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 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 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# 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.
<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>
Efficient approach: Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Create a DP to store the solution of the subproblems and initialize it with 0.
- Initialize the DP with base cases
- Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP.
- After filling the DP now in Main function call every query and get the answer stored in DP.
Implementation :
C++
#include <bits/stdc++.h>
#define C_MAX 30
#define max_arr_len 10
using namespace std;
// Function to compute the maximum value that can be obtained
// from the knapsack within the given capacity
int dp[max_arr_len][C_MAX + 1] = {0};
void findMax(int w[], int n, int c)
{
// Initializing the DP table with 0
// Filling the DP table bottom-up
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= c; j++) {
if (w[i - 1] <= j) {
dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - w[i - 1]] + w[i - 1]);
}
else {
dp[i][j] = dp[i - 1][j];
}
}
}
// Returning the maximum value that can be obtained
return ;
}
// Driver code
int main()
{
// Input array
int w[] = { 3, 8, 9 };
int n = sizeof(w) / sizeof(int);
// all given queries
int queries[] = { 11, 10, 4 };
int q = sizeof(queries) / sizeof(queries[0]);
// find maximum
int ma = *max_element(queries, queries + q);
// function call
findMax(w, n , ma);
// Perform queries
for (int i = 0; i < q; i++)
cout << dp[n][queries[i]] << endl;
return 0;
}
public class MaxSubsetSum {
static int[][] findMax(int[] w, int n, int c) {
// Initializing the DP table with 0
int[][] dp = new int[n + 1][c + 1];
// Filling the DP table bottom-up
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= c; j++) {
if (w[i - 1] <= j) {
dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - w[i - 1]] + w[i - 1]);
} else {
dp[i][j] = dp[i - 1][j];
}
}
}
return dp;
}
// Driver code
public static void main(String[] args) {
int[] w = {3, 8, 9};
int n = w.length;
// All given queries
int[] queries = {11, 10, 4};
int q = queries.length;
// Find maximum
int ma = Integer.MIN_VALUE;
for (int query : queries) {
ma = Math.max(ma, query);
}
// Function call
int[][] dp = findMax(w, n, ma);
// Perform queries
for (int i = 0; i < q; i++) {
System.out.println(dp[n][queries[i]]);
}
}
}
import sys
# Function to compute the maximum value that can be obtained
# from the knapsack within the given capacity
def findMax(w, n, c):
# Initializing the DP table with 0
dp = [[0 for j in range(c + 1)] for i in range(n + 1)]
# Filling the DP table bottom-up
for i in range(1, n + 1):
for j in range(1, c + 1):
if w[i - 1] <= j:
dp[i][j] = max(dp[i - 1][j], dp[i - 1]
[j - w[i - 1]] + w[i - 1])
else:
dp[i][j] = dp[i - 1][j]
# Returning the maximum value that can be obtained
return dp
# Driver code
if __name__ == '__main__':
# Input array
w = [3, 8, 9]
n = len(w)
# all given queries
queries = [11, 10, 4]
q = len(queries)
# find maximum
ma = max(queries)
# function call
dp = findMax(w, n, ma)
# Perform queries
for i in range(q):
print(dp[n][queries[i]])
using System;
public class MaxSubsetSum
{
static int[,] findMax(int[] w, int n, int c)
{
// Initializing the DP table with 0
int[,] dp = new int[n + 1, c + 1];
// Filling the DP table bottom-up
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= c; j++)
{
if (w[i - 1] <= j)
{
dp[i, j] = Math.Max(dp[i - 1, j], dp[i - 1, j - w[i - 1]] + w[i - 1]);
}
else
{
dp[i, j] = dp[i - 1, j];
}
}
}
return dp;
}
// Driver code
public static void Main(string[] args)
{
int[] w = { 3, 8, 9 };
int n = w.Length;
// All given queries
int[] queries = { 11, 10, 4 };
int q = queries.Length;
// Find maximum
int ma = int.MinValue;
foreach (int query in queries)
{
ma = Math.Max(ma, query);
}
// Function call
int[,] dp = findMax(w, n, ma);
// Perform queries
for (int i = 0; i < q; i++)
{
Console.WriteLine(dp[n, queries[i]]);
}
}
}
function findMax(w, n, c) {
// Initializing the DP table with 0
let dp = new Array(n + 1).fill(null).map(() => new Array(c + 1).fill(0));
// Filling the DP table bottom-up
for (let i = 1; i <= n; i++) {
for (let j = 1; j <= c; j++) {
if (w[i - 1] <= j) {
dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - w[i - 1]] + w[i - 1]);
} else {
dp[i][j] = dp[i - 1][j];
}
}
}
// Returning the maximum value that can be obtained
return dp;
}
// Driver code
const w = [3, 8, 9];
const n = w.length;
// all given queries
const queries = [11, 10, 4];
const q = queries.length;
// find maximum
const ma = Math.max(...queries);
// function call
const dp = findMax(w, n, ma);
// Perform queries
for (let i = 0; i < q; i++) {
console.log(dp[n][queries[i]]);
}
Time complexity: O(n*c)
Auxiliary Space: O(n*C)
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