Binary Number with Alternating Bits

Binary Number with Alternating Bits - Problem

Bit Manipulation Easy

Given a positive integer, determine if it has alternating bits. A number has alternating bits if no two adjacent bits in its binary representation have the same value.

For example:

5 has binary representation 101 → alternating bits ✓
7 has binary representation 111 → adjacent 1's ✗
10 has binary representation 1010 → alternating bits ✓

Return true if the number has alternating bits, false otherwise.

Input & Output

Example 1 — Alternating Pattern

$ Input: n = 5

› Output: true

💡 Note: 5 in binary is 101. Adjacent bits are different: 1→0→1, so return true

Example 2 — Non-Alternating Pattern

$ Input: n = 7

› Output: false

💡 Note: 7 in binary is 111. Adjacent bits 1 and 1 are the same, so return false

Example 3 — Longer Alternating Pattern

$ Input: n = 10

› Output: true

💡 Note: 10 in binary is 1010. All adjacent bits alternate: 1→0→1→0, so return true

Constraints

1 ≤ n ≤ 2³¹ - 1

Visualization

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Asked in

G Google 15 f Facebook 12 M Microsoft 8

The key insight is using XOR to detect alternating patterns. When we XOR a number with itself shifted right by 1, alternating bits produce all 1's. Best approach uses bit manipulation with XOR operation. Time: O(1), Space: O(1)

Common Approaches

✓ Bit Manipulation - XOR Pattern

⏱️ Time: O(1) Space: O(1)

XOR the number with itself shifted right by 1. If alternating, this creates all 1's. Then check if result is a power of 2 minus 1.

Brute Force - Check Adjacent Bits

⏱️ Time: O(log n) Space: O(1)

Convert number to binary representation and compare each bit with the next bit. If any two adjacent bits are the same, return false.

Bit Manipulation - XOR Pattern — Algorithm Steps

Step 1: XOR number with itself shifted right by 1 (n ^ (n >> 1))
Step 2: Result should be all 1's if alternating (like 111, 1111)
Step 3: Check if result is power of 2 minus 1 using (x & (x + 1)) == 0
Step 4: Return true if condition met

Visualization

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Step-by-Step Walkthrough

XOR Operation

XOR n with n shifted right by 1

Pattern Check

Result should be all 1's if alternating

Validation

Check if result is power of 2 minus 1

Code -

solution.c — C

#include <stdio.h>
#include <stdbool.h>

bool solution(int n) {
    // XOR n with n shifted right by 1
    int xorResult = n ^ (n >> 1);
    
    // Check if result is all 1's (power of 2 minus 1)
    // For all 1's: (x & (x + 1)) == 0
    return (xorResult & (xorResult + 1)) == 0;
}

int main() {
    int n;
    scanf("%d", &n);
    bool result = solution(n);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Only performs a constant number of bit operations regardless of input size

⚡ Linearithmic

Space Complexity

O(1)

Uses only a single variable to store intermediate result

✓ Linear Space

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Medium Frequency

~10 min Avg. Time

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Bit Manipulation - XOR Pattern — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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