len = 10

// To store states of DP
var dp = Array.from(Array(max_arr_len), ()=>Array(C_MAX+1));

// To check if a state has
// been solved
var v = Array.from(Array(max_arr_len), ()=>Array(C_MAX+1));

// Function to compute the states
function findMax(i, r, w, n)
{

    // Base case
    if (r < 0)
        return -1000000000;
    if (i == n)
        return 0;

    // Check if a state has
    // been solved
    if (v[i][r])
        return dp[i][r];

    // Setting a state as solved
    v[i][r] = 1;
    dp[i][r] = Math.max(w[i] + findMax(i + 1, r - w[i], w, n),
                   findMax(i + 1, r, w, n));

    // Returning the solved state
    return dp[i][r];
}

// Function to pre-compute the states
// dp[0][0], dp[0][1], .., dp[0][C_MAX]
function preCompute(w, n)
{
    for (var i = C_MAX; i >= 0; i--)
        findMax(0, i, w, n);
}

// Function to answer a query in O(1)
function ansQuery(w)
{
    return dp[0][w];
}

// Driver code
var w = [3, 8, 9];
var n = w.length;
// Performing required
// pre-computation
preCompute(w, n);
var queries = [11, 10, 4];
var q = queries.length;
// Perform queries
for (var i = 0; i < q; i++)
    document.write( ansQuery(queries[i])+ "<br>");

</script>

Output:

11
9
3