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Buy minimum items without change and given coins
  • Last Updated : 19 May, 2021

You have an unlimited number of 10-rupee coins and exactly one coin of r rupee and you need to buy minimum items each of cost k such that you do not ask for change.
Examples: 
 

Input: k = 15, r = 2 
Output: 2 
You should buy two cables and pay 2*15=30 rupees. It is obvious that you can pay this sum without any change.
Input: k = 237, r = 7 
Output:1 
It is enough for you to buy one cable.

 

It is obvious that we can pay for 10 items without any change (by paying the required amount of 10-rupee coins and not using the coin of r rupee). But perhaps you can buy fewer hammers and pay without any change. Note that you should buy at least one item. 
 

C++




#include <bits/stdc++.h>
using namespace std;
 
int minItems(int k, int r)
{
   // See if we can buy less than 10 items
   // Using 10 Rs coins and one r Rs coin
   for (int i = 1; i < 10; i++)
        if ((i * k - r) % 10 == 0 ||
            (i * k) % 10 == 0)
            return i;
 
    // We can always buy 10 items
    return 10;
}
 
int main()
{
    int k = 15;
    int r = 2;
    cout << minItems(k, r);
    return 0;
}

Java




// Java implementation of above approach
import java.util.*;
 
class GFG
{
static int minItems(int k, int r)
{
     
// See if we can buy less than 10 items
// Using 10 Rs coins and one r Rs coin
for (int i = 1; i < 10; i++)
        if ((i * k - r) % 10 == 0 ||
            (i * k) % 10 == 0)
            return i;
 
    // We can always buy 10 items
    return 10;
}
 
// Driver Code
public static void main(String args[])
{
    int k = 15;
    int r = 2;
    System.out.println(minItems(k, r));
}
}
 
// This code is contributed
// by SURENDRA_GANGWAR

Python3




# Python3 implementation of above approach
 
def minItems(k, r) :
 
    # See if we can buy less than 10 items
    # Using 10 Rs coins and one r Rs coin
    for i in range(1, 10) :
            if ((i * k - r) % 10 == 0 or
                (i * k) % 10 == 0) :
                return i
     
    # We can always buy 10 items
    return 10;
     
# Driver Code
if __name__ == "__main__" :
 
    k, r = 15 , 2;
    print(minItems(k, r))
 
# This code is contributed by Ryuga

C#




// C# implementation of above approach
using System;
 
class GFG
{
static int minItems(int k, int r)
{
     
// See if we can buy less than 10 items
// Using 10 Rs coins and one r Rs coin
for (int i = 1; i < 10; i++)
        if ((i * k - r) % 10 == 0 ||
            (i * k) % 10 == 0)
            return i;
 
    // We can always buy 10 items
    return 10;
}
 
// Driver Code
public static void Main()
{
    int k = 15;
    int r = 2;
    Console.WriteLine(minItems(k, r));
}
}
 
// This code is contributed
// by inder_verma

PHP




<?php
// See if we can buy less than 10 items
// Using 10 Rs coins and one r Rs coin
function minItems($k, $r)
{
    for ($i = 1; $i < 10; $i++)
    if (($i * $k - $r) % 10 == 0 ||
        ($i * $k) % 10 == 0)
        return $i;
         
    // We can always buy 10 items
    return 10;
}
 
// Driver Code
$k = 15;
$r = 2;
echo minItems($k, $r);
 
// This code is contributed by Rajput-Ji
?>

Javascript




<script>
 
// Javascript program of the above approach
 
function minItems(k, r)
{
     
// See if we can buy less than 10 items
// Using 10 Rs coins and one r Rs coin
for (let i = 1; i < 10; i++)
        if ((i * k - r) % 10 == 0 ||
            (i * k) % 10 == 0)
            return i;
 
    // We can always buy 10 items
    return 10;
}
 
// Driver code
 
    let k = 15;
    let r = 2;
    document.write(minItems(k, r));
 
</script>
Output: 
2

 

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