# Minimum cost to buy N kilograms of sweet for M persons

• Difficulty Level : Expert
• Last Updated : 21 May, 2021

Given an array of n sizes, it includes 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 sweets for the ‘m’ people.
Since sweets are available in packets, you have to buy at most a ‘m’ packet of sweets for your ‘m’ relatives. You cannot buy more than a ‘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 indexes, but since you require at most 2 packets to obtain the 7 sweets packets sweet answer, which is not possible. Hence, the answer is 0 as it is formed by 3 packets, not 2.

Approach:

1. Create a matrix sweet[m+1][n+1][n+1], where m is the number of relatives and n is the total kg of sweets to be bought, and the number of packages of sweets.
2. Initialize sweet[i][0][j] element with 0 and sweet[i][j][0] with -1.
3. Now fill the matrix according to the following rules –
• Buy the ‘k’ package and assign it to dp array. If i>0 and j>=Number of current packages and the 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]
4. If dp[m][n][n] is -1, the answer is 0. Otherwise, print dp[m][n][n]

Below is the implementation of the above approach:

## C++

 `// C++ program to minimum cost to buy``// N kilograms of sweet for M persons``#include ``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 ``=` `3``adj ``=` `[``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`

## Javascript

 ``
Output:
`3`

The time complexity of the above algorithm is O(m*n*n).

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