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:

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

C++

// C++ program to find minimum 
// number of denominations 
#include <bits/stdc++.h> 
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[0]); 
  
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 <stdio.h> 
#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<Integer> 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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

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

C++

C

Java

Python3

Recommended Posts: