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++
// 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 ?> |
10
Time Complexity: O(1)
Space Complexity: O(1)
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