How do you solve Self Crossing on LeetCode?

Self Crossing (LeetCode #335) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The path crosses itself if any of the last four segments intersect with previous segments.

What is the brute force solution for Self Crossing?

The brute force approach for Self Crossing is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves simulating the entire path and checking for intersections at every step. By storing all visited points, we can determine if the current position has been visited before. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Self Crossing?

Self Crossing uses the following concepts: Array, Math, Geometry. The recommended patterns to study are: Geometric properties of movements, Array manipulation.

What are the interview tips for Self Crossing?

Start with a brute-force solution to demonstrate your understanding of the problem. Clearly explain your thought process and how you derived conditions for crossing. Practice visualizing the movements on a grid to better understand the geometric relationships.

What are common mistakes when solving Self Crossing?

Not considering edge cases where the path length is less than 4. Over-complicating the problem by trying to track all visited points instead of focusing on the last few movements.

#335

Self Crossing

Hard

ArrayMathGeometryGeometric properties of movementsArray manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages geometric properties to determine if the path crosses itself without explicitly tracking all visited points. By analyzing the movement pattern, we can identify crossing conditions based on the last few movements.

⚙️

Algorithm

3 steps

1Step 1: Check if the path crosses based on the last four movements.
2Step 2: For each distance, update the current position and check if it crosses the previous segments.
3Step 3: Return true if any crossing condition is met, otherwise return false.

solution.py14 lines

1# Full working Python code
2
3def isSelfCrossing(distance):
4    n = len(distance)
5    if n < 4:
6        return False
7    for i in range(3, n):
8        if distance[i] >= distance[i - 2] and distance[i - 1] <= distance[i - 3]:
9            return True
10        if i >= 4 and distance[i - 1] == distance[i - 3] and distance[i] + distance[i - 4] >= distance[i - 2]:
11            return True
12        if i >= 5 and distance[i - 2] >= distance[i - 4] and distance[i] + distance[i - 4] >= distance[i - 2] and distance[i - 1] + distance[i - 3] >= distance[i - 1]:
13            return True
14    return False

ℹ

Complexity note: The time complexity is O(n) because we only traverse the distance array once, checking conditions based on the last few movements, leading to linear growth in time complexity.

1The path crosses itself if any of the last four segments intersect with previous segments.
2Understanding the geometric properties of the movements helps in deriving conditions for crossings.

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