Pairs with same Manhattan and Euclidean distance

In a given Cartesian plane, there are N points. The task is to find the Number of Pairs of points(A, B) such that 

• Point A and Point B do not coincide.
• 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).

Examples:

Input: N = 3, Points = {{1, 2}, {2, 3}, {1, 3}} 
Output: 2 
Pairs are: 
1) (1, 2) and (1, 3) 
Euclidean distance of (1, 2) and (1, 3) = &root;((1 – 1)² + (3 – 2)²) = 1 
Manhattan distance of (1, 2) and (1, 3) = |(1 – 1)| + |(2 – 3)| = 1
2) (1, 3) and (2, 3) 
Euclidean distance of (1, 3) and (2, 3) = &root;((1 – 2)² + (3 – 3)²) = 1 
Manhattan distance of (1, 3) and (2, 3) = |(1 – 2)| + |(3 – 3)| = 1
Input: N = 3, Points = { {1, 1}, {2, 3}, {1, 1} } 
Output: 0 
Here none of the pairs satisfy the above two conditions

Approach: On solving the equation 

|x2-x1|+|y2-y1| = sqrt((x2-x1)^2+(y2-y1)^2)

we get , x2 = x1 or y2 = y1.
Consider 3 maps, 
1) Map X, where X[x_i] stores the number of points having their x-coordinate equal to x_i
2) Map Y, where Y[y_i] stores the number of points having their y-coordinate equal to y_i
3) Map XY, where XY[(X_i, Y_i)] stores the number of points coincident with point (x_i, y_i)
Now, 
Let Xans be the Number of pairs with same X-coordinates = ^X[x_i]₂ for all distinct x_i = 
Let Yans be the Number of pairs with same Y-coordinates = ^Y[x_i]₂ for all distinct y_i
Let XYans be the Number of coincident points = ^{XY[{x_i, y_i}]}₂ for all distinct points (x_i, y_i)
Thus the required answer = Xans + Yans – XYans
Below is the implementation of the above approach:

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the number of non coincident
// pairs of points with manhattan distance
// equal to euclidean distance
int findManhattanEuclidPair(pair<int, int> arr[], int n)
{
    // To store frequency of all distinct Xi
    map<int, int> X;

    // To store Frequency of all distinct Yi
    map<int, int> Y;

    // To store Frequency of all distinct
    // points (Xi, Yi);
    map<pair<int, int>, int> XY;

    for (int i = 0; i < n; i++) {
        int xi = arr[i].first;
        int yi = arr[i].second;

        // Hash xi coordinate
        X[xi]++;

        // Hash yi coordinate
        Y[yi]++;

        // Hash the point (xi, yi)
        XY[arr[i]]++;
    }

    int xAns = 0, yAns = 0, xyAns = 0;

    // find pairs with same Xi
    for (auto xCoordinatePair : X) {
        int xFrequency = xCoordinatePair.second;

        // calculate ((xFrequency) C2)
        int sameXPairs
            = (xFrequency * (xFrequency - 1)) / 2;
        xAns += sameXPairs;
    }

    // find pairs with same Yi
    for (auto yCoordinatePair : Y) {
        int yFrequency = yCoordinatePair.second;

        // calculate ((yFrequency) C2)
        int sameYPairs
            = (yFrequency * (yFrequency - 1)) / 2;
        yAns += sameYPairs;
    }

    // find pairs with same (Xi, Yi)
    for (auto XYPair : XY) {
        int xyFrequency = XYPair.second;

        // calculate ((xyFrequency) C2)
        int samePointPairs
            = (xyFrequency * (xyFrequency - 1)) / 2;
        xyAns += samePointPairs;
    }

    return (xAns + yAns - (2 * xyAns));
    /*    we are subtracting 2 * xyAns because we have counted
       let say A,B coinciding points two times in xAns and
       yAns which should not be add to the final answer so
       we are subtracting xyAns 2 times. */
}

// Driver Code
int main()
{
    pair<int, int> arr[]
        = { { 1, 2 }, { 1, 2 }, { 4, 3 }, { 1, 3 } };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << findManhattanEuclidPair(arr, n) << endl;
    return 0;
}

Python3

# Python3 implementation of the
# above approach
from collections import defaultdict

# Function to return the number of
# non coincident pairs of points with
# manhattan distance equal to
# euclidean distance
def findManhattanEuclidPair(arr, n):

    # To store frequency of all distinct Xi
    X = defaultdict(lambda:0)

    # To store Frequency of all distinct Yi
    Y = defaultdict(lambda:0)

    # To store Frequency of all distinct
    # points (Xi, Yi)
    XY = defaultdict(lambda:0)

    for i in range(0, n):
        xi = arr[i][0]
        yi = arr[i][1]

        # Hash xi coordinate
        X[xi] += 1

        # Hash yi coordinate
        Y[yi] += 1

        # Hash the point (xi, yi)
        XY[tuple(arr[i])] += 1

    xAns, yAns, xyAns = 0, 0, 0

    # find pairs with same Xi
    for xCoordinatePair in X:
        xFrequency = X[xCoordinatePair]

        # calculate ((xFrequency) C2)
        sameXPairs = (xFrequency *
                     (xFrequency - 1)) // 2
        xAns += sameXPairs

    # find pairs with same Yi
    for yCoordinatePair in Y:
        yFrequency = Y[yCoordinatePair]

        # calculate ((yFrequency) C2)
        sameYPairs = (yFrequency *
                     (yFrequency - 1)) // 2
        yAns += sameYPairs

    # find pairs with same (Xi, Yi)
    for XYPair in XY:
        xyFrequency = XY[XYPair]

        # calculate ((xyFrequency) C2)
        samePointPairs = (xyFrequency *
                         (xyFrequency - 1)) // 2
        xyAns += samePointPairs

    return (xAns + yAns - 2 * xyAns)
  #    we are subtracting 2 * xyAns because we have counted let say A,B coinciding points two times
   #  in xAns and yAns which should not be add to the final answer so we are subtracting xyAns 2 times.

# Driver Code
if __name__ == "__main__":

    arr = [[1, 2], [1,2], [4, 3], [1, 3]]

    n = len(arr)

    print(findManhattanEuclidPair(arr, n))

# This code is contributed by Rituraj Jain

Output

3

Time Complexity: O(NlogN), where N is the number of points 
Space Complexity: O(N), since N extra space has been taken.