Comparing leading zeros in binary representations of two numbers

Given two Integer numbers x and y. Compare and print which one of them has more leading zeros using Bitwise operation. If both the no. has same no. of leading zeros, print “Equal”.

Note:- A leading zero is any 0 digit that comes before the first nonzero digit in the binary notation of the number.

Examples:



Input : 10, 16
Output :10
Explanation: If we represent the no.s using 8 bit only then 
Binary(10) = 00001010
Binary(16) = 00010000
Clearly, 10 has 4 leading zeros and 16 has 3 leading zeros

Input : 10, 12
Output : Equal
Binary(10) = 00001010
Binary(12) = 00001100
Both have equal no. of leading zeros.

Solution 1 : The Naive approach is to first find the binary representation of the numbers and then count the no. of leading zeros.

Solution 2 : Find largest power of twos smaller than given numbers, and compare these powers of twos to decide answer.

Solution 3: An efficient approach is to bitwise XOR and AND operators.

Case 1: If both have same no. of leading zeros then (x^y) <= (x & y) because same number of leading 0s would cause a 1 at higher position in x & y.

Case 2 : If we do negation of y and do bitwise AND with x, we get a one at higher position than in y when y has more number of leading 0s.

Case 3: Else x has more leading zeros

C++

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// CPP program to find the number with more
// leading zeroes.
#include <bits/stdc++.h>
using namespace std;
  
// Function to compare the no. of leading zeros
void LeadingZeros(int x, int y)
{
    // if both have same no. of leading zeros
    if ((x ^ y) <= (x & y)) 
        cout << "\nEqual";
  
    // if y has more leading zeros
    else if ((x & (~y)) > y) 
        cout << y;
  
    else 
        cout << x;
}
  
// Main Function
int main()
{
    int x = 10, y = 16;
    LeadingZeros(x, y);
    return 0;
}

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Java

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// Java program to find the number 
// with more leading zeroes. 
class GFG
{
// Function to compare the no. 
// of leading zeros 
static void LeadingZeros(int x, int y) 
    // if both have same no. of 
    // leading zeros 
    if ((x ^ y) <= (x & y)) 
        System.out.print("\nEqual");
  
    // if y has more leading zeros 
    else if ((x & (~y)) > y) 
        System.out.print(y); 
  
    else
        System.out.print(x); 
  
// Driver Code
public static void main (String[] args) 
{
    int x = 10, y = 16
    LeadingZeros(x, y); 
}
  
// This code is contributed by Smitha

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Python3

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# Python 3 program to find the number 
# with more leading zeroes.
  
# Function to compare the no. of
# leading zeros
def LeadingZeros(x, y):
      
    # if both have same no. of
    # leading zeros
    if ((x ^ y) <= (x & y)): 
        print("Equal")
  
    # if y has more leading zeros
    elif ((x & (~y)) > y) :
        print(y)
  
    else:
        print(x)
  
# Driver Code
if __name__ == '__main__':
    x = 10
    y = 16
    LeadingZeros(x, y)
  
# This code is contributed 
# by Surendra_Gangwar

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C#

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// C# program to find the number 
// with more leading zeroes. 
using System;
  
class GFG
{
// Function to compare the no. 
// of leading zeros 
static void LeadingZeros(int x, int y) 
    // if both have same no. of 
    // leading zeros 
    if ((x ^ y) <= (x & y)) 
        Console.WriteLine("\nEqual"); 
  
    // if y has more leading zeros 
    else if ((x & (~y)) > y) 
        Console.WriteLine(y); 
  
    else
        Console.WriteLine(x); 
  
// Driver Code 
static public void Main ()
{
    int x = 10, y = 16; 
    LeadingZeros(x, y); 
}
  
// This code is contributed by ajit

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PHP

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<?php
// PHP program to find the number
// with more leading zeroes.
  
// Function to compare the no.
// of leading zeros
function LeadingZeros($x, $y)
{
    // if both have same no. of
    // leading zeros
    if (($x ^ $y) <= ($x & $y)) 
        echo "\nEqual";
  
    // if y has more leading zeros
    else if (($x & (~$y)) > $y
        echo $y;
  
    else
        echo $x;
}
  
// Driver Code
$x = 10;
$y = 16;
LeadingZeros($x, $y);
      
// This code is contributed by ajit
?>

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Output:

10

Time Complexity: O(1)
Space Complexity: O(1)



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