How do you solve Count Binary Palindromic Numbers on LeetCode?

Count Binary Palindromic Numbers (LeetCode #3677) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n * 2^(log n / 2)) time complexity and O(1) space complexity. Key insight: Binary representation length is crucial.

What is the time complexity of Count Binary Palindromic Numbers?

The optimal solution for Count Binary Palindromic Numbers runs in O(log n * 2^(log n / 2)) time and uses O(1) space. The outer loop runs log n times, and the inner loop generates palindromes based on half-length.

What is the brute force solution for Count Binary Palindromic Numbers?

The brute force approach for Count Binary Palindromic Numbers is Brute Force, with O(n log n) time complexity and O(1) space complexity. Check each number from 0 to n and verify if its binary representation is a palindrome. This is straightforward but inefficient for large n. The optimal Optimal Solution approach improves this to O(log n * 2^(log n / 2)).

What algorithm or data structure is used to solve Count Binary Palindromic Numbers?

Count Binary Palindromic Numbers uses the following concepts: Math, Bit Manipulation. The recommended patterns to study are: Bit Manipulation, Dynamic Programming.

What are the interview tips for Count Binary Palindromic Numbers?

Explain your thought process clearly. Use examples to illustrate your approach. Practice coding under time constraints.

What are common mistakes when solving Count Binary Palindromic Numbers?

Forgetting to check leading zeros. Not considering all lengths of binary numbers.

#3677

Count Binary Palindromic Numbers

Hard

MathBit ManipulationBit ManipulationDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n log n)	O(log n * 2^(log n / 2))
Space	O(1)	O(1)

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Intuition

Time O(log n * 2^(log n / 2))Space O(1)

Instead of checking each number, count palindromic binaries based on their lengths. The first half determines the whole binary palindrome.

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Algorithm

3 steps

1Step 1: Determine the maximum length of binary representation for n.
2Step 2: For each length, generate palindromic numbers by constructing the first half and mirroring it.
3Step 3: Count valid palindromic numbers that are <= n.

solution.py7 lines

1def countBinaryPalindromes(n):
2    count = 0
3    for length in range(1, n.bit_length() + 1):
4        for half in range(1 << (length // 2)):
5            pal = (half << (length // 2)) | int(bin(half)[:length // 2][::-1], 2)
6            if pal <= n: count += 1
7    return count

ℹ

Complexity note: The outer loop runs log n times, and the inner loop generates palindromes based on half-length.

1Binary representation length is crucial.
2First half of the binary number defines the palindrome.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.