Binary Number with Alternating Bits - Bit Manipulation - Easy - LeetCode

1 min read 153 words

sumitc916 October 2020Updated: 6 October 2020 1 min read

ShareWhatsApp Post on X

1A positive integer has alternating bits if no two adjacent bits in its binary representation are the same.
2The function checks each bit of the integer to determine if it alternates, returning true or false accordingly.
3The algorithm operates with a time complexity of O(log n) and a space complexity of O(1).

AI-generated summary · May not capture all nuances

Key Insight

AskGif

"A positive integer has alternating bits if no two adjacent bits in its binary representation are the same."

— Binary Number with Alternating Bits - Bit Manipulation - Easy - LeetCode

Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values.

Example 1:

Input: n = 5 Output: true Explanation: The binary representation of 5 is: 101 Example 2:

Input: n = 7 Output: false Explanation: The binary representation of 7 is: 111. Example 3:

Input: n = 11 Output: false Explanation: The binary representation of 11 is: 1011. Example 4:

Input: n = 2 Output: true Explanation: The binary representation of 10 is: 1010. Example 5:

Input: n = 3 Output: false

Constraints:

1 <= n <= 231 - 1

public class Solution {
 public bool HasAlternatingBits(int n) {
 int? prev = null;
 for(int i=0;i<32 && n!=0;i++){
 if(prev==null){
 prev = (n&1);
 }
 else{
 if(prev == (n&1)){
 return false;
 }
 prev = (n&1);
 }
 
 n=n>>1;
 }
 
 return true;
 }
}

Time Complexity: O(logn) which corresponds to the count of binary bits.

Space Complexity: O(1)

🛒 You'll Need This

Products related to this article