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++ program to find minimum // number of denominations #include <bits/stdc++.h> using namespace std;
// All denominations of Indian Currency int 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 Code int main()
{ int n = 93;
cout << "Following is minimal"
<< " number of change for " << n << ": " ;
// Function Call
findMin(n);
return 0;
} |
// 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 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 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 |
// C# program to find minimum // number of denominations using 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 |
<script> // Javascript program to find minimum // number of denominations // All denominations of Indian Currency let 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 code n = 93; document.write( "Following is minimal number " + "of change for " + n + ": " );
findMin(n); // This code is contributed by rag2127 </script> |
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.
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++ 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 Code int 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 program to find minimum import 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 |
# Python Code # Function to find minimal number of coins def 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 Code v = 93
print ( "Following is minimal number of change for " , v, ":" )
# Function Call vec = findMin(v)
# print the vector for it in vec:
print (it, end = " " )
|
// C# program to find minimum using 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 |
// Function to find minimal number of coins function 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 code let v = 93; console.log( "Following is minimal number of change for " + v + " :" );
// Function call let vec = findMin(v); // Print the array console.log(vec.join( " " ));
|
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