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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ implementtaion 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 - xyAns); 
} 
  
// Driver Code 
int main() 
{ 
    pair<int, int> arr[] = { 
        { 1, 2 }, 
        { 2, 3 }, 
        { 1, 3 } 
    }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    cout << findManhattanEuclidPair(arr, n) << endl; 
    return 0; 
} 

Python3

# Python3 implementtaion 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 - xyAns)  
  
# Driver Code  
if __name__ == "__main__": 
  
    arr = [[1, 2], [2, 3], [1, 3]]  
      
    n = len(arr)  
  
    print(findManhattanEuclidPair(arr, n))  
      
# This code is contributed by Rituraj Jain 

Output:

Time Complexity: O(NlogN), where N is the number of points
Space Complexity: O(N)

My Personal Notes

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Python3

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