Given a number you have to check whether there is pair of adjacent set bit or not.
Examples :
Input : N = 67 Output : Yes There is a pair of adjacent set bit The binary representation is 100011 Input : N = 5 Output : No
A simple solution is to traverse all bits. For every set bit, check if next bit is also set.
An efficient solution is to shift number by 1 and then do bitwise AND. If bitwise AND is non-zero then there are two adjacent set bits. Else not.
C++
// CPP program to check // if there are two // adjacent set bits. #include <iostream> using namespace std; bool adjacentSet(int n) { return (n & (n >> 1)); } // Driver Code int main() { int n = 3; adjacentSet(n) ? cout << "Yes" : cout << "No"; return 0; } |
Java
// Java program to check // if there are two // adjacent set bits. class GFG { static boolean adjacentSet(int n) { int x = (n & (n >> 1)); if(x > 0) return true; else return false; } // Driver code public static void main(String args[]) { int n = 3; if(adjacentSet(n)) System.out.println("Yes"); else System.out.println("No"); } } // This code is contributed by Sam007. |
Python3
# Python 3 program to check if there # are two adjacent set bits. def adjacentSet(n): return (n & (n >> 1)) # Driver Code if __name__ == '__main__': n = 3 if (adjacentSet(n)): print("Yes") else: print("No") # This code is contributed by # Shashank_Sharma |
C#
// C# program to check // if there are two // adjacent set bits. using System; class GFG { static bool adjacentSet(int n) { int x = (n & (n >> 1)); if(x > 0) return true; else return false; } // Driver code public static void Main () { int n = 3; if(adjacentSet(n)) Console.WriteLine("Yes"); else Console.WriteLine("No"); } } // This code is contributed by Sam007. |
php
<?php // PHP program to check // if there are two // adjacent set bits. function adjacentSet($n) { return ($n & ($n >> 1)); } // Driver Code $n = 3; adjacentSet($n) ? print("Yes") : print("No"); // This code is contributed by Sam007. ?> |
Javascript
<script> // Javascript program to check // if there are two // adjacent set bits. function adjacentSet(n) { let x = (n & (n >> 1)); if(x > 0) return true; else return false; } // driver program let n = 3; if(adjacentSet(n)) document.write("Yes"); else document.write("No"); </script> |
Output :
Yes
Time Complexity : O(1)
Auxiliary Space : O(1)
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