## The challenge

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 `p<sub>i</sub>` is represented as `[x<sub>i</sub>, y<sub>i</sub>]`. The input is not given in any order.

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`
• `-10<sup>4</sup>&nbsp;<= x<sub>i</sub>, y<sub>i</sub>&nbsp;<= 10<sup>4</sup>`

## The solution in Java code

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 `````` ``````class Solution { public double dist(int[] p1, int[] p2) { return (p2 - p1) * (p2 - p1) + (p2 - p1) * (p2 - p1); } public boolean validSquare(int[] p1, int[] p2, int[] p3, int[] p4) { int[][] p={p1,p2,p3,p4}; Arrays.sort(p, (l1, l2) -> l2 == l1 ? l1 - l2 : l1 - l2); return dist(p, p) != 0 && dist(p, p) == dist(p, p) && dist(p, p) == dist(p, p) && dist(p, p) == dist(p, p) && dist(p,p)==dist(p,p); } } ``````