Self Crossing LeetCode Solution

Problem – Self Crossing LeetCode Solution

You are given an array of integers distance.

You start at point (0,0) on an X-Y plane and you move distance[0] meters to the north, then distance[1] meters to the west, distance[2] meters to the south, distance[3] meters to the east, and so on. In other words, after each move, your direction changes counter-clockwise.

Return true if your path crosses itself, and false if it does not.

Example 1:

Input: distance = [2,1,1,2]
Output: true

Example 2:

Input: distance = [1,2,3,4]
Output: false

Example 3:

Input: distance = [1,1,1,1]
Output: true

Constraints:

  • 1 <= distance.length <= 105
  • 1 <= distance[i] <= 105

Self Crossing LeetCode Solution in Java

// Categorize the self-crossing scenarios, there are 3 of them: 
// 1. Fourth line crosses first line and works for fifth line crosses second line and so on...
// 2. Fifth line meets first line and works for the lines after
// 3. Sixth line crosses first line and works for the lines after
public class Solution {
    public boolean isSelfCrossing(int[] x) {
        int l = x.length;
        if(l <= 3) return false;
        
        for(int i = 3; i < l; i++){
            if(x[i] >= x[i-2] && x[i-1] <= x[i-3]) return true;  //Fourth line crosses first line and onward
            if(i >=4)
            {
                if(x[i-1] == x[i-3] && x[i] + x[i-4] >= x[i-2]) return true; // Fifth line meets first line and onward
            }
            if(i >=5)
            {
                if(x[i-2] - x[i-4] >= 0 && x[i] >= x[i-2] - x[i-4] && x[i-1] >= x[i-3] - x[i-5] && x[i-1] <= x[i-3]) return true;  // Sixth line crosses first line and onward
            }
        }
        return false;
    }
}

Self Crossing LeetCode Solution in Python

def isSelfCrossing(self, x):
    return any(d >= b > 0 and (a >= c or a >= c-e >= 0 and f >= d-b)
               for a, b, c, d, e, f in ((x[i:i+6] + [0] * 6)[:6]
                                        for i in xrange(len(x))))

Self Crossing LeetCode Solution in C++

class Solution
{
public:
    bool isSelfCrossing(vector<int>& x)
    {
        x.insert(x.begin(), 4, 0);

        int len = x.size();
        int i = 4;

        // outer spiral
        for (; i < len && x[i] > x[i - 2]; i++);

        if (i == len) return false;

        // check border
        if (x[i] >= x[i - 2] - x[i - 4])
        {
            x[i - 1] -= x[i - 3];
        }

        // inner spiral
        for (i++; i < len && x[i] < x[i - 2]; i++);

        return i != len;
    }
};
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