335-self-crossing

Question

https://leetcode.com/problems/self-crossing/description/

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:

Given x = [2, 1, 1, 2],
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?   ?
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    ?

Return true (self crossing)

Given x = [1, 2, 3, 4],
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?      ?
?
?
?????????????>

Return false (not self crossing)

Given x = [1, 1, 1, 1],
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Return true (self crossing)

Thought Process

1. Spiral Directions
   1. Spiraling outward: If the parallel lines' keep increasing, there will not be any cross, x[i] > x[i - 2]
   2. Spiraling inward: once the direction switch, we need to make sure the distance keep decreasing
   3. Time complexity O(n)
   4. Space complexity O(1)

Solution

class Solution {
    public boolean isSelfCrossing(int[] x) {
        if (x.length < 4) return false;
        int i = 2;
        // spiraling outward
        while (i < x.length && x[i] > x[i - 2]) i++;
        if (i == x.length) return false;
        // transition
        if ((i == 3 && x[i] == x[i - 2]) || (i >= 4 && x[i] >= x[i - 2] - x[i - 4])) {
            x[i - 1] -= x[i - 3];
        }
        i++;
        // spiraling inward
        while (i < x.length) {
            if (x[i] >= x[i - 2]) return true;
            i++;
        }
        return false;
    }
}

class Solution {
    public boolean isSelfCrossing(int[] x) {
        if (x.length < 4) return false;
        for (int i = 3; i < x.length; i++) {
            // first case: fourth line cross the first line
            if (x[i] >= x[i - 2] && x[i - 3] >= x[i - 1]) return true;
            // second case: fifth line meet the first line
            if (i >= 4 && x[i - 1] == x[i - 3] && x[i] >= x[i - 2] - x[i - 4]) return true;
            // third case: sixth line cross the first line
            if (i >= 5 && x[i - 2] >= x[i - 4] && x[i - 3] >= x[i - 1] && x[i - 1] >= x[i - 3] - x[i - 5] && x[i] >= x[i - 2] - x[i - 4]) return true;
        }
        return false;
    }
}

Additional