# Buy minimum items without change and given coins

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 ` `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

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## Javascript

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Output:

`2`

Time Complexity : O(10), as 10 is constant so => O(1)
Auxiliary Space : O(1)