Leetcode #868: Binary Gap
In this guide, we solve Leetcode #868 Binary Gap in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a positive integer n, find and return the longest distance between any two adjacent 1's in the binary representation of n. If there are no two adjacent 1's, return 0.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Bit Manipulation
Intuition
The problem structure lets us track state with bitwise operations.
Bit operations are constant time and avoid extra memory.
Approach
Apply XOR/AND/OR and shifts to maintain the required invariant.
Aggregate the result in a single pass.
Steps:
- Identify a bitwise invariant.
- Combine values with bit operations.
- Return the aggregated result.
Example
Input: n = 22
Output: 2
Explanation: 22 in binary is "10110".
The first adjacent pair of 1's is "10110" with a distance of 2.
The second adjacent pair of 1's is "10110" with a distance of 1.
The answer is the largest of these two distances, which is 2.
Note that "10110" is not a valid pair since there is a 1 separating the two 1's underlined.
Python Solution
class Solution:
def binaryGap(self, n: int) -> int:
ans = 0
pre, cur = inf, 0
while n:
if n & 1:
ans = max(ans, cur - pre)
pre = cur
cur += 1
n >>= 1
return ans
Complexity
The time complexity is , where is the given integer. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.