Given a value of V Rs and an infinite supply of each of the denominations {1, 2, 5, 10, 20, 50, 100, 500, 1000} valued coins/notes, The task is to find the minimum number of coins and/or notes needed to make the change?
Examples:
Input: V = 70
Output: 2
Explanation: We need a 50 Rs note and a 20 Rs note.Input: V = 121
Output: 3
Explanation: We need a 100 Rs note, a 20 Rs note, and a 1 Rs coin.
Approach:
The 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.
Follow the steps below to implement the idea:
- Sort the array of coins in decreasing order.
- Initialize ans vector as empty.
- Find the largest denomination that is smaller than remaining amount and while it is smaller than the remaining amount:
- Add found denomination to ans. Subtract value of found denomination from amount.
- If amount becomes 0, then print ans.
Below is the implementation of above approach.
C++
// C++ program to find minimum// number of denominations#include <bits/stdc++.h>using namespace std;// All denominations of Indian Currencyint denomination[] = { 1, 2, 5, 10, 20, 50, 100, 500, 1000 };int n = sizeof(denomination) / sizeof(denomination[0]);void findMin(int V){ sort(denomination, denomination + n); // Initialize result vector<int> ans; // Traverse through all denomination for (int i = n - 1; i >= 0; i--) { // Find denominations while (V >= denomination[i]) { V -= denomination[i]; ans.push_back(denomination[i]); } } // Print result for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";}// Driver Codeint main(){ int n = 93; cout << "Following is minimal" << " number of change for " << n << ": "; // Function Call 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 Currencyint 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 denominationsimport 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
# Python3 program to find minimum # number of denominationsdef 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 Codeif __name__ == '__main__': n = 93 print("Following is minimal number", "of change for", n, ": ", end = "") findMin(n) # This code is contributed by# Surendra_Gangwar |
C#
// C# program to find minimum// number of denominationsusing System;using System.Collections.Generic;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 List<int> ans = new List<int>(); // 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.Count; i++) { Console.Write(" " + ans[i]); }}// Driver code public static void Main(String[] args) { int n = 93; Console.Write("Following is minimal number " + "of change for " + n + ": "); findMin(n);}}// This code is contributed by gauravrajput1 |
Javascript
<script>// Javascript program to find minimum// number of denominations// All denominations of Indian Currencylet deno=[1, 2, 5, 10, 20, 50, 100, 500, 1000];let n = deno.length;function findMin(V){ // Initialize result let ans = []; // Traverse through all denomination for (let i = n - 1; i >= 0; i--) { // Find denominations while (V >= deno[i]) { V -= deno[i]; ans.push(deno[i]); } } // Print result for (let i = 0; i < ans.length; i++) { document.write( " " + ans[i]); }}// Driver coden = 93;document.write("Following is minimal number "+"of change for " + n + ": ");findMin(n);// This code is contributed by rag2127</script> |
Output
Following is minimal number of change for 93: 50 20 20 2 1
Time Complexity: O(V).
Auxiliary Space: O(V).
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.
Minimum number of Coins using Ladder If-Else approach:
In this approach, we will simply iterate through the greater to smaller coins until the n is greater to that coin and decrement that value from n afterward using ladder if-else and will push back that coin value in the vector.
Follow the steps below to implement the idea:
- Declare a vector that store the coins.
- while n is greater than 0 iterate through greater to smaller coins:
- if n is greater than equal to 2000 than push 2000 into the vector and decrement its value from n.
- else if n is greater than equal to 500 than push 500 into the vector and decrement its value from n.
- And so on till the last coin using ladder if else.
- return the vector/array
C++
// C++ program to find minimum// number of coins#include <bits/stdc++.h>using namespace std;vector<int> findMin(int n){ // initialize vector to store the coins vector<int> v; // iterate till n>0 and check condition according to the // greatest coin possible while (n > 0) { if (n >= 2000) { v.push_back(2000); n -= 2000; } else if (n >= 500) { v.push_back(500); n -= 500; } else if (n >= 200) { v.push_back(200); n -= 200; } else if (n >= 100) { v.push_back(100); n -= 100; } else if (n >= 50) { v.push_back(50); n -= 50; } else if (n >= 20) { v.push_back(20); n -= 20; } else if (n >= 10) { v.push_back(10); n -= 10; } else if (n >= 5) { v.push_back(5); n -= 5; } else if (n >= 2) { v.push_back(2); n -= 2; } else if (n >= 1) { v.push_back(1); n -= 1; } } // return the ans that stores in the vector return v;}// Driver Codeint main(){ int v = 93; cout << "Following is minimal" << " number of change for " << v << ": "; // Function Call vector<int> vec = findMin(v); // print the vector for (auto it : vec) cout << it << " "; return 0;}// this code is contributed by Prateek Kumar Singh |
Java
// Java program to find minimumimport java.util.Vector; class GFG{ static Vector<Integer> findMin(int n){ // initialize vector to store the coins Vector<Integer> vec = new Vector<>(); // iterate till n>0 and check condition according to the // greatest coin possible while (n > 0) { if (n >= 2000) { vec.addElement(2000); n -= 2000; } else if (n >= 500) { vec.addElement(500); n -= 500; } else if (n >= 200) { vec.addElement(200); n -= 200; } else if (n >= 100) { vec.addElement(100); n -= 100; } else if (n >= 50) { vec.addElement(50); n -= 50; } else if (n >= 20) { vec.addElement(20); n -= 20; } else if (n >= 10) { vec.addElement(10); n -= 10; } else if (n >= 5) { vec.addElement(5); n -= 5; } else if (n >= 2) { vec.addElement(2); n -= 2; } else if (n >= 1) { vec.addElement(1); n -= 1; } } // return the ans that stores in the vector return vec;} // Driver Code public static void main(String[] args) { int n = 93; System.out.print( "Following is minimal number " +"of change for " + n + ": "); // Function Call Vector<Integer> vec = findMin(n); // print the vector for (Integer i = 0; i < vec.size(); i++) { System.out.print(vec.get(i) + " "); }}}// this code is contributed by Shivam Miglani |
Python3
# Python Code# Function to find minimal number of coinsdef findMin(n): # Vector to store the coins v = [] # Iterate till n > 0 while(n > 0): # check condition if(n >= 2000): v.append(2000) n -= 2000 elif(n >= 500): v.append(500) n -= 500 elif(n >= 200): v.append(200) n -= 200 elif(n >= 100): v.append(100) n -= 100 elif(n >= 50): v.append(50) n -= 50 elif(n >= 20): v.append(20) n -= 20 elif(n >= 10): v.append(10) n -= 10 elif(n >= 5): v.append(5) n -= 5 elif(n >= 2): v.append(2) n -= 2 elif(n >= 1): v.append(1) n -= 1 # Return the ans return v# Driver Codev = 93print("Following is minimal number of change for ", v, ":")# Function Callvec = findMin(v)# print the vectorfor it in vec: print(it, end=" ") |
C#
// C# program to find minimumusing System;using System.Collections.Generic;class GFG { static List<int> findMin(int n) { // initialize list to store the coins List<int> lst = new List<int>(); // iterate till n>0 and check condition according to // the greatest coin possible while (n > 0) { if (n >= 2000) { lst.Add(2000); n -= 2000; } else if (n >= 500) { lst.Add(500); n -= 500; } else if (n >= 200) { lst.Add(200); n -= 200; } else if (n >= 100) { lst.Add(100); n -= 100; } else if (n >= 50) { lst.Add(50); n -= 50; } else if (n >= 20) { lst.Add(20); n -= 20; } else if (n >= 10) { lst.Add(10); n -= 10; } else if (n >= 5) { lst.Add(5); n -= 5; } else if (n >= 2) { lst.Add(2); n -= 2; } else if (n >= 1) { lst.Add(1); n -= 1; } } // return the ans that stores in the list return lst; } // Driver Code public static void Main() { int n = 93; Console.Write("Following is minimal number " + "of change for " + n + ": "); // Function Call List<int> lst = findMin(n); // print the list for (int i = 0; i < lst.Count; i++) { Console.Write(lst[i] + " "); } }}// This code is contributed by phasing17 |
Javascript
// Function to find minimal number of coinsfunction findMin(n) { // Array to store the coins let v = []; // Iterate till n > 0 while (n > 0) { // Check condition if (n >= 2000) { v.push(2000); n -= 2000; } else if (n >= 500) { v.push(500); n -= 500; } else if (n >= 200) { v.push(200); n -= 200; } else if (n >= 100) { v.push(100); n -= 100; } else if (n >= 50) { v.push(50); n -= 50; } else if (n >= 20) { v.push(20); n -= 20; } else if (n >= 10) { v.push(10); n -= 10; } else if (n >= 5) { v.push(5); n -= 5; } else if (n >= 2) { v.push(2); n -= 2; } else if (n >= 1) { v.push(1); n -= 1; } } // Return the array return v;}// Driver codelet v = 93;console.log("Following is minimal number of change for " + v + " :");// Function calllet vec = findMin(v);// Print the arrayconsole.log(vec.join(" ")); |
Output
Following is minimal number of change for 93: 50 20 20 2 1
Time Complexity: O(N) that is equal to the amount v.
Auxiliary Space: O(1) that is optimized
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