LeetCode #335 — HARD

Self Crossing

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

[2,1,1,2]

Example 2

[1,2,3,4]

Example 3

[1,1,1,2,1]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #335: Self Crossing
class Solution {
    public boolean isSelfCrossing(int[] distance) {
        int[] d = distance;
        for (int i = 3; i < d.length; ++i) {
            if (d[i] >= d[i - 2] && d[i - 1] <= d[i - 3]) {
                return true;
            }
            if (i >= 4 && d[i - 1] == d[i - 3] && d[i] + d[i - 4] >= d[i - 2]) {
                return true;
            }
            if (i >= 5 && d[i - 2] >= d[i - 4] && d[i - 1] <= d[i - 3]
                && d[i] >= d[i - 2] - d[i - 4] && d[i - 1] + d[i - 5] >= d[i - 3]) {
                return true;
            }
        }
        return false;
    }
}

// Accepted solution for LeetCode #335: Self Crossing
func isSelfCrossing(distance []int) bool {
	d := distance
	for i := 3; i < len(d); i++ {
		if d[i] >= d[i-2] && d[i-1] <= d[i-3] {
			return true
		}
		if i >= 4 && d[i-1] == d[i-3] && d[i]+d[i-4] >= d[i-2] {
			return true
		}
		if i >= 5 && d[i-2] >= d[i-4] && d[i-1] <= d[i-3] && d[i] >= d[i-2]-d[i-4] && d[i-1]+d[i-5] >= d[i-3] {
			return true
		}
	}
	return false
}

# Accepted solution for LeetCode #335: Self Crossing
class Solution:
    def isSelfCrossing(self, distance: List[int]) -> bool:
        d = distance
        for i in range(3, len(d)):
            if d[i] >= d[i - 2] and d[i - 1] <= d[i - 3]:
                return True
            if i >= 4 and d[i - 1] == d[i - 3] and d[i] + d[i - 4] >= d[i - 2]:
                return True
            if (
                i >= 5
                and d[i - 2] >= d[i - 4]
                and d[i - 1] <= d[i - 3]
                and d[i] >= d[i - 2] - d[i - 4]
                and d[i - 1] + d[i - 5] >= d[i - 3]
            ):
                return True
        return False

// Accepted solution for LeetCode #335: Self Crossing
/**
 * [335] Self Crossing
 *
 * 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:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/03/14/selfcross1-plane.jpg" style="width: 400px; height: 435px;" />
 * Input: distance = [2,1,1,2]
 * Output: true
 *
 * Example 2:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/03/14/selfcross2-plane.jpg" style="width: 400px; height: 435px;" />
 * Input: distance = [1,2,3,4]
 * Output: false
 *
 * Example 3:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/03/14/selfcross3-plane.jpg" style="width: 400px; height: 435px;" />
 * Input: distance = [1,1,1,1]
 * Output: true
 *
 *  
 * Constraints:
 *
 * 	1 <= distance.length <= 10^5
 * 	1 <= distance[i] <= 10^5
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/self-crossing/
// discuss: https://leetcode.com/problems/self-crossing/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    // Credit: https://leetcode.com/problems/self-crossing/discuss/79131/Java-Oms-with-explanation
    pub fn is_self_crossing(distance: Vec<i32>) -> bool {
        let l = distance.len();
        if l <= 3 {
            return false;
        }

        for i in 3..l {
            if distance[i] >= distance[i - 2] && distance[i - 1] <= distance[i - 3] {
                //Fourth line crosses first line and onward
                return true;
            }
            if i >= 4 {
                if distance[i - 1] == distance[i - 3]
                    && distance[i] + distance[i - 4] >= distance[i - 2]
                {
                    // Fifth line meets first line and onward
                    return true;
                }
            }
            if i >= 5 {
                if distance[i - 2] - distance[i - 4] >= 0
                    && distance[i] >= distance[i - 2] - distance[i - 4]
                    && distance[i - 1] >= distance[i - 3] - distance[i - 5]
                    && distance[i - 1] <= distance[i - 3]
                {
                    // Sixth line crosses first line and onward
                    return true;
                }
            }
        }
        false
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_0335_example_1() {
        let distance = vec![2, 1, 1, 2];
        let result = true;

        assert_eq!(Solution::is_self_crossing(distance), result);
    }

    #[test]
    fn test_0335_example_2() {
        let distance = vec![1, 2, 3, 4];
        let result = false;

        assert_eq!(Solution::is_self_crossing(distance), result);
    }

    #[test]
    fn test_0335_example_3() {
        let distance = vec![1, 1, 1, 1];
        let result = true;

        assert_eq!(Solution::is_self_crossing(distance), result);
    }
}

// Accepted solution for LeetCode #335: Self Crossing
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #335: Self Crossing
// class Solution {
//     public boolean isSelfCrossing(int[] distance) {
//         int[] d = distance;
//         for (int i = 3; i < d.length; ++i) {
//             if (d[i] >= d[i - 2] && d[i - 1] <= d[i - 3]) {
//                 return true;
//             }
//             if (i >= 4 && d[i - 1] == d[i - 3] && d[i] + d[i - 4] >= d[i - 2]) {
//                 return true;
//             }
//             if (i >= 5 && d[i - 2] >= d[i - 4] && d[i - 1] <= d[i - 3]
//                 && d[i] >= d[i - 2] - d[i - 4] && d[i - 1] + d[i - 5] >= d[i - 3]) {
//                 return true;
//             }
//         }
//         return false;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.