Self Crossing

You are given an array x of n positive numbers. You start at point (0,0) and moves x[0] metres to the north, then x[1] metres to the west, x[2] metres to the south, x[3] metres to the east and so on. In other words, after each move your direction changes counter-clockwise.

Write a one-pass algorithm with O(1) extra space to determine, if your path crosses itself, or not.

Example 1:
Given x = [2, 1, 1, 2]
Return true (self crossing)
Example 2:
Given x = [1, 2, 3, 4]
Return false (not self crossing)
Example 3:
Given x = [1, 1, 1, 1]
Return true (self crossing)

Solution. The solution is to solve the below 3 scenarios.

Accordingly, the code is like below:

public static boolean isSelfCrossing(int[] x) {
    if (x == null || x.length <= 3) {
        return false;
    }
    for (int i = 3; i < x.length; i++) {
        if (x[i - 1] <= x[i - 3] && x[i] >= x[i - 2]) {
            return true;
        }
        else if (i >= 4 && x[i - 1] == x[i - 3] && x[i] + x[i - 4] >= x[i - 2]) {
            return true;
        }
        else if (i >= 5 && x[i - 1] <= x[i - 3] && x[i - 2] >= x[i - 4]
                && x[i] + x[i - 4] >= x[i - 2] && x[i - 5] + x[i - 1] >= x[i - 3]) {
            return true;
        }
    }
    return false;
}

The solution is found here.

Check my code on github: link