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Count trailing zero bits using lookup table

  • Difficulty Level : Hard
  • Last Updated : 22 Sep, 2021

Given an integer, count the number of trailing zeroes. For example, for n = 12, its binary representation is 1100 and number of trailing zero bits is 2. 

Examples : 

Input : 8
Output : 3
Binary of 8 is 1000, so there are three
trailing zero bits.

Input : 18
Output : 1
Binary of 18 is 10010, so there is one
trailing zero bit.

A simple solution is to traverse bits from LSB (Least Significant Bit) and increment count while bit is 0.  

C++




// Simple C++ code for counting trailing zeros
// in binary representation of a number
#include<bits/stdc++.h>
using namespace std;
 
int countTrailingZero(int x)
{
  int count = 0;
  while ((x & 1) == 0)
  {
      x = x >> 1;
      count++;
  }
  return count;
}
 
// Driver Code
int main()
{
    cout << countTrailingZero(11) << endl;
    return 0;
}

Java




// Simple Java code for counting
// trailing zeros in binary
// representation of a number
import java.io.*;
 
class GFG
{
    public static int countTrailingZero(int x)
    {
        int count = 0;
         
        while ((x & 1) == 0)
        {
            x = x >> 1;
            count++;
        }
        return count;
    }
     
    // Driver Code
    public static void main (String[] args)
    {
         
        System.out.println(countTrailingZero(11));
    }
}
 
// This code is contributed by ajit

Python3




# Python 3 code for counting trailing zeros
# in binary representation of a number
def countTrailingZero(x):
    count = 0
    while ((x & 1) == 0):
        x = x >> 1
        count += 1
     
    return count
 
# Driver Code
if __name__ == '__main__':
    print(countTrailingZero(11))
     
# This code is contributed by
# Sanjit_Prasad

C#




// Simple C# code for counting
// trailing zeros in binary
// representation of a number
using System;
 
class GFG
{
    public static int countTrailingZero(int x)
    {
        int count = 0;
         
        while ((x & 1) == 0)
        {
            x = x >> 1;
            count++;
        }
        return count;
    }
     
    // Driver Code
    static public void Main ()
    {
        Console.WriteLine(countTrailingZero(11));
    }
}
 
// This code is contributed by aj_36

PHP




<?php
// Simple PHP code for counting trailing zeros
// in binary representation of a number
 
function countTrailingZero($x)
{
    $count = 0;
    while (($x & 1) == 0)
    {
        $x = $x >> 1;
        $count++;
    }
    return $count;
}
 
    // Driver Code
    echo countTrailingZero(11),"\n";
     
// This code is contributed by ajit
?>

Javascript




<script>
 
// Simple Javascript code for counting trailing zeros
// in binary representation of a number
 
 function countTrailingZero(x)
    {
        let count = 0;
         
        while ((x & 1) == 0)
        {
            x = x >> 1;
            count++;
        }
        return count;
    }
 
// Driver Code
 
    document.write(countTrailingZero(11));
 
</script>

Output : 

0

Time Complexity : O(Log n)
 
The lookup table solution is based on following concepts : 



  1. The solution assumes that negative numbers are stored in 2’s complement form which is true for most of the devices. If numbers are represented in 2’s complement form, then (x & -x) [Bitwise and of x and minus x] produces a number with only last set bit.
  2. Once we get a number with only one bit set, we can find its position using lookup table. It makes use of the fact that the first 32 bit position values are relatively prime with 37, so performing a modulus division with 37 gives a unique number from 0 to 36 for each. These numbers may then be mapped to the number of zeros using a small lookup table.

C++




// C++ code for counting trailing zeros
// in binary representation of a number
#include<bits/stdc++.h>
using namespace std;
 
int countTrailingZero(int x)
{
     // Map a bit value mod 37 to its position
     static const int lookup[] = {32, 0, 1,
     26, 2, 23, 27, 0, 3, 16, 24, 30, 28, 11,
     0, 13, 4, 7, 17, 0, 25, 22, 31, 15, 29,
     10, 12, 6, 0, 21, 14, 9, 5, 20, 8, 19,
     18};
 
     // Only difference between (x and -x) is
     // the value of signed magnitude(leftmostbit)
     // negative numbers signed bit is 1
     return lookup[(-x & x) % 37];
}
 
// Driver Code
int main()
{
    cout << countTrailingZero(48) << endl;
    return 0;
}

Java




// Java code for counting
// trailing zeros in binary
// representation of a number
import java.io.*;
 
class GFG
{
static int countTrailingZero(int x)
{
     
    // Map a bit value mod
    // 37 to its position
    int lookup[] = {32, 0, 1, 26, 2, 23
                    27, 0, 3, 16, 24, 30,
                    28, 11, 0, 13, 4, 7,
                    17, 0, 25, 22, 31, 15,
                    29, 10, 12, 6, 0, 21,
                    14, 9, 5, 20, 8, 19, 18};
 
    // Only difference between
    // (x and -x) is the value
    // of signed magnitude
    // (leftmostbit) negative
    // numbers signed bit is 1
    return lookup[(-x & x) % 37];
}
 
// Driver Code
public static void main (String[] args)
{
    System.out.println(countTrailingZero(48));
}
}
 
// This code is contributed
// by ajit

Python3




# Python3 code for counting trailing zeros
# in binary representation of a number
 
def countTrailingZero(x):
 
    # Map a bit value mod 37 to its position
    lookup = [32, 0, 1, 26, 2, 23, 27, 0,
              3, 16, 24, 30, 28, 11, 0, 13,
              4, 7, 17, 0, 25, 22, 31, 15,
              29, 10, 12, 6, 0, 21, 14, 9,
              5, 20, 8, 19, 18]
 
    # Only difference between (x and -x) is
    # the value of signed magnitude(leftmostbit)
    # negative numbers signed bit is 1
    return lookup[(-x & x) % 37]
 
# Driver Code
if __name__ == "__main__":
 
    print(countTrailingZero(48))
 
# This code is contributed
# by Rituraj Jain

C#




// C# code for counting
// trailing zeros in binary
// representation of a number
using System;
 
class GFG
{
static int countTrailingZero(int x)
{
     
    // Map a bit value mod
    // 37 to its position
    int []lookup = {32, 0, 1, 26, 2, 23,
                    27, 0, 3, 16, 24, 30,
                    28, 11, 0, 13, 4, 7,
                    17, 0, 25, 22, 31, 15,
                    29, 10, 12, 6, 0, 21,
                    14, 9, 5, 20, 8, 19, 18};
 
    // Only difference between
    // (x and -x) is the value
    // of signed magnitude
    // (leftmostbit) negative
    // numbers signed bit is 1
    return lookup[(-x & x) % 37];
}
 
// Driver Code
static public void Main ()
{
    Console.WriteLine(countTrailingZero(48));
}
}
 
// This code is contributed
// by m_kit

PHP




<?php
// PHP code for counting
// trailing zeros in binary
// representation of a number
 
function countTrailingZero($x)
{
    // Map a bit value mod
    // 37 to its position
    $lookup = array (32, 0, 1, 26, 2, 23,
                     27, 0, 3, 16, 24, 30,
                     28, 11, 0, 13, 4, 7,
                     17, 0, 25, 22, 31, 15,
                     29, 10, 12, 6, 0, 21,
                     14, 9, 5, 20, 8, 19, 18);
 
    // Only difference between
    // (x and -x) is the value
    // of signed magnitude
    // (leftmostbit) negative
    // numbers signed bit is 1
    return $lookup[(-$x &
                     $x) % 37];
}
 
// Driver Code
echo countTrailingZero(48), "\n";
     
// This code is contributed
// by akt_mit
?>

Javascript




<script>
 
// Javascript code for counting
// trailing zeros in binary
// representation of a number
function countTrailingZero(x)
{
     
    // Map a bit value mod
    // 37 to its position
    let lookup = [ 32, 0, 1, 26, 2, 23,
                   27, 0, 3, 16, 24, 30,
                   28, 11, 0, 13, 4, 7,
                   17, 0, 25, 22, 31, 15,
                   29, 10, 12, 6, 0, 21,
                   14, 9, 5, 20, 8, 19, 18 ];
 
    // Only difference between
    // (x and -x) is the value
    // of signed magnitude
    // (leftmostbit) negative
    // numbers signed bit is 1
    return lookup[(-x & x) % 37];
}
 
// Driver code
document.write(countTrailingZero(48));
 
// This code is contributed by divyesh072019
 
</script>

Output : 

4

Time Complexity : O(1)
Source : 
https://graphics.stanford.edu/~seander/bithacks.html

This article is contributed by Sumit Sudhakar(Sam). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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