Decimal to Binary using recursion and without using power operator
Given an integer N, the task is convert and print the binary equaiva;ent of N.
Examples:
Input: N = 13
Output: 1101
Input: N = 15
Output: 1111
Approach Write a recursive function that takes an argument N and recursively calls itself with the value N / 2 as the new argument and prints N % 2 after the call. The base condition will be when N = 0, simply print 0 and return out of the function in that case.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void decimalToBinary( int n)
{
if (n == 0) {
cout << "0" ;
return ;
}
decimalToBinary(n / 2);
cout << n % 2;
}
int main()
{
int n = 13;
decimalToBinary(n);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static void decimalToBinary( int n)
{
if (n == 0 )
{
System.out.print( "0" );
return ;
}
decimalToBinary(n / 2 );
System.out.print( n % 2 );
}
public static void main (String[] args)
{
int n = 13 ;
decimalToBinary(n);
}
}
|
Python3
def decimalToBinary(n) :
if (n = = 0 ) :
print ( "0" ,end = "");
return ;
decimalToBinary(n / / 2 );
print (n % 2 ,end = "");
if __name__ = = "__main__" :
n = 13 ;
decimalToBinary(n);
|
C#
using System;
class GFG
{
static void decimalToBinary( int n)
{
if (n == 0)
{
Console.Write( "0" );
return ;
}
decimalToBinary(n / 2);
Console.Write(n % 2);
}
public static void Main(String[] args)
{
int n = 13;
decimalToBinary(n);
}
}
|
Javascript
<script>
function decimalToBinary(n) {
if (n == 0) {
document.write( "0" );
return ;
}
decimalToBinary(parseInt(n / 2));
document.write(n % 2);
}
var n = 13;
decimalToBinary(n);
</script>
|
Time Complexity: O(logN)
Auxiliary Space: O(logN)
Last Updated :
04 Aug, 2021
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