335. 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

Example 2:

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

Return false (not self

Example 3:

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

Return true (self

Solution:
When it will be sefl crossing? the graph goes in spiral way, either alway growing outside, or always shrinking inside, both case will never self crossing. It only happens when from growing to shrinking. Graph default should be growing in former two steps, self crossing can only happens after third step.

One edge case is that once the graph not growing, there are two cases

1. it reaches the edge x[i] + x[i-4] >= x[i-2] (i>=4)

x[i] ==x[i-2] (i==3)

we need to resize the x[i-1] to x[i-1] - x[i-3], because next x[i+1] should compare x[i-1]

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