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 theree trailing zero bits. Input : 18 Output : 1 Binary of 10 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.

// 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; }

Output :

4

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.

// 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; }

Output :

4

Time Complexity = O(1)

**Source : **

https://graphics.stanford.edu/~seander/bithacks.html

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