# 18.02.2026 [693. Binary Number with Alternating Bits]

Alternating bits #easy #bits

18.02.2026

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Problem TLDR

Alternating bits #easy #bits

Intuition

Check 31 bits with brute force.

The “clever solutions”:

• shift right by 2 positions, should match; shift right by 1 positions, should be opposite

• shift right by 1 position, do or: should be all ones

Approach

• regex: (00|11)

Complexity

Time complexity: $$O(1)$$

Space complexity: $$O(1)$$

Code

// 0ms
    fun hasAlternatingBits(n: Int) = 
    n or n/4 == n && n and n/2 == 0
    /*
    Regex("(11|00)") !in n.toString(2)
    (n xor n/2).toString(2).all { it == '1' }
     */

// 0ms
    pub fn has_alternating_bits(mut n: i32) -> bool {
        n ^= n/2; n & n+1 == 0
    }