Given an integer n, output the no. of leading zeros in its binary form.

A leading zero is any 0 digit that comes before the first nonzero digit in a number’s binary form. **Examples:**

Input : 16 Output :27 As Binary(16) = (00000000000000000000000000010000) Input :33 Output :26 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 = 101;` ` ` `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

**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

**Solution 3: Using the GCC __builtin_clz(x):** This function is used to count the leading zeros of the integer where * clz stands for count leading zero’s*. It counts a number of zeros before the first occurrence of one(set bit).

## C

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

**Output:** Count of leading zeros before the first occurrence: 27

**Time Complexity:** The time complexity of this approach is O(1) **Space Complexity:** The space complexity of this approach is O(1)

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