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. have the 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++
// 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; } |
Java
// 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 |
Python3
# 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 |
C#
// 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 |
PHP
<?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 ?> |
Javascript
<script> // Javascript 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)) document.write( "<br>Equal" ); // if y has more leading zeros else if ((x & (~y)) > y) document.write(y); else document.write(x); } // Main Function let x = 10, y = 16; LeadingZeros(x, y); </script> |
10
Time Complexity: O(1)
Space Complexity: O(1)
Approach#2: Using format
Take two integer inputs from the user. Convert the input numbers to 8-bit binary strings using the format() function. Count the number of leading zeros in each binary string using the lstrip() and len() functions. Compare the number of leading zeros of the two binary strings. Print the number which has more leading zeros, or a message saying that both numbers have equal number of leading zeros.
Algorithm
1. Read the two integers num1 and num2 from the user.
2. Convert num1 and num2 to 8-bit binary strings using the format() function.
3. Count the number of leading zeros in bin1 and bin2 using the lstrip() and len() functions.
4. Compare leading_zeros1 and leading_zeros2.
5. If leading_zeros1 is greater than leading_zeros2, print num1 has more leading zeros.
6. If leading_zeros2 is greater than leading_zeros1, print num2 has more leading zeros.
7. If leading_zeros1 is equal to leading_zeros2, print both numbers have equal number of leading zeros.
Python3
num1 = 10 num2 = 12 bin1 = format (num1, '08b' ) # convert to 8-bit binary string bin2 = format (num2, '08b' ) leading_zeros1 = len (bin1) - len (bin1.lstrip( '0' )) leading_zeros2 = len (bin2) - len (bin2.lstrip( '0' )) if leading_zeros1 > leading_zeros2: print (num1) elif leading_zeros2 > leading_zeros1: print (num2) else : print ( "Equal" ) |
Equal
Time Complexity: O(1) because the program executes a fixed number of instructions regardless of the input values. The execution time of the program is constant and does not depend on the input size.
Space Complexity: O(1) because the program uses a fixed amount of memory to store the input variables, binary strings, and variables to count the leading zeros. The amount of memory used by the program does not depend on the input size.
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