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Minimum cost to fill given weight in a bag

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You are given a bag of size W kg and you are provided costs of packets different weights of oranges in array cost[] where cost[i] is basically the cost of ‘i’ kg packet of oranges. Where cost[i] = -1 means that ‘i’ kg packet of orange is unavailable
Find the minimum total cost to buy exactly W kg oranges and if it is not possible to buy exactly W kg oranges then print -1. It may be assumed that there is an infinite supply of all available packet types.
Note: array starts from index 1. 

Examples: 

Input  : W = 5, cost[] = {20, 10, 4, 50, 100}
Output : 14
We can choose two oranges to minimize cost. First 
orange of 2Kg and cost 10. Second orange of 3Kg
and cost 4. 

Input  : W = 5, cost[] = {1, 10, 4, 50, 100}
Output : 5
We can choose five oranges of weight 1 kg.

Input  : W = 5, cost[] = {1, 2, 3, 4, 5}
Output : 5
Costs of 1, 2, 3, 4 and 5 kg packets are 1, 2, 3, 
4 and 5 Rs respectively. 
We choose packet of 5kg having cost 5 for minimum
cost to get 5Kg oranges.

Input  : W = 5, cost[] = {-1, -1, 4, 5, -1}
Output : -1
Packets of size 1, 2 and 5 kg are unavailable
because they have cost -1. Cost of 3 kg packet 
is 4 Rs and of 4 kg is 5 Rs. Here we have only 
weights 3 and 4 so by using these two we can  
not make exactly W kg weight, therefore answer 
is -1.

Method 1: 

This problem is can be reduced to Unbounded Knapsack. So in the cost array, we first ignore those packets which are not available i.e; cost is -1 and then traverse the cost array and create two array val[] for storing the cost of ‘i’ kg packet of orange and wt[] for storing weight of the corresponding packet. Suppose cost[i] = 50 so the weight of the packet will be i and the cost will be 50. 
Algorithm :

  • Create matrix min_cost[n+1][W+1], where n is number of distinct weighted packets of orange and W is the maximum capacity of the bag.
  • Initialize the 0th row with INF (infinity) and 0th Column with 0.
  • Now fill the matrix
    • if wt[i-1] > j then min_cost[i][j] = min_cost[i-1][j] ;
    • if wt[i-1] <= j then min_cost[i][j] = min(min_cost[i-1][j], val[i-1] + min_cost[i][j-wt[i-1]]);
  • If min_cost[n][W]==INF then output will be -1 because this means that we cant not make weight W by using these weights else output will be min_cost[n][W].

Below is the implementation of the above algorithm:

C++

// C++ program to find minimum cost to get exactly
// W Kg with given packets
#include<bits/stdc++.h>
#define INF 1000000
using namespace std;
 
// cost[] initial cost array including unavailable packet
// W capacity of bag
int MinimumCost(int cost[], int n, int W)
{
    // val[] and wt[] arrays
    // val[] array to store cost of 'i' kg packet of orange
    // wt[] array weight of packet of orange
    vector<int> val, wt;
 
    // traverse the original cost[] array and skip
    // unavailable packets and make val[] and wt[]
    // array. size variable tells the available number
    // of distinct weighted packets
    int size = 0;
    for (int i=0; i<n; i++)
    {
        if (cost[i]!= -1)
        {
            val.push_back(cost[i]);
            wt.push_back(i+1);
            size++;
        }
    }
 
    n = size;
    int min_cost[n+1][W+1];
 
    // fill 0th row with infinity
    for (int i=0; i<=W; i++)
        min_cost[0][i] = INF;
 
    // fill 0'th column with 0
    for (int i=1; i<=n; i++)
        min_cost[i][0] = 0;
 
    // now check for each weight one by one and fill the
    // matrix according to the condition
    for (int i=1; i<=n; i++)
    {
        for (int j=1; j<=W; j++)
        {
            // wt[i-1]>j means capacity of bag is
            // less than weight of item
            if (wt[i-1] > j)
                min_cost[i][j] = min_cost[i-1][j];
 
            // here we check we get minimum cost either
            // by including it or excluding it
            else
                min_cost[i][j] = min(min_cost[i-1][j],
                     min_cost[i][j-wt[i-1]] + val[i-1]);
        }
    }
 
    // exactly weight W can not be made by given weights
    return (min_cost[n][W]==INF)? -1: min_cost[n][W];
}
 
// Driver program to run the test case
int main()
{
    int cost[] = {1, 2, 3, 4, 5}, W = 5;
    int n = sizeof(cost)/sizeof(cost[0]);
 
    cout << MinimumCost(cost, n, W);
    return 0;
}

                    

Java

// Java Code for Minimum cost to
// fill given weight in a bag
import java.util.*;
 
class GFG {
     
    // cost[] initial cost array including
    // unavailable packet W capacity of bag
    public static int MinimumCost(int cost[], int n,
                                             int W)
    {
        // val[] and wt[] arrays
        // val[] array to store cost of 'i' kg
        // packet of orange wt[] array weight of
        // packet of orange
        Vector<Integer> val = new Vector<Integer>();
        Vector<Integer> wt = new Vector<Integer>();
      
        // traverse the original cost[] array and skip
        // unavailable packets and make val[] and wt[]
        // array. size variable tells the available
        // number of distinct weighted packets
        int size = 0;
        for (int i = 0; i < n; i++)
        {
            if (cost[i] != -1)
            {
                val.add(cost[i]);
                wt.add(i + 1);
                size++;
            }
        }
      
        n = size;
        int min_cost[][] = new int[n+1][W+1];
      
        // fill 0th row with infinity
        for (int i = 0; i <= W; i++)
            min_cost[0][i] = Integer.MAX_VALUE;
      
        // fill 0'th column with 0
        for (int i = 1; i <= n; i++)
            min_cost[i][0] = 0;
      
        // now check for each weight one by one and
        // fill the matrix according to the condition
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= W; j++)
            {
                // wt[i-1]>j means capacity of bag is
                // less than weight of item
                if (wt.get(i-1) > j)
                    min_cost[i][j] = min_cost[i-1][j];
      
                // here we check we get minimum cost
                // either by including it or excluding
                // it
                else
                    min_cost[i][j] = Math.min(min_cost[i-1][j],
                                  min_cost[i][j-wt.get(i-1)] +
                                              val.get(i-1));
            }
        }
      
        // exactly weight W can not be made by
        // given weights
        return (min_cost[n][W] == Integer.MAX_VALUE)? -1:
                                        min_cost[n][W];
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
         int cost[] = {1, 2, 3, 4, 5}, W = 5;
            int n = cost.length;
          
        System.out.println(MinimumCost(cost, n, W));
    }
}
// This code is contributed by Arnav Kr. Mandal.

                    

Python3

# Python program to find minimum cost to get exactly
# W Kg with given packets
 
INF = 1000000
 
# cost[] initial cost array including unavailable packet
# W capacity of bag
def MinimumCost(cost, n, W):
 
    # val[] and wt[] arrays
    # val[] array to store cost of 'i' kg packet of orange
    # wt[] array weight of packet of orange
    val = list()
    wt= list()
 
    # traverse the original cost[] array and skip
    # unavailable packets and make val[] and wt[]
    # array. size variable tells the available number
    # of distinct weighted packets.
    size = 0
    for i in range(n):
        if (cost[i] != -1):
            val.append(cost[i])
            wt.append(i+1)
            size += 1
 
    n = size
    min_cost = [[0 for i in range(W+1)] for j in range(n+1)]
 
    # fill 0th row with infinity
    for i in range(W+1):
        min_cost[0][i] = INF
 
    # fill 0th column with 0
    for i in range(1, n+1):
        min_cost[i][0] = 0
 
    # now check for each weight one by one and fill the
    # matrix according to the condition
    for i in range(1, n+1):
        for j in range(1, W+1):
            # wt[i-1]>j means capacity of bag is
            # less than weight of item
            if (wt[i-1] > j):
                min_cost[i][j] = min_cost[i-1][j]
 
            # here we check we get minimum cost either
            # by including it or excluding it
            else:
                min_cost[i][j] = min(min_cost[i-1][j],
                    min_cost[i][j-wt[i-1]] + val[i-1])
 
    # exactly weight W can not be made by given weights
    if(min_cost[n][W] == INF):
        return -1
    else:
        return min_cost[n][W]
 
# Driver program to run the test case
cost = [1, 2, 3, 4, 5]
W = 5
n = len(cost)
 
print(MinimumCost(cost, n, W))
 
# This code is contributed by Soumen Ghosh.

                    

C#

// C# Code for Minimum cost to
// fill given weight in a bag
 
using System;
using System.Collections.Generic;
 
class GFG {
     
    // cost[] initial cost array including
    // unavailable packet W capacity of bag
    public static int MinimumCost(int []cost, int n,
                                            int W)
    {
        // val[] and wt[] arrays
        // val[] array to store cost of 'i' kg
        // packet of orange wt[] array weight of
        // packet of orange
        List<int> val = new List<int>();
        List<int> wt = new List<int>();
     
        // traverse the original cost[] array and skip
        // unavailable packets and make val[] and wt[]
        // array. size variable tells the available
        // number of distinct weighted packets
        int size = 0;
        for (int i = 0; i < n; i++)
        {
            if (cost[i] != -1)
            {
                val.Add(cost[i]);
                wt.Add(i + 1);
                size++;
            }
        }
     
        n = size;
        int [,]min_cost = new int[n+1,W+1];
     
        // fill 0th row with infinity
        for (int i = 0; i <= W; i++)
            min_cost[0,i] = int.MaxValue;
     
        // fill 0'th column with 0
        for (int i = 1; i <= n; i++)
            min_cost[i,0] = 0;
     
        // now check for each weight one by one and
        // fill the matrix according to the condition
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= W; j++)
            {
                // wt[i-1]>j means capacity of bag is
                // less than weight of item
                if (wt[i-1] > j)
                    min_cost[i,j] = min_cost[i-1,j];
     
                // here we check we get minimum cost
                // either by including it or excluding
                // it
                else
                    min_cost[i,j] = Math.Min(min_cost[i-1,j],
                                min_cost[i,j-wt[i-1]] + val[i-1]);
            }
        }
     
        // exactly weight W can not be made by
        // given weights
        return (min_cost[n,W] == int.MaxValue )? -1: min_cost[n,W];
    }
     
    /* Driver program to test above function */
    public static void Main()
    {
            int []cost = {1, 2, 3, 4, 5};
            int W = 5;
            int n = cost.Length;
         
            Console.WriteLine(MinimumCost(cost, n, W));
    }
}
// This code is contributed by Ryuga

                    

PHP

<?php
// PHP program to find minimum cost to
// get exactly W Kg with given packets
$INF = 1000000;
 
// cost[] initial cost array including
// unavailable packet W capacity of bag
function MinimumCost(&$cost, $n, $W)
{
    global $INF;
     
    // val[] and wt[] arrays
    // val[] array to store cost of 'i'
    // kg packet of orange
    // wt[] array weight of packet of orange
    $val = array();
    $wt = array();
 
    // traverse the original cost[] array
    // and skip unavailable packets and
    // make val[] and wt[] array. size
    // variable tells the available number
    // of distinct weighted packets
    $size = 0;
    for ($i = 0; $i < $n; $i++)
    {
        if ($cost[$i] != -1)
        {
            array_push($val, $cost[$i]);
            array_push($wt, $i + 1);
            $size++;
        }
    }
 
    $n = $size;
    $min_cost = array_fill(0, $n + 1,
                array_fill(0, $W + 1, NULL));
 
    // fill 0th row with infinity
    for ($i = 0; $i <= $W; $i++)
        $min_cost[0][$i] = $INF;
 
    // fill 0'th column with 0
    for ($i = 1; $i <= $n; $i++)
        $min_cost[$i][0] = 0;
 
    // now check for each weight one by
    // one and fill the matrix according
    // to the condition
    for ($i = 1; $i <= $n; $i++)
    {
        for ($j = 1; $j <= $W; $j++)
        {
            // wt[i-1]>j means capacity of bag
            // is less than weight of item
            if ($wt[$i - 1] > $j)
                $min_cost[$i][$j] = $min_cost[$i - 1][$j];
 
            // here we check we get minimum
            // cost either by including it
            // or excluding it
            else
                $min_cost[$i][$j] = min($min_cost[$i - 1][$j],
                                        $min_cost[$i][$j - $wt[$i - 1]] +
                                                           $val[$i - 1]);
        }
    }
 
    // exactly weight W can not be made
    // by given weights
    if ($min_cost[$n][$W] == $INF)
            return -1;
    else
        return $min_cost[$n][$W];
}
 
// Driver Code
$cost = array(1, 2, 3, 4, 5);
$W = 5;
$n = sizeof($cost);
echo MinimumCost($cost, $n, $W);
 
// This code is contributed by ita_c
?>

                    

Javascript

<script>
    // Javascript program to find minimum cost to get exactly
    // W Kg with given packets
     
    let INF = 1000000;
     
    // cost[] initial cost array including unavailable packet
    // W capacity of bag
    function MinimumCost(cost, n, W)
    {
        // val[] and wt[] arrays
        // val[] array to store cost of 'i' kg packet of orange
        // wt[] array weight of packet of orange
        let val = [], wt = [];
 
        // traverse the original cost[] array and skip
        // unavailable packets and make val[] and wt[]
        // array. size variable tells the available number
        // of distinct weighted packets
        let size = 0;
        for (let i=0; i<n; i++)
        {
            if (cost[i]!= -1)
            {
                val.push(cost[i]);
                wt.push(i+1);
                size++;
            }
        }
 
        n = size;
        let min_cost = new Array(n+1);
        for(let i = 0; i < n + 1; i++)
        {
            min_cost[i] = new Array(W + 1);
        }
 
        // fill 0th row with infinity
        for (let i=0; i<=W; i++)
            min_cost[0][i] = INF;
 
        // fill 0'th column with 0
        for (let i=1; i<=n; i++)
            min_cost[i][0] = 0;
 
        // now check for each weight one by one and fill the
        // matrix according to the condition
        for (let i=1; i<=n; i++)
        {
            for (let j=1; j<=W; j++)
            {
                // wt[i-1]>j means capacity of bag is
                // less than weight of item
                if (wt[i-1] > j)
                    min_cost[i][j] = min_cost[i-1][j];
 
                // here we check we get minimum cost either
                // by including it or excluding it
                else
                    min_cost[i][j] = Math.min(min_cost[i-1][j],
                         min_cost[i][j-wt[i-1]] + val[i-1]);
            }
        }
 
        // exactly weight W can not be made by given weights
        return (min_cost[n][W]==INF)? -1: min_cost[n][W];
    }
     
    // Driver code
    let cost = [1, 2, 3, 4, 5], W = 5;
    let n = cost.length;
  
    document.write(MinimumCost(cost, n, W));
     
    // This code is contributed by suresh07.
</script>

                    

Output
5

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

Space Optimized Solution If we take a closer look at this problem, we may notice that this is a variation of Rod Cutting Problem. Instead of doing maximization, here we need to do minimization.

C++

// C++ program to find minimum cost to
// get exactly W Kg with given packets
#include<bits/stdc++.h>
using namespace std;
 
/* Returns the best obtainable price for
   a rod of length n and price[] as prices
   of different pieces */
int minCost(int cost[], int n)
{
   int dp[n+1];
   dp[0] = 0;
   
   // Build the table val[] in bottom up
   // manner and return the last entry
   // from the table
   for (int i = 1; i<=n; i++)
   {
       int min_cost = INT_MAX;
       for (int j = 0; j < i; j++)
         if(cost[j]!=-1)
             min_cost = min(min_cost, cost[j] + dp[i-j-1]);
       dp[i] = min_cost;
   }
   
   return dp[n];
}
 
/* Driver code */
int main()
{
   int cost[] = {20, 10, 4, 50, 100};
   int W = sizeof(cost)/sizeof(cost[0]);
   cout << minCost(cost, W);
   return 0;
}

                    

Java

// Java program to find minimum cost to
// get exactly W Kg with given packets
import java.util.*;
class Main
{
   
    /* Returns the best obtainable price for
    a rod of length n and price[] as prices
    of different pieces */
    public static int minCost(int cost[], int n)
    {
       int dp[] = new int[n + 1];
       dp[0] = 0;
        
       // Build the table val[] in bottom up
       // manner and return the last entry
       // from the table
       for (int i = 1; i <= n; i++)
       {
           int min_cost = Integer.MAX_VALUE;
           for (int j = 0; j < i; j++)
               if(cost[j]!=-1) {
                 min_cost = Math.min(min_cost, cost[j] + dp[i - j - 1]);
            }
            dp[i] = min_cost;
       }
        
       return dp[n];
    }
 
    public static void main(String[] args) {
       int cost[] = {10,-1,-1,-1,-1};
       int W = cost.length;
       System.out.print(minCost(cost, W));
    }
}
 
// This code is contributed by divyeshrabadiya07

                    

Python3

# Python3 program to find minimum cost to
# get exactly W Kg with given packets
import sys
 
# Returns the best obtainable price for
# a rod of length n and price[] as prices
# of different pieces
def minCost(cost, n):
     
    dp = [0 for i in range(n + 1)]
 
    # Build the table val[] in bottom up
       # manner and return the last entry
       # from the table
    for i in range(1, n + 1):
        min_cost = sys.maxsize
 
        for j in range(i):
            if cost[j]!=-1:
                min_cost = min(min_cost,
                           cost[j] + dp[i - j - 1])
             
        dp[i] = min_cost
         
    return dp[n]
 
# Driver code
cost = [ 10,-1,-1,-1,-1 ]
W = len(cost)
 
print(minCost(cost, W))
 
# This code is contributed by rag2127

                    

C#

// C# program to find minimum cost to
// get exactly W Kg with given packets
using System;
class GFG {
     
    /* Returns the best obtainable price for
    a rod of length n and price[] as prices
    of different pieces */
    static int minCost(int[] cost, int n)
    {
       int[] dp = new int[n + 1];
       dp[0] = 0;
         
       // Build the table val[] in bottom up
       // manner and return the last entry
       // from the table
       for (int i = 1; i <= n; i++)
       {
           int min_cost = Int32.MaxValue;
           for (int j = 0; j < i; j++)
             if(cost[j]!=-1)
                 min_cost = Math.Min(min_cost,
                                    cost[j] + dp[i - j - 1]);
           dp[i] = min_cost;
       }
         
       return dp[n];
    }
    
  // Driver code
  static void Main() {
   int[] cost = {20, 10, 4, 50, 100};
   int W = cost.Length;
   Console.Write(minCost(cost, W));
  }
}
 
// This code is contributed by divyesh072019

                    

Javascript

<script>
 
    // Javascript program to find minimum cost to
    // get exactly W Kg with given packets
     
    /* Returns the best obtainable price for
     a rod of length n and price[] as prices
     of different pieces */
    function minCost(cost, n)
    {
       let dp = new Array(n+1);
       dp[0] = 0;
 
       // Build the table val[] in bottom up
       // manner and return the last entry
       // from the table
       for (let i = 1; i<=n; i++)
       {
           let min_cost = Number.MAX_VALUE;
           for (let j = 0; j < i; j++)
             if(j < n)
                 min_cost = Math.min(min_cost, cost[j] + dp[i-j-1]);
           dp[i] = min_cost;
       }
 
       return dp[n];
    }
 
    let cost = [20, 10, 4, 50, 100];
    let W = cost.length;
    document.write(minCost(cost, W));
     
</script>

                    

Output
14

Time Complexity: O(N2)
Auxiliary Space: O(N)

Top Down Approach: We can also solve the problem using memoization.

C++

// C++ program to find minimum cost to
// get exactly W Kg with given packets
#include <bits/stdc++.h>
using namespace std;
 
int helper(vector<int>& cost, vector<int>& weight, int n,
           int w, vector<vector<int> >& dp)
{
    // base cases
    if (w == 0)
        return 0;
    if (w < 0 or n <= 0)
        return INT_MAX;
 
    if (dp[n][w] != -1)
        return dp[n][w];
 
    if (cost[n - 1] < 0)
        return dp[n][w]
               = min(INT_MAX,
                     helper(cost, weight, n - 1, w, dp));
 
    if (weight[n - 1] <= w) {
        return dp[n][w]
               = min(cost[n - 1]
                         + helper(cost, weight, n,
                                  w - weight[n - 1], dp),
                     helper(cost, weight, n - 1, w, dp));
    }
    return dp[n][w] = helper(cost, weight, n - 1, w, dp);
}
int minCost(vector<int>& cost, int W)
{
    int N = cost.size();
    // Your code goes here
    vector<int> weight(N);
 
    // create the weight array
    for (int i = 1; i <= N; i++) {
        weight[i - 1] = i;
    }
 
    // initialize the dp array
    vector<vector<int> > dp(N + 1, vector<int>(W + 1, -1));
 
    int res = helper(cost, weight, N, W, dp);
 
    // return -1 if result is MAX
    return (res == INT_MAX) ? -1 : res;
}
 
/* Driver code */
int main()
{
    vector<int> cost = { 20, 10, 4, 50, 100 };
    int W = cost.size();
    cout << minCost(cost, W);
    return 0;
}

                    

Java

// Java program to find minimum cost to
// get exactly W Kg with given packets
import java.io.*;
 
class GFG {
 
    public static int helper(int cost[], int weight[],
                             int n, int w, int dp[][])
    {
        // base cases
        if (w == 0)
            return 0;
        if (w < 0 || n <= 0)
            return Integer.MAX_VALUE;
 
        if (dp[n][w] != -1)
            return dp[n][w];
 
        if (cost[n - 1] < 0)
            return dp[n][w] = Math.min(
                       Integer.MAX_VALUE,
                       helper(cost, weight, n - 1, w, dp));
 
        if (weight[n - 1] <= w) {
            return dp[n][w] = Math.min(
                       cost[n - 1]
                           + helper(cost, weight, n,
                                    w - weight[n - 1], dp),
                       helper(cost, weight, n - 1, w, dp));
        }
        return dp[n][w]
            = helper(cost, weight, n - 1, w, dp);
    }
    public static int minCost(int cost[], int W)
    {
        int N = cost.length;
        int weight[] = new int[N];
 
        // create the weight array
        for (int i = 1; i <= N; i++) {
            weight[i - 1] = i;
        }
 
        // initialize the dp array
        int dp[][] = new int[N + 1][W + 1];
        for (int i = 0; i < N + 1; i++)
            for (int j = 0; j < W + 1; j++)
                dp[i][j] = -1;
 
        int res = helper(cost, weight, N, W, dp);
 
        // return -1 if result is MAX
        return (res == Integer.MAX_VALUE) ? -1 : res;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int cost[] = { 20, 10, 4, 50, 100 };
        int W = cost.length;
        System.out.print(minCost(cost, W));
    }
}
 
// This code is contributed by Rohit Pradhan

                    

Python3

# Python3 program to find minimum cost to
# get exactly W Kg with given packets
import sys
 
def helper(cost, weight, n, w, dp):
 
    # base cases
    if (w == 0):
        return 0
    if (w < 0 or n <= 0):
        return sys.maxsize
 
    if (dp[n][w] != -1):
        return dp[n][w]
 
    if (cost[n - 1] < 0):
        dp[n][w] = min(sys.maxsize, helper(cost, weight, n - 1, w, dp))
        return dp[n][w]
 
    if (weight[n - 1] <= w):
        dp[n][w] = min(cost[n - 1] + helper(cost, weight, n, w - weight[n - 1], dp), helper(cost, weight, n - 1, w, dp))
        return dp[n][w]
 
     
    dp[n][w] = helper(cost, weight, n - 1, w, dp)
    return dp[n][w]
 
def minCost(cost, W):
 
    N = len(cost)
     
    weight = [0 for i in range(N)]
 
    # create the weight array
    for i in range(1,N + 1):
        weight[i - 1] = i
 
    # initialize the dp array
    dp = [[-1 for i in range(W + 1)]for j in range(N + 1)]
 
    res = helper(cost, weight, N, W, dp)
 
    # return -1 if result is MAX
    return -1 if(res == sys.maxsize) else res
 
# Driver code
cost = [ 20, 10, 4, 50, 100 ]
W = len(cost)
print(minCost(cost, W))
 
# This code is contributed by shinjanpatra

                    

C#

// C# program to find minimum cost to
// get exactly W Kg with given packets
using System;
 
class GFG
{
  static int helper(int[] cost, int[] weight,
                    int n, int w, int[,] dp)
  {
    // base cases
    if (w == 0)
      return 0;
    if (w < 0 || n <= 0)
      return Int32.MaxValue;
 
    if (dp[n,w] != -1)
      return dp[n,w];
 
    if (cost[n - 1] < 0)
      return dp[n,w] = Math.Min(
      Int32.MaxValue,
      helper(cost, weight, n - 1, w, dp));
 
    if (weight[n - 1] <= w)
    {
      return dp[n,w] = Math.Min(
        cost[n - 1]
        + helper(cost, weight, n,
                 w - weight[n - 1], dp),
        helper(cost, weight, n - 1, w, dp));
    }
    return dp[n,w]
      = helper(cost, weight, n - 1, w, dp);
  }
 
  static int minCost(int[] cost, int W)
  {
    int N = cost.Length;
    int[] weight = new int[N];
 
    // create the weight array
    for (int i = 1; i <= N; i++)
    {
      weight[i - 1] = i;
    }
 
    // initialize the dp array
    int[,] dp = new int[N + 1, W + 1];
    for (int i = 0; i < N + 1; i++)
      for (int j = 0; j < W + 1; j++)
        dp[i,j] = -1;
 
    int res = helper(cost, weight, N, W, dp);
 
    // return -1 if result is MAX
    return (res == Int32.MaxValue) ? -1 : res;
  }
 
  // Driver Code
  static public void Main()
  {
    int[] cost = { 20, 10, 4, 50, 100 };
    int W = cost.Length;
    Console.Write(minCost(cost, W));
  }
}
 
// This code is contributed by kothavvsaakash

                    

Javascript

<script>
 
// JavaScript program to find minimum cost to
// get exactly W Kg with given packets
function helper(cost, weight, n, w, dp)
{
    // base cases
    if (w == 0)
        return 0;
    if (w < 0 || n <= 0)
        return Number.MAX_VALUE;
 
    if (dp[n][w] != -1)
        return dp[n][w];
 
    if (cost[n - 1] < 0)
        return dp[n][w]
               = Math.min(Number.MAX_VALUE,
                     helper(cost, weight, n - 1, w, dp));
 
    if (weight[n - 1] <= w) {
        return dp[n][w]
               = Math.min(cost[n - 1]
                         + helper(cost, weight, n,
                                  w - weight[n - 1], dp),
                     helper(cost, weight, n - 1, w, dp));
    }
    return dp[n][w] = helper(cost, weight, n - 1, w, dp);
}
function minCost(cost,W)
{
    let N = cost.length;
    // Your code goes here
    let weight = new Array(N);
 
    // create the weight array
    for (let i = 1; i <= N; i++) {
        weight[i - 1] = i;
    }
 
    // initialize the dp array
    let dp = new Array(N + 1).fill(-1).map(()=>new Array(W + 1).fill(-1));
 
    let res = helper(cost, weight, N, W, dp);
 
    // return -1 if result is MAX
    return (res == Number.MAX_VALUE) ? -1 : res;
}
 
/* Driver code */
let cost = [ 20, 10, 4, 50, 100 ];
let W = cost.length;
document.write(minCost(cost, W),"</br>");
 
// This code is contributed by shinjanpatra
 
</script>

                    

Output
14

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

This article is reviewed by team GeeksForGeeks. 

 



Last Updated : 17 Apr, 2023
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