Leetcode #335: Self Crossing
In this guide, we solve Leetcode #335 Self Crossing 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
You are given an array of integers distance. You start at the point (0, 0) on an X-Y plane, and you move distance[0] meters to the north, then distance[1] meters to the west, distance[2] meters to the south, distance[3] meters to the east, and so on.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Geometry, Array, Math
Intuition
There is a mathematical invariant or formula that directly leads to the result.
Using math avoids unnecessary loops and reduces complexity.
Approach
Derive the formula or update rule, then compute the answer directly.
Handle edge cases like overflow or zero carefully.
Steps:
- Identify the math relationship.
- Compute the result with a loop or formula.
- Handle edge cases.
Example
Input: distance = [2,1,1,2]
Output: true
Explanation: The path crosses itself at the point (0, 1).
Python Solution
class Solution:
def isSelfCrossing(self, distance: List[int]) -> bool:
d = distance
for i in range(3, len(d)):
if d[i] >= d[i - 2] and d[i - 1] <= d[i - 3]:
return True
if i >= 4 and d[i - 1] == d[i - 3] and d[i] + d[i - 4] >= d[i - 2]:
return True
if (
i >= 5
and d[i - 2] >= d[i - 4]
and d[i - 1] <= d[i - 3]
and d[i] >= d[i - 2] - d[i - 4]
and d[i - 1] + d[i - 5] >= d[i - 3]
):
return True
return False
Complexity
The time complexity is O(n) or O(1). The space complexity is O(1).
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.