593. Valid Square LeetCode Solution

In this guide, you will get 593. Valid Square LeetCode Solution with the best time and space complexity. The solution to Valid Square 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
Valid Square solution in C++
Valid Square solution in Java
Valid Square solution in Python
Additional Resources

593. Valid SquareLeetCode Solution image

Problem Statement of Valid Square

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]. The input is not given in any order.
A valid square has four equal sides with positive length and four equal angles (90-degree angles).

Example 1:

Input: p1 = [0,0], p2 = [1,1], p3 = [1,0], p4 = [0,1]
Output: true

Example 2:

Input: p1 = [0,0], p2 = [1,1], p3 = [1,0], p4 = [0,12]
Output: false

Example 3:

Input: p1 = [1,0], p2 = [-1,0], p3 = [0,1], p4 = [0,-1]
Output: true

Constraints:

p1.length == p2.length == p3.length == p4.length == 2
-104 <= xi, yi <= 104

Complexity Analysis

Time Complexity: O(4^2) = O(1)
Space Complexity: O(1)

593. Valid Square LeetCode Solution in C++

class Solution {
 public:
  bool validSquare(vector<int>& p1, vector<int>& p2,  //
                   vector<int>& p3, vector<int>& p4) {
    unordered_set<int> distSet;
    vector<vector<int>> points{p1, p2, p3, p4};

    for (int i = 0; i < 4; ++i)
      for (int j = i + 1; j < 4; ++j)
        distSet.insert(dist(points[i], points[j]));

    return !distSet.contains(0) && distSet.size() == 2;
  }

 private:
  int dist(vector<int>& p1, vector<int>& p2) {
    return (p1[0] - p2[0]) * (p1[0] - p2[0]) +
           (p1[1] - p2[1]) * (p1[1] - p2[1]);
  }
};
/* code provided by PROGIEZ */

593. Valid Square LeetCode Solution in Java

class Solution {
  public boolean validSquare(int[] p1, int[] p2, int[] p3, int[] p4) {
    Set<Integer> distSet = new HashSet<>();
    int[][] points = {p1, p2, p3, p4};

    for (int i = 0; i < 4; ++i)
      for (int j = i + 1; j < 4; ++j)
        distSet.add(dist(points[i], points[j]));

    return !distSet.contains(0) && distSet.size() == 2;
  }

  private int dist(int[] p1, int[] p2) {
    return (p1[0] - p2[0]) * (p1[0] - p2[0]) + (p1[1] - p2[1]) * (p1[1] - p2[1]);
  }
}
// code provided by PROGIEZ

593. Valid Square LeetCode Solution in Python

class Solution:
  def validSquare(
      self,
      p1: list[int],
      p2: list[int],
      p3: list[int],
      p4: list[int],
  ) -> bool:
    def dist(p1: list[int], p2: list[int]) -> int:
      return (p1[0] - p2[0])**2 + (p1[1] - p2[1])**2

    distSet = set([dist(*pair)
                   for pair in list(
                       itertools.combinations([p1, p2, p3, p4], 2))])
    return 0 not in distSet and len(distSet) == 2
 # code by PROGIEZ