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; } |
chevron_right
filter_none
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 } |
chevron_right
filter_none
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 # Prthe 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 |
chevron_right
filter_none
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 |
chevron_right
filter_none
Output:
3
The time complexity of the above algorithm is O(m*n*n).
Recommended Posts:
- Minimum cost to make two strings same
- Minimum cost to fill given weight in a bag
- Minimum cost to merge numbers from 1 to N
- Minimum odd cost path in a matrix
- Minimum Cost Polygon Triangulation
- Minimum cost to sort a matrix of numbers from 0 to n^2 - 1
- Minimum cost of choosing the array element
- Minimum cost to cover the given positions in a N*M grid
- Minimum cost to partition the given binary string
- Minimum cost to build N blocks from one block
- Minimum Cost To Make Two Strings Identical
- Minimum cost to merge all elements of List
- Find minimum adjustment cost of an array
- Find the minimum cost to cross the River
- Minimum cost to convert 3 X 3 matrix into magic square
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
- Count of subarrays of size K which is a permutation of numbers from 1 to K
- Length of the smallest substring which contains all vowels
- Construct an Array of size N in which sum of odd elements is equal to sum of even elements
- Construct an Array such that cube sum of all element is a perfect square
- Minimum circular rotations to obtain a given numeric string by avoiding a set of given strings