EasyNo. 0868
Binary Gap — LeetCode 868 Python Solution
- Pattern
- Bit Manipulation
- Topics
- Bit Manipulation
- Reading time
- 2 min
- Source
- leetcode.com
The problem
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.
Example
- Input
- n = 22
- Output
- 2
- Explanation
- 22 in binary is "10110".
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 ansComplexity
- Time
- O(log n)where n is the given integer
- Space · auxiliary
- O(1)
Pattern: Bit Manipulation
Use XOR, masks and the low-bit trick to replace whole data structures with an integer. LeetCode 868. Binary Gap is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Bit Manipulation.
The bit manipulation guide has the Python template for the pattern and the 194 LeetCode problems that use it.
Related problems
| No. | Problem | Difficulty |
|---|---|---|
| 0029 | Divide Two Integers | Medium |
| 0067 | Add Binary | Easy |
| 0136 | Single Number | Easy |
| 0137 | Single Number II | Medium |
| 0190 | Reverse Bits | Easy |
| 0191 | Number of 1 Bits | Easy |
Frequently asked questions
- How hard is LeetCode 868. Binary Gap?
LeetCode 868. Binary Gap is rated Easy on LeetCode.
- What is the time complexity of LeetCode 868. Binary Gap?
The Python solution on this page runs in O(log n), where n is the given integer.
- What is the space complexity of LeetCode 868. Binary Gap?
The Python solution on this page uses O(1) auxiliary space.
- What topics does LeetCode 868. Binary Gap cover?
LeetCode 868. Binary Gap is tagged Bit Manipulation on LeetCode.