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 approachimport 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 coinfor (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 Codepublic 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 approachdef 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 Codeif __name__ == "__main__" : k, r = 15 , 2; print(minItems(k, r))# This code is contributed by Ryuga |
C#
// C# implementation of above approachusing 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 coinfor (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 Codepublic 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 coinfunction 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 approachfunction minItems(k, r){ // See if we can buy less than 10 items// Using 10 Rs coins and one r Rs coinfor (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
Time Complexity : O(10), as 10 is constant so => O(1)
Auxiliary Space : O(1)
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