Count trailing zero bits using lookup table - GeeksforGeeks

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++
Java
Python3
C#
PHP
Javascript
// 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 Codeint main(){    cout << countTrailingZero(11) << endl;    return 0;}
// Simple Java code for counting// trailing zeros in binary// representation of a numberimport 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
# Python 3 code for counting trailing zeros# in binary representation of a numberdef countTrailingZero(x):    count = 0    while ((x & 1) == 0):        x = x >> 1        count += 1         return count # Driver Codeif __name__ == '__main__':    print(countTrailingZero(11))     # This code is contributed by# Sanjit_Prasad
// Simple C# code for counting// trailing zeros in binary// representation of a numberusing 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// 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?>
<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 : 
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.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.
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.
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++
Java
Python3
C#
PHP
Javascript
// 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 Codeint main(){    cout << countTrailingZero(48) << endl;    return 0;}
// Java code for counting// trailing zeros in binary// representation of a numberimport 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 Codepublic static void main (String[] args){    System.out.println(countTrailingZero(48));}} // This code is contributed// by ajit
# 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 Codeif __name__ == "__main__":     print(countTrailingZero(48)) # This code is contributed# by Rituraj Jain
// C# code for counting// trailing zeros in binary// representation of a numberusing 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 Codestatic public void Main (){    Console.WriteLine(countTrailingZero(48));}} // This code is contributed// by m_kit
<?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 Codeecho countTrailingZero(48), "\n";     // This code is contributed// by akt_mit?>
<script> // Javascript code for counting// trailing zeros in binary// representation of a numberfunction 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 codedocument.write(countTrailingZero(48)); // This code is contributed by divyesh072019 </script>
Output : 
4
Time Complexity : O(1)Source : https://graphics.stanford.edu/~seander/bithacks.html
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