335. Self Crossing LeetCode Solution

In this guide, you will get 335. Self Crossing LeetCode Solution with the best time and space complexity. The solution to Self Crossing problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Self Crossing solution in C++
Self Crossing solution in Java
Self Crossing solution in Python
Additional Resources

335. Self Crossing LeetCode Solution image

Problem Statement of Self Crossing

You are given an array of integers distance.
You start at the 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 or false if it does not.

Example 1:

Input: distance = [2,1,1,2]
Output: true
Explanation: The path crosses itself at the point (0, 1).

Example 2:

Input: distance = [1,2,3,4]
Output: false
Explanation: The path does not cross itself at any point.

Example 3:

Input: distance = [1,1,1,2,1]
Output: true
Explanation: The path crosses itself at the point (0, 0).

Constraints:

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

Complexity Analysis

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

335. Self Crossing LeetCode Solution in C++

class Solution {
 public:
  bool isSelfCrossing(vector<int>& x) {
    if (x.size() <= 3)
      return false;

    for (int i = 3; i < x.size(); ++i) {
      if (x[i - 2] <= x[i] && x[i - 1] <= x[i - 3])
        return true;
      if (i >= 4 && x[i - 1] == x[i - 3] && x[i - 2] <= x[i] + x[i - 4])
        return true;
      if (i >= 5 && x[i - 4] <= x[i - 2] && x[i - 2] <= x[i] + x[i - 4] &&
          x[i - 1] <= x[i - 3] && x[i - 3] <= x[i - 1] + x[i - 5])
        return true;
    }

    return false;
  }
};
/* code provided by PROGIEZ */

335. Self Crossing LeetCode Solution in Java

class Solution {
  public boolean isSelfCrossing(int[] x) {
    if (x.length <= 3)
      return false;

    for (int i = 3; i < x.length; ++i) {
      if (x[i - 2] <= x[i] && x[i - 1] <= x[i - 3])
        return true;
      if (i >= 4 && x[i - 1] == x[i - 3] && x[i - 2] <= x[i] + x[i - 4])
        return true;
      if (i >= 5 && x[i - 4] <= x[i - 2] && x[i - 2] <= x[i] + x[i - 4] && x[i - 1] <= x[i - 3] &&
          x[i - 3] <= x[i - 1] + x[i - 5])
        return true;
    }

    return false;
  }
}
// code provided by PROGIEZ

335. Self Crossing LeetCode Solution in Python

class Solution:
  def isSelfCrossing(self, x: list[int]) -> bool:
    if len(x) <= 3:
      return False

    for i in range(3, len(x)):
      if x[i - 2] <= x[i] and x[i - 1] <= x[i - 3]:
        return True
      if i >= 4 and x[i - 1] == x[i - 3] and x[i - 2] <= x[i] + x[i - 4]:
        return True
      if i >= 5 and x[i - 4] <= x[i - 2] and x[i - 2] <= x[i] + x[i - 4] and x[i - 1] <= x[i - 3] and x[i - 3] <= x[i - 1] + x[i - 5]:
        return True

    return False
 # code by PROGIEZ