Pairs of Non Coinciding Points

In a given cartesian plane, there are N points. We need to find the Number of Pairs of points(A,B) such that

1. Point A and Point B do not coincide. ​

2. Manhattan Distance and the Euclidean Distance between the points should be equal.

Note : Pair of 2 points(A,B) is considered same as Pair of 2 points(B ,A). Manhattan Distance = |x2-x1|+|y2-y1|

Euclidean Distance = ((x2-x1)^2 + (y2-y1)^2)^0.5 where points are (x1,y1) and (x2,y2).

Input:

First Line Consist of T - number of test cases. For each Test case:- First Line consist of N , Number of points Next line contains N pairs contains two integers Xi and Yi i.e, X coordinate and the Y coordinate of a Point Output:

Print the number of pairs as asked above.

Constraints:

1<=T <= 50

1<=N <= 2*10 ^ 5

0<=(|Xi|, |Yi| ) <= 10^9

Example:

Sample Input :

1
2
1 1
7 5

Sample Output :

0

// A+B=Root(A^2+B^2) => 2AB=0
// A => |x1-x2| & B => |y1-y2|
class GFG {
    public static int coincidingPoints(int[][] points, int n) {
        Map<Integer, Integer> mapX = new HashMap<>();
        Map<Integer, Integer> mapY = new HashMap<>();
        Map<String, Integer> set = new HashMap<>();
        int ans = 0;
        for (int[] point : points) {
            String key = Integer.toString(point[0]) + " " + Integer.toString(point[1]);
            set.put(key, set.getOrDefault(key, 0) + 1);
            mapX.put(point[0], mapX.getOrDefault(point[0], 0) + 1);
            // Removing coinciding pairs
            ans += mapX.get(point[0]) - set.get(key);
            mapY.put(point[1], mapY.getOrDefault(point[1], 0) + 1);
            // Removing coinciding pairs
            ans += mapY.get(point[1]) - set.get(key);
        }
        return ans;
    }
}