# Greedy Algorithm to find Minimum number of Coins

Given a value V, if we want to make a change for V Rs, and we have an infinite supply of each of the denominations in Indian currency, i.e., we have an 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.

Solution: Greedy Approach.

Approach: A common intuition would be to take coins with greater value first. This can reduce the total number of coins needed. Start from the largest possible denomination and keep adding denominations while the remaining value is greater than 0.

Algorithm:

1. Sort the array of coins in decreasing order.
2. Initialize result as empty.
3. Find the largest denomination that is smaller than current amount.
4. Add found denomination to result. Subtract value of found denomination from amount.
5. If amount becomes 0, then print result.
6. Else repeat steps 3 and 4 for new value of V.

## 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) ` `{ ` `    ``sort(deno, deno + n); ` ` `  `    ``// 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]; ` `            ``// Add coin in the list ` `            ``coinList[k++] = coins[i]; ` `        ``} ` `    ``} ` ` `  `    ``for` `(i = 0; i < k; i++) { ` `        ``// Print ` `        ``printf``(``"%d "``, coinList[i]); ` `    ``} ` `    ``return``; ` `} ` ` `  `int` `main(``void``) ` `{ ` `    ``// input value ` `    ``int` `n = 93; ` ` `  `    ``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```

Complexity Analysis:

• Time Complexity: O(N*logN).
• Auxiliary Space: O(1) as no additional space is used.

Note: The 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, below dynamic programming approach can be used:
Find minimum number of coins that make a given value

Thanks to Utkarsh for providing the above solution here.

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