Leetcode #593: Valid Square
In this guide, we solve Leetcode #593 Valid Square in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given the coordinates of four points in 2D space p1, p2, p3 and p4, return true if the four points construct a square. The coordinate of a point pi is represented as [xi, yi].
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Geometry, Math
Intuition
There is a mathematical invariant or formula that directly leads to the result.
Using math avoids unnecessary loops and reduces complexity.
Approach
Derive the formula or update rule, then compute the answer directly.
Handle edge cases like overflow or zero carefully.
Steps:
- Identify the math relationship.
- Compute the result with a loop or formula.
- Handle edge cases.
Example
Input: p1 = [0,0], p2 = [1,1], p3 = [1,0], p4 = [0,1]
Output: true
Python Solution
class Solution:
def validSquare(
self, p1: List[int], p2: List[int], p3: List[int], p4: List[int]
) -> bool:
def check(a, b, c):
(x1, y1), (x2, y2), (x3, y3) = a, b, c
d1 = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)
d2 = (x1 - x3) * (x1 - x3) + (y1 - y3) * (y1 - y3)
d3 = (x2 - x3) * (x2 - x3) + (y2 - y3) * (y2 - y3)
return any(
[
d1 == d2 and d1 + d2 == d3 and d1,
d2 == d3 and d2 + d3 == d1 and d2,
d1 == d3 and d1 + d3 == d2 and d1,
]
)
return (
check(p1, p2, p3)
and check(p2, p3, p4)
and check(p1, p3, p4)
and check(p1, p2, p4)
)
Complexity
The time complexity is O(n) or O(1). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.