# 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 metres to the north, then x metres to the west, x metres to the south, x 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],
?????
?   ?
???????>
?

Return true (self crossing)
``````
``````Given x = [1, 2, 3, 4],
????????
?      ?
?
?
?????????????>

Return false (not self crossing)
``````
``````Given x = [1, 1, 1, 1],
?????
?   ?
?????>

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;
}
}
``````