LeetCode 335: Self Crossing Solution

Master LeetCode problem 335 (Self Crossing), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

335. Self Crossing

Hard

Array Math Geometry

Problem Explanation

Explanation:

To solve this problem, we need to keep track of the boundary lines that the path can cross. We define four cases where a self-crossing can happen based on the distances moved. By checking these cases, we can determine if a self-crossing occurs.

The current line crosses the line 3 steps ahead.
The current line crosses the line 4 steps ahead.
The current line crosses the line 5 steps ahead.
The current line crosses the line 6 steps ahead.

We iterate through the array of distances, keeping track of the current location and the previous locations to determine if a self-crossing occurs.

Time complexity: O(n) where n is the length of the input array distance.
Space complexity: O(1)

Solution Code

class Solution {
    public boolean isSelfCrossing(int[] distance) {
        int n = distance.length;
        if (n < 4) return false;
        
        for (int i = 3; i < n; i++) {
            if (distance[i] >= distance[i - 2] && distance[i - 1] <= distance[i - 3]) return true;
            if (i >= 4 && distance[i - 1] == distance[i - 3] && distance[i] + distance[i - 4] >= distance[i - 2]) return true;
            if (i >= 5 && distance[i - 2] >= distance[i - 4] && distance[i] + distance[i - 4] >= distance[i - 2] && distance[i - 1] <= distance[i - 3] && distance[i - 1] + distance[i - 5] >= distance[i - 3]) return true;
        }
        return false;
    }
}

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