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 codeint main(){ int n = 28; cout << multiplyBy2(n) << endl; n = 255; cout << multiplyBy2(n) << endl; return 0;} |
Java
// JAVA code for finding // number of operationsimport 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 codepublic 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 operationsdef 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 coden = 28;print(multiplyBy2(n));n = 255;print(multiplyBy2(n)); # This code is contributed by mits |
C#
// C# code for finding // number of operationsusing 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 codepublic 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 operationsfunction 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?> |
Javascript
<script>// Javascript code for finding // number of operations function multiplyBy2(n) { var 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 var n = 28; document.write(multiplyBy2(n)+"<br/>"); n = 255; document.write(multiplyBy2(n));// This code is contributed by todaysgaurav </script> |
Output
-1 1
Time Complexity: O(1), because it is performing constant operations
Auxiliary Space: O(1)
Method: Using nested if-else
C++
// C++ code to multiply given number// by 2 such that it is divisible by 10// using nested if else#include <iostream>using namespace std;int main() {int n = 10; // checking if the number is divisible// by 10 or not if divisible print 0if (n % 10 == 0) cout << 0; // if not divisible by 10 then multiply// it by 2 and check againelse{ n = n*2; if (n % 10 == 0) cout<<0; else cout<<-1;} return 0;} // This code is contributed by akashish__ |
Java
// Java code to multiply given number// by 2 such that it is divisible by 10// using nested if elseimport java.io.*;class GFG { public static void main (String[] args) { int n = 10; // checking if the number is divisible // by 10 or not if divisible print 0 if (n % 10 == 0) System.out.println("0"); // if not divisible by 10 then multiply // it by 2 and check again else{ n = n*2; if (n % 10 == 0) System.out.println("0"); else System.out.println("-1"); } }}// This code is contributed by laxmigangarajula03. |
Python3
# Python code to multiply given number# by 2 such that it is divisible by 10# using nested if elsen = 10# checking if the number is divisible# by 10 or not if divisible print 0if n % 10 == 0: print(0)# if not divisible by 10 then multiply# it by 2 and check againelse: n = n*2 if n % 10 == 0: print(0) else: print(-1) # this code is contributed by gangarajula laxmi |
C#
// C# code to multiply given number// by 2 such that it is divisible by 10// using nested if elseusing System;public class GFG{ static public void Main (){ // Code int n = 10; // checking if the number is divisible // by 10 or not if divisible print 0 if (n % 10 == 0) { Console.WriteLine("0"); } // if not divisible by 10 then multiply // it by 2 and check again else{ n = n*2; if (n % 10 == 0) { Console.WriteLine("0"); } else { Console.WriteLine("-1"); } } }}// This code is contributed by akashish__ |
Javascript
<script> // JavaScript code for the above approach let n = 10; // checking if the number is divisible // by 10 or not if divisible print 0 if (n % 10 == 0) document.write(0); // if not divisible by 10 then multiply // it by 2 and check again else { n = n * 2; if (n % 10 == 0) document.write(0); else document.write(-1); } // This code is contributed by Potta Lokesh </script> |
Output
0
Time Complexity: O(1), because it is performing constant operations
Auxiliary Space: O(1)
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