self-crossing.py

# Time:  O(n)
# Space: O(1)

# 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)

class Solution(object):
    def isSelfCrossing(self, x):
        """
        :type x: List[int]
        :rtype: bool
        """
        if len(x) >= 5 and x[3] == x[1] and x[4] + x[0] >= x[2]:
            # Crossing in a loop:
            #     2
            # 3 ┌────┐
            #   └─══>┘1
            #   4  0  (overlapped)  
            return True

        for i in xrange(3, len(x)):
            if x[i] >= x[i - 2] and x[i - 3] >= x[i - 1]:
                # Case 1:
                #    i-2
                # i-1┌─┐
                #    └─┼─>i
                #     i-3
                return True
            elif i >= 5 and x[i - 4] <= x[i - 2] and x[i] + x[i - 4] >= x[i - 2] and \
                            x[i - 1] <= x[i - 3] and x[i - 5] + x[i - 1] >= x[i - 3]:
                # Case 2:
                #    i-4
                #    ┌──┐ 
                #    │i<┼─┐
                # i-3│ i-5│i-1
                #    └────┘
                #      i-2
                return True
        return False