# Greedy Algorithm to find Minimum number of Coins

Given a value V, if we want to make change for V Rs, and we have infinite supply of each of the denominations in Indian currency, i.e., we have infinite supply of { 1, 2, 5, 10, 20, 50, 100, 500, 1000} valued coins/notes, what is the minimum number of coins and/or notes needed to make the change?

Examples:

```Input: V = 70
Output: 2
We need a 50 Rs note and a 20 Rs note.

Input: V = 121
Output: 3
We need a 100 Rs note, a 20 Rs note and a
1 Rs coin. ```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

The idea is simple Greedy Algorithm. Start from largest possible denomination and keep adding denominations while remaining value is greater than 0. Below is complete algorithm.

```1) Initialize result as empty.
2) find the largest denomination that is
smaller than V.
3) Add found denomination to result. Subtract
value of found denomination from V.
4) If V becomes 0, then print result.
Else repeat steps 2 and 3 for new value of V
```

Below is the implementation of above algorithm.

## C++

 `// C++ program to find minimum number of denominations ` `#include ` `using` `namespace` `std; ` ` `  `// All denominations of Indian Currency ` `int` `deno[] = { 1, 2, 5, 10, 20, 50, 100, 500, 1000 }; ` `int` `n = ``sizeof``(deno) / ``sizeof``(deno); ` ` `  `void` `findMin(``int` `V) ` `{ ` `    ``// Initialize result ` `    ``vector<``int``> ans; ` ` `  `    ``// Traverse through all denomination ` `    ``for` `(``int` `i = n - 1; i >= 0; i--) { ` `        ``// Find denominations ` `        ``while` `(V >= deno[i]) { ` `            ``V -= deno[i]; ` `            ``ans.push_back(deno[i]); ` `        ``} ` `    ``} ` ` `  `    ``// Print result ` `    ``for` `(``int` `i = 0; i < ans.size(); i++) ` `        ``cout << ans[i] << ``"  "``; ` `} ` ` `  `// Driver program ` `int` `main() ` `{ ` `    ``int` `n = 93; ` `    ``cout << ``"Following is minimal number of change for "` `<< n << ``": "``; ` `    ``findMin(n); ` `    ``return` `0; ` `} `

## C

 `// C program to find minimum number of denominations ` `#include ` `#define COINS 9 ` `#define MAX 20 ` ` `  `// All denominations of Indian Currency ` `int` `coins[COINS] = { 1, 2, 5, 10, 20, 50, 100, 200, 2000 }; ` ` `  `void` `findMin(``int` `cost) ` `{ ` `    ``int` `coinList[MAX] = { 0 }; ` `    ``int` `i, k = 0; ` ` `  `    ``for` `(i = COINS - 1; i >= 0; i--) { ` `        ``while` `(cost >= coins[i]) { ` `            ``cost -= coins[i]; ` `            ``coinList[k++] = coins[i]; ``// add coin in the list ` `        ``} ` `    ``} ` ` `  `    ``for` `(i = 0; i < k; i++) { ` `        ``printf``(``"%d "``, coinList[i]); ``// print ` `    ``} ` `    ``return``; ` `} ` ` `  `int` `main(``void``) ` `{ ` `    ``int` `n = 93; ``// input value ` ` `  `    ``printf``(``"Following is minimal number of change for %d: "``, n); ` `    ``findMin(n); ` `    ``return` `0; ` `} ` `// Code by Munish Bhardwaj `

## Java

 `// Java program to find minimum number of denominations ` `import` `java.util.Vector; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// All denominations of Indian Currency  ` `    ``static` `int` `deno[] = {``1``, ``2``, ``5``, ``10``, ``20``, ``50``, ``100``, ``500``, ``1000``}; ` `    ``static` `int` `n = deno.length; ` ` `  `    ``static` `void` `findMin(``int` `V) ` `    ``{ ` `        ``// Initialize result  ` `        ``Vector ans = ``new` `Vector<>(); ` ` `  `        ``// Traverse through all denomination  ` `        ``for` `(``int` `i = n - ``1``; i >= ``0``; i--) ` `        ``{ ` `            ``// Find denominations  ` `            ``while` `(V >= deno[i])  ` `            ``{ ` `                ``V -= deno[i]; ` `                ``ans.add(deno[i]); ` `            ``} ` `        ``} ` ` `  `        ``// Print result  ` `        ``for` `(``int` `i = ``0``; i < ans.size(); i++) ` `        ``{ ` `            ``System.out.print(``" "` `+ ans.elementAt(i)); ` `        ``} ` `    ``} ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `n = ``93``; ` `        ``System.out.print(``"Following is minimal number of change for "` `+ n + ``": "``); ` `        ``findMin(n); ` ` `  `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python 3 program to find minimum  ` `# number of denominations ` ` `  `def` `findMin(V): ` `     `  `    ``# All denominations of Indian Currency ` `    ``deno ``=` `[``1``, ``2``, ``5``, ``10``, ``20``, ``50``,  ` `            ``100``, ``500``, ``1000``] ` `    ``n ``=` `len``(deno) ` `     `  `    ``# Initialize Result ` `    ``ans ``=` `[] ` ` `  `    ``# Traverse through all denomination ` `    ``i ``=` `n ``-` `1` `    ``while``(i >``=` `0``): ` `         `  `        ``# Find denominations ` `        ``while` `(V >``=` `deno[i]): ` `            ``V ``-``=` `deno[i] ` `            ``ans.append(deno[i]) ` ` `  `        ``i ``-``=` `1` ` `  `    ``# Print result ` `    ``for` `i ``in` `range``(``len``(ans)): ` `        ``print``(ans[i], end ``=` `" "``) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n ``=` `93` `    ``print``(``"Following is minimal number"``, ` `          ``"of change for"``, n, ``": "``, end ``=` `"") ` `    ``findMin(n) ` `     `  `# This code is contributed by ` `# Surendra_Gangwar `

Output:

`Following is minimal number of change for 93: 50  20  20  2  1`

Note that above approach may not work for all denominations. For example, it doesn’t work for denominations {9, 6, 5, 1} and V = 11. The above approach would print 9, 1 and 1. But we can use 2 denominations 5 and 6.
For general input, we use below dynamic programming approach.

Find minimum number of coins that make a given value

Thanks to Utkarsh for providing above solution here.