Minimum cost to buy N kilograms of sweet for M persons
Given an array of n size contains the cost of sweets in such a way that that sweet[i] is the cost of i sweets. The task is to find the minimum cost to spend to buy exactly ‘n’ kilograms of sweet for the ‘m’ persons.
Since sweets are available in the packets, you have to buy at most ‘m’ packet of sweets for your ‘m’ relatives. You cannot buy more than ‘m’ packet of sweets. Also cost[i] = 0, represents that the sweet with packet size i is unavailable. Also, there is an infinite number of packets with i size of sweets.
Examples:
Input: m = 3, n = 6, arr[] = {2, 1, 3, 0, 4, 10}
Output: 3
We can choose at most 3 packets. We choose 3 packets of size 2 having cost 1 each.Thus output is 3.Input: m = 2, n = 7, arr[] = {1, 3, 0, 5, 0, 0, 0}
Output : 0
We can choose at most 2 packets. 7 can be formed by 1 2 and 4 index but since you require at most 2 packets to obtain the 7 sweets packets answer which is not possible. Hence the answer is 0 as it is formed by 3 packets, not 2.
Approach:
- Create matrix sweet[m+1][n+1][n+1], where m is number of relatives and n is the total kg of sweets to be bought and number of packages of sweet.
- Initialize sweet[i][0][j] element with 0 and sweet[i][j][0] with -1.
- Now fill the matrix according to following rules –
- Buy the ‘k’ package and assigning it to dp array. If i>0 and j>=Number of current packages and Price of k sweets is greater than 0. Define dp as dp [i-1][j-k][k] + sweet[k]
- If dp is undefined select from previous k-1 packages -> dp[i][j][k]=dp[i][j][k-1]
- If dp[m][n][n] is -1, answer is 0 otherwise print dp[m][n][n]
Below is the implementation of above approach:
C++
// C++ program to minimum cost to buy // N kilograms of sweet for M persons #include <bits/stdc++.h> using namespace std; // Function to find the minimum cost of sweets int find(int m, int n, int adj[]) { // Defining the sweet array int sweet[n + 1]; // DP array to store the values int dp[n + 1][n + 1][n + 1]; sweet[0] = 0; // Since index starts from 1 we // reassign the array into sweet for (int i = 1; i <= m; ++i) sweet[i] = adj[i - 1]; // Assigning base cases for dp array for (int i = 0; i <= m; ++i) { for (int k = 0; k <= n; ++k) // At 0 it is free dp[i][0][k] = 0; // Package not available for desirable amount of sweets for (int k = 1; k <= n; ++k) dp[i][k][0] = -1; } for (int i = 0; i <= m; ++i) { for (int j = 1; j <= n; ++j) { for (int k = 1; k <= n; ++k) { dp[i][j][k] = -1; // Buying the 'k' kg package and // assigning it to dp array if (i > 0 && j >= k && sweet[k] > 0 && dp[i - 1][j - k][k] != -1) dp[i][j][k] = dp[i - 1][j - k][k] + sweet[k]; // If no solution, select from previous k-1 packages if (dp[i][j][k] == -1 || (dp[i][j][k - 1] != -1 && dp[i][j][k] > dp[i][j][k - 1])) dp[i][j][k] = dp[i][j][k - 1]; } } } // If solution does not exist if (dp[m][n][n] == -1) return 0; // Print the solution else return dp[m][n][n]; } // Driver Function int main() { int m = 3; int adj[] = { 2, 1, 3, 0, 4, 10 }; int n = sizeof(adj) / sizeof(adj[0]); // Calling the desired function cout << find(m, n, adj); return 0; } |
Java
// Java program to minimum cost to buy // N kilograms of sweet for M persons public class GFG { // Function to find the minimum cost of sweets static int find(int m, int n, int adj[]) { // Defining the sweet array int sweet[] = new int [n + 1] ; // DP array to store the values int dp[][][] = new int [n + 1][n + 1][n + 1] ; sweet[0] = 0; // Since index starts from 1 we // reassign the array into sweet for (int i = 1; i <= m; ++i) sweet[i] = adj[i - 1]; // Assigning base cases for dp array for (int i = 0; i <= m; ++i) { for (int k = 0; k <= n; ++k) // At 0 it is free dp[i][0][k] = 0; // Package not available for desirable amount of sweets for (int k = 1; k <= n; ++k) dp[i][k][0] = -1; } for (int i = 0; i <= m; ++i) { for (int j = 1; j <= n; ++j) { for (int k = 1; k <= n; ++k) { dp[i][j][k] = -1; // Buying the 'k' kg package and // assigning it to dp array if (i > 0 && j >= k && sweet[k] > 0 && dp[i - 1][j - k][k] != -1) dp[i][j][k] = dp[i - 1][j - k][k] + sweet[k]; // If no solution, select from previous k-1 packages if (dp[i][j][k] == -1 || (dp[i][j][k - 1] != -1 && dp[i][j][k] > dp[i][j][k - 1])) dp[i][j][k] = dp[i][j][k - 1]; } } } // If solution does not exist if (dp[m][n][n] == -1) return 0; // Print the solution else return dp[m][n][n]; } // Driver code public static void main(String args[]) { int m = 3; int adj[] = { 2, 1, 3, 0, 4, 10 }; int n = adj.length ; System.out.println( find(m, n, adj)); } // This Code is contributed by ANKITRAI1 } |
Python
# Python3 program to minimum cost to buy # N kilograms of sweet for M persons # Function to find the minimum cost of sweets def find(m, n, adj): # Defining the sweet array sweet = [0] * (n + 1) # DP array to store the values dp = [[[ 0 for i in range(n + 1)] for i in range(n + 1)] for i in range(n + 1)] sweet[0] = 0 # Since index starts from 1 we # reassign the array into sweet for i in range(1, m + 1): sweet[i] = adj[i - 1] # Assigning base cases for dp array for i in range(m + 1): for k in range(n + 1): # At 0 it is free dp[i][0][k] = 0 # Package not available for desirable amount of sweets for k in range(1, n + 1): dp[i][k][0] = -1 for i in range(m + 1): for j in range(1, n + 1): for k in range(1, n + 1): dp[i][j][k] = -1 # Buying the 'k' kg package and # assigning it to dp array if (i > 0 and j >= k and sweet[k] > 0 and dp[i - 1][j - k][k] != -1): dp[i][j][k] = dp[i - 1][j - k][k] + sweet[k] # If no solution, select from previous k-1 packages if (dp[i][j][k] == -1 or (dp[i][j][k - 1] != -1 and dp[i][j][k] > dp[i][j][k - 1])): dp[i][j][k] = dp[i][j][k - 1] # If solution does not exist if (dp[m][n][n] == -1): return 0 # Print the solution else: return dp[m][n][n] # Driver Function m = 3adj = [2, 1, 3, 0, 4, 10] n = len(adj) # Calling the desired function print(find(m, n, adj)) # This code is contributed by mohit kumar 29 |
C#
// C# program to minimum cost to buy // N kilograms of sweet for M persons using System; class GFG { // Function to find the minimum // cost of sweets static int find(int m, int n, int[] adj) { // Defining the sweet array int[] sweet = new int [n + 1] ; // DP array to store the values int[,,] dp = new int [n + 1, n + 1, n + 1]; sweet[0] = 0; // Since index starts from 1 we // reassign the array into sweet for (int i = 1; i <= m; ++i) sweet[i] = adj[i - 1]; // Assigning base cases // for dp array for (int i = 0; i <= m; ++i) { for (int k = 0; k <= n; ++k) // At 0 it is free dp[i, 0, k] = 0; // Package not available for // desirable amount of sweets for (int k = 1; k <= n; ++k) dp[i, k, 0] = -1; } for (int i = 0; i <= m; ++i) { for (int j = 1; j <= n; ++j) { for (int k = 1; k <= n; ++k) { dp[i, j, k] = -1; // Buying the 'k' kg package and // assigning it to dp array if (i > 0 && j >= k && sweet[k] > 0 && dp[i - 1, j - k, k] != -1) dp[i, j, k] = dp[i - 1, j - k, k] + sweet[k]; // If no solution, select from // previous k-1 packages if (dp[i, j, k] == -1 || (dp[i, j, k - 1] != -1 && dp[i, j, k] > dp[i, j, k - 1])) dp[i, j, k] = dp[i, j, k - 1]; } } } // If solution does not exist if (dp[m, n, n] == -1) return 0; // Print the solution else return dp[m, n, n]; } // Driver code public static void Main() { int m = 3; int[] adj = { 2, 1, 3, 0, 4, 10 }; int n = adj.Length ; Console.Write(find(m, n, adj)); } } // This code is contributed // by ChitraNayal // C# program to minimum cost to buy // N kilograms of sweet for M persons using System; class GFG { // Function to find the minimum // cost of sweets static int find(int m, int n, int[] adj) { // Defining the sweet array int[] sweet = new int [n + 1] ; // DP array to store the values int[,,] dp = new int [n + 1, n + 1, n + 1]; sweet[0] = 0; // Since index starts from 1 we // reassign the array into sweet for (int i = 1; i <= m; ++i) sweet[i] = adj[i - 1]; // Assigning base cases // for dp array for (int i = 0; i <= m; ++i) { for (int k = 0; k <= n; ++k) // At 0 it is free dp[i, 0, k] = 0; // Package not available for // desirable amount of sweets for (int k = 1; k <= n; ++k) dp[i, k, 0] = -1; } for (int i = 0; i <= m; ++i) { for (int j = 1; j <= n; ++j) { for (int k = 1; k <= n; ++k) { dp[i, j, k] = -1; // Buying the 'k' kg package and // assigning it to dp array if (i > 0 && j >= k && sweet[k] > 0 && dp[i - 1, j - k, k] != -1) dp[i, j, k] = dp[i - 1, j - k, k] + sweet[k]; // If no solution, select from // previous k-1 packages if (dp[i, j, k] == -1 || (dp[i, j, k - 1] != -1 && dp[i, j, k] > dp[i, j, k - 1])) dp[i, j, k] = dp[i, j, k - 1]; } } } // If solution does not exist if (dp[m, n, n] == -1) return 0; // Print the solution else return dp[m, n, n]; } // Driver code public static void Main() { int m = 3; int[] adj = { 2, 1, 3, 0, 4, 10 }; int n = adj.Length ; Console.Write(find(m, n, adj)); } } // This code is contributed // by ChitraNayal |
Output:
3
The time complexity of the above algorithm is O(m*n*n).
