Given a number, the only operation allowed is to multiply the number by 2. Calculate the minimum number of operations to make the number divisible by 10.
NOTE: If it is not possible to convert then print -1.
Examples:
Input: 10
Output: 0
As the given number is itself divisible by 10,
the answer is 0.Input: 1
Output: -1
As by multiplying with 2, given no. can’t be
converted into a number that is divisible by 10,
therefore the answer is -1.
Approach: Any given number is divisible by 10 only if the last digit of the number is 0. For this problem, extract the last digit of the input number and check it in the following ways :
1) If the last digit is 0 then it is already divisible by 10 , so the minimum number of steps is 0.
2) If the last digit is 5 then multiplying it by 2 one time will make it divisible by 10, so the minimum number of steps is 1.
3) If the last digit is an even or odd number (apart from 0 and 5) then multiplying it by 2 any number of times will only produce even number so we can never make it divisible by 10. Therefore the number of steps is -1.
C++
// C++ code for finding // number of operations #include <bits/stdc++.h> using namespace std; int multiplyBy2(int n) { int rem, value; // Find the last digit or remainder rem = n % 10; switch (rem) { // If the last digit is 0 case 0: value = 0; break; // If the last digit is 5 case 5: value = 1; break; // If last digit is other // than 0 and 5. default: value = -1; } return value; } // Driver code int main() { int n = 28; cout << multiplyBy2(n) << endl; n = 255; cout << multiplyBy2(n) << endl; return 0; } |
Java
// JAVA code for finding // number of operations import java.io.*; class GFG { static int multiplyBy2(int n) { int rem, value; // Find the last digit // or remainder rem = n % 10; switch (rem) { // If the last digit is 0 case 0: value = 0; break; // If the last digit is 5 case 5: value = 1; break; // If last digit is other // than 0 and 5. default: value = -1; } return value; } // Driver code public static void main (String[] args) { int n = 28; System.out.println(multiplyBy2(n)); n = 255; System.out.println(multiplyBy2(n)); } } // This code is contributed // by shiv_bhakt. |
Python3
# Python3 code for finding number # of operations def dig(argu): switcher = { 0: 0, 5: 1, } return switcher.get(argu, -1) def multiplyBy2(n): # Find the last digit or remainder rem = n % 10; return dig(rem); # Driver code n = 28; print(multiplyBy2(n)); n = 255; print(multiplyBy2(n)); # This code is contributed by mits |
C#
// C# code for finding // number of operations using System; class GFG { static int multiplyBy2(int n) { int rem, value; // Find the last digit // or remainder rem = n % 10; switch (rem) { // If the last // digit is 0 case 0: value = 0; break; // If the last // digit is 5 case 5: value = 1; break; // If last digit is // other than 0 and 5. default: value = -1; break; } return value; } // Driver code public static void Main () { int n = 28; Console.WriteLine(multiplyBy2(n)); n = 255; Console.WriteLine(multiplyBy2(n)); } } // This code is contributed // by shiv_bhakt. |
PHP
<?php // PHP code for finding // number of operations function multiplyBy2($n) { $rem; $value; // Find the last digit or remainder $rem = $n % 10; switch ($rem) { // If the last digit is 0 case 0: $value = 0; break; // If the last digit is 5 case 5: $value = 1; break; // If last digit is other // than 0 and 5. default: $value = -1; } return $value; } // Driver code $n = 28; echo multiplyBy2($n),"\n"; $n = 255; echo multiplyBy2($n),"\n"; // This code is contributed by aj_36 ?> |
-1 1
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
