Number of leading zeros in binary representation of a given number

Given a positive integer N, the task is to find the number of leading zeros in its binary representation.
A leading zero is any 0 digit that comes before the first nonzero digit in a number’s binary form.

Examples:

Input : N = 16
Output : 27
Explanation: As Binary(16) = (00000000000000000000000000010000)

Input : N = 33
Output : 26
Explanation: As Binary(16)=(00000000000000000000000000100001)

Solution 1: A naive approach is to convert the no. into its binary form and then count the no. of leading zeros. It uses expensive divide operations. 

C++

// C++ program of number of leading zeros in
// binary representation of a given number
#include <bits/stdc++.h>
using namespace std;

// Function to count the no. of leading zeros
int countZeros(unsigned int x)
{

    // Keep shifting x by one until leftmost bit
    // does not become 1.
    int total_bits = sizeof(x) * 8;

    int res = 0;
    while (!(x & (1 << (total_bits - 1)))) {
        x = (x << 1);
        res++;
    }

    return res;
}

// Main function
int main()
{
    int x = 0;
    cout << countZeros(x);
    return 0;
}

Java

// Java program of number of leading zeros in
// binary representation of a given number
class GFG {
    static byte sizeofInt = 8;

    // Function to count the no. of leading zeros
    static int countZeros(int x)
    {
        // Keep shifting x by one until leftmost bit
        // does not become 1.
        int total_bits = sizeofInt * 8;
        int res = 0;
        while ((x & (1 << (total_bits - 1))) == 0) {
            x = (x << 1);
            res++;
        }

        return res;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int x = 101;
        System.out.println(countZeros(x));
    }
}

// This code is contributed by PrinciRaj1992

Python3

# Python3 program of number of
# leading zeros in binary
# representation of a given number

# Function to count the
# no. of leading zeros
def countZeros(x):

    # Keep shifting x by one until
    # leftmost bit does not become 1.
    total_bits = 32
    res = 0
    while ((x & (1 << (total_bits - 1))) == 0):
        x = (x << 1)
        res += 1

    return res

# Driver Code
x = 101
print(countZeros(x))

# This code is contributed
# by Mohit Kumar

C#

// C# program of number of leading zeros in
// binary representation of a given number
using System;

class GFG {
    static byte sizeofInt = 8;

    // Function to count the
    // no. of leading zeros
    static int countZeros(int x)
    {
        // Keep shifting x by one until
        // leftmost bit does not become 1.
        int total_bits = sizeofInt * 8;
        int res = 0;
        while ((x & (1 << (total_bits - 1))) == 0) {
            x = (x << 1);
            res++;
        }
        return res;
    }

    // Driver Code
    public static void Main(String[] args)
    {
        int x = 101;
        Console.WriteLine(countZeros(x));
    }
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// JavaScript program of number of leading zeros in
// binary representation of a given number

let sizeofInt = 8;

// Function to count the no. of leading zeros
function countZeros(x)
{
    // Keep shifting x by one until leftmost bit
    // does not become 1.
    let total_bits = sizeofInt * 8;
    let res = 0;
    while ((x & (1 << (total_bits - 1))) == 0)
    {
        x = (x << 1);
        res++;
    }

    return res;
}

// Driver Code
let x = 101;
document.write(countZeros(x));


// This code is contributed by unknown2108

</script>

Output

25

Time complexity: O(1)
Auxiliary space: O(1)

Solution 2: An efficient approach is to use Bitwise right shift operation to achieve the same. The steps in the algorithm are: 
Let x be our no. then

    unsigned y;
    int n = 32;
    y = x >>16; if (y != 0) {n = n -16; x = y;}
    y = x >> 8; if (y != 0) {n = n - 8; x = y;}
    y = x >> 4; if (y != 0) {n = n - 4; x = y;}
    y = x >> 2; if (y != 0) {n = n - 2; x = y;}
    y = x >> 1; if (y != 0) return n - 2;
    return n - x;

The above approach executes in only 12 to 20 instructions.

C++

// C++ program of number of leading zeros in
// binary representation of a given number
#include <bits/stdc++.h>
using namespace std;

// Function to count the no. of leading zeros
int countZeros(int x)
{
    unsigned y;
    int n = 32;
    y = x >> 16;
    if (y != 0) {
        n = n - 16;
        x = y;
    }
    y = x >> 8;
    if (y != 0) {
        n = n - 8;
        x = y;
    }
    y = x >> 4;
    if (y != 0) {
        n = n - 4;
        x = y;
    }
    y = x >> 2;
    if (y != 0) {
        n = n - 2;
        x = y;
    }
    y = x >> 1;
    if (y != 0)
        return n - 2;
    return n - x;
}

// Main function
int main()
{
    int x = 101;
    cout << countZeros(x);
    return 0;
}

Java

// Java program of number of leading zeros in
// binary representation of a given number
import java.io.*;

class GFG {
    // Function to count the no. of leading zeros
static int countZeros(int x)
{
    int y;
    int n = 32;
    y = x >> 16;
    if (y != 0) {
        n = n - 16;
        x = y;
    }
    y = x >> 8;
    if (y != 0) {
        n = n - 8;
        x = y;
    }
    y = x >> 4;
    if (y != 0) {
        n = n - 4;
        x = y;
    }
    y = x >> 2;
    if (y != 0) {
        n = n - 2;
        x = y;
    }
    y = x >> 1;
    if (y != 0)
        return n - 2;
    return n - x;
}

// Main function
    public static void main (String[] args) {
    int x = 101;
    System.out.println (countZeros(x));
    }
//This code is contributed by @Tushil.
}

Python3

# Python3 program of number of leading zeros in
# binary representation of a given number


# Function to count the no. of leading zeros
def countZeros(x):
    n = 32;
    y = x >> 16;
    if (y != 0):
        n = n - 16;
        x = y;

    y = x >> 8;
    if (y != 0):
        n = n - 8;
        x = y;

    y = x >> 4;
    if (y != 0):
        n = n - 4;
        x = y;

    y = x >> 2;
    if (y != 0):
        n = n - 2;
        x = y;

    y = x >> 1;
    if (y != 0):
        return n - 2;
    return n - x;


# Main function
def main():
    x = 101;
    print(countZeros(x))


if __name__ == '__main__':
    main()

C#

// C# program of number of leading zeros in
// binary representation of a given number
using System;

class GFG
{
// Function to count the no. of
// leading zeros
static int countZeros(int x)
{
    int y;
    int n = 32;
    y = x >> 16;

    if (y != 0)
    {
        n = n - 16;
        x = y;
    }
    y = x >> 8;

    if (y != 0)
    {
        n = n - 8;
        x = y;
    }
    y = x >> 4;

    if (y != 0)
    {
        n = n - 4;
        x = y;
    }
    y = x >> 2;

    if (y != 0)
    {
        n = n - 2;
        x = y;
    }
    y = x >> 1;

    if (y != 0)
        return n - 2;
    return n - x;
}

// Driver Code
static public void Main ()
{
    int x = 101;
    Console.WriteLine(countZeros(x));
}
}

// This code is contributed by ajit

PHP

<?php
// PHP program of number of leading zeros in
// binary representation of a given number

// Function to count the no. of leading zeros
function countZeros($x)
{
    $y;
    $n = 32;
    $y = $x >> 16;
    if ($y != 0)
    {
        $n = $n - 16;
        $x = $y;
    }
    $y = $x >> 8;
    if ($y != 0)
    {
        $n = $n - 8;
        $x = $y;
    }
    $y = $x >> 4;
    if ($y != 0)
    {
        $n = $n - 4;
        $x = $y;
    }
    $y = $x >> 2;
    if ($y != 0) {
        $n = $n - 2;
        $x = $y;
    }
    $y = $x >> 1;
    if ($y != 0)
        return $n - 2;
    return $n - $x;
}

// Driver Code
$x = 101;
echo countZeros($x);

// This code is contributed
// by Akanksha Rai

Javascript

<script>

// JavaScript program of number of leading zeros in
// binary representation of a given number

 // Function to count the no. of leading zeros
function countZeros(x)
{
    let y;
    let n = 32;
    y = x >> 16;
    if (y != 0) {
        n = n - 16;
        x = y;
    }
    y = x >> 8;
    if (y != 0) {
        n = n - 8;
        x = y;
    }
    y = x >> 4;
    if (y != 0) {
        n = n - 4;
        x = y;
    }
    y = x >> 2;
    if (y != 0) {
        n = n - 2;
        x = y;
    }
    y = x >> 1;
    if (y != 0)
        return n - 2;
    return n - x;
}

// Main function
let x = 101;
document.write(countZeros(x));


// This code is contributed by patel2127

</script>

Output

25

Time complexity: O(1)
Auxiliary space: O(1)

Solution 3: Using predefined functions

In Java, the numberOfLeadingZeros() method of the Integer and Long class is used to count the leading zeros of the integer.

C

#include <stdio.h>
int main()
{
    int n = 19; // 00000000 00000000 00000000 010011
    printf("Count of leading zeros before first "
           "occurrence: %d",
           __builtin_clz(n));
    return 0;
}

C++

#include <iostream>
using namespace std;

int main()
{
    int n = 19; // 00000000 00000000 00000000 010011
    cout << "Count of leading zeros before first "
            "occurrence: "
         << __builtin_clz(n) << endl;
    return 0;
}

Java

// Java Program that counts the number of leading zeroes

class GFG {
    public static void main(String[] args)
    {
        int n = 19; // 00000000 00000000 00000000 010011
        System.out.println(
            "Count of leading zeros before first occurrence: "
            + Integer.numberOfLeadingZeros(n));
    }
}

Python3

# Python program to count the number of leading zeroes

n = 19  # 00000000 00000000 00000000 010011
print("Count of leading zeros before first occurrence: ", bin(n)[2:].zfill(32).index('1'))


# This code is contributed by rishabmilhdijo

C#

// C# program to count the number of leading zeros

using System;

public class GFG {
    public static void Main(string[] args)
    {

        int n = 19; // 00000000 00000000 00000000 010011
        string binary = Convert.ToString(n, 2).PadLeft(
            32, '0'); // PadLeft method adds 0s to the start
                      // of the string until it reaches 32
                      // characters
        Console.WriteLine(
            "Count of leading zeros before first occurrence: "
            + binary.IndexOf(
                '1')); // returns the index of the first
                       // occurrence of 1 which is the count
                       // of leading zeros
    }
}

Javascript

// JavaScript program to count the number of leading zeros
let n = 19; // 00000000 00000000 00000000 010011
let binary = n.toString(2).padStart(32, '0'); // padStart method adds 0s to the start of the string until it reaches 32 characters
console.log("Count of leading zeros before first occurrence: ",binary.indexOf('1')); // returns the index of the first occurence of 1 which is the count of leading zeros

Output

Count of leading zeros before first occurrence: 27

Time complexity: O(1)
Auxiliary space: O(1)