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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0301_0400.s0335_self_crossing;

// #Hard #Array #Math #Geometry #2022_07_10_Time_5_ms_(79.38%)_Space_47.7_MB_(92.78%)

/**
 * 335 - Self Crossing\.
 *
 * Hard
 *
 * You are given an array of integers `distance`.
 *
 * You start at 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. In other words, after each move, your direction changes counter-clockwise.
 *
 * Return `true` if your path crosses itself, and `false` if it does not.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/03/14/selfcross1-plane.jpg)
 *
 * **Input:** distance = [2,1,1,2]
 *
 * **Output:** true 
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/03/14/selfcross2-plane.jpg)
 *
 * **Input:** distance = [1,2,3,4]
 *
 * **Output:** false 
 *
 * **Example 3:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/03/14/selfcross3-plane.jpg)
 *
 * **Input:** distance = [1,1,1,1]
 *
 * **Output:** true 
 *
 * **Constraints:**
 *
 * *   1 <= distance.length <= 10⁵
 * *   1 <= distance[i] <= 10⁵
**/
public class Solution {
    public boolean isSelfCrossing(int[] x) {
        if (x.length < 4) {
            return false;
        }
        for (int i = 3; i < x.length; i++) {
            if (x[i - 1] <= x[i - 3] && x[i] >= x[i - 2]) {
                return true;
            }
            if (i > 4
                    && x[i] >= x[i - 2] - x[i - 4]
                    && x[i - 1] >= x[i - 3] - x[i - 5]
                    && x[i - 1] <= x[i - 3]
                    && x[i - 2] >= x[i - 4]) {
                return true;
            }
            if (i > 3 && x[i - 1] == x[i - 3] && x[i] >= x[i - 2] - x[i - 4]) {
                return true;
            }
        }
        return false;
    }
}