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:
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:
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[xi] stores the number of points having their x-coordinate equal to xi
2. Map Y, where Y[yi] stores the number of points having their y-coordinate equal to yi
3. Map XY, where XY[(Xi, Yi)] stores the number of points coincident with point (xi, yi)

Now,
Let Xans be the Number of pairs with same X-coordinates = X[xi]2 for all distinct xi
Let Yans be the Number of pairs with same Y-coordinates = Y[xi]2 for all distinct yi
Let XYans be the Number of coincident points = XY[{xi, yi}]2 for all distinct points (xi, yi)
Thus the required answer = Xans + Yans – XYans

Below is the implementation of the above approach:

C++

 `// C++ implementation of the above approach` `#include ` `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, ``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;` `}`

Java

 `// Java implementation of the above approach` `import` `java.util.*;`   `public` `class` `GFG` `{` `  `  `  ``// Function to return the number of non coincident` `  ``// pairs of points with manhattan distance` `  ``// equal to euclidean distance` `  ``static` `int` `findManhattanEuclidPair(``int``[][] arr, ``int` `n)` `  ``{` `    `  `    ``// To store frequency of all distinct Xi` `    ``Map X = ``new` `HashMap();`   `    ``// To store Frequency of all distinct Yi` `    ``Map Y = ``new` `HashMap();`   `    ``// To store Frequency of all distinct` `    ``// points (Xi, Yi);` `    ``Map, Integer> XY = ``new` `HashMap, Integer>();`   `    ``for` `(``int` `i = ``0``; i < n; i++) {` `      ``int` `xi = arr[i][``0``];` `      ``int` `yi = arr[i][``1``];`   `      ``// Hash xi coordinate` `      ``if` `(!X.containsKey(xi))` `        ``X.put(xi, ``0``);` `      ``X.put(xi, X.get(xi) + ``1``);`   `      ``// Hash yi coordinate` `      ``if` `(!Y.containsKey(yi))` `        ``Y.put(yi, ``0``);` `      ``Y.put(yi, Y.get(yi) + ``1``);`   `      ``// Hash the point (xi, yi)` `      ``if` `(!XY.containsKey(Arrays.asList(xi, yi)))` `        ``XY.put(Arrays.asList(xi, yi), ``0``);` `      ``XY.put( Arrays.asList(xi, yi), XY.get(Arrays.asList(xi, yi)) + ``1``);` `    ``}`   `    ``int` `xAns = ``0``, yAns = ``0``, xyAns = ``0``;`   `    ``// find pairs with same Xi` `    ``for` `(Map.Entry xCoordinatePair : X.entrySet())` `    ``{` `      ``int` `xFrequency = xCoordinatePair.getValue();`   `      ``// calculate ((xFrequency) C2)` `      ``int` `sameXPairs` `        ``= (xFrequency * (xFrequency - ``1``)) / ``2``;` `      ``xAns += sameXPairs;` `    ``}`   `    ``// find pairs with same Yi` `    ``for` `(Map.Entry yCoordinatePair : Y.entrySet()) {` `      ``int` `yFrequency = yCoordinatePair.getValue();`   `      ``// calculate ((yFrequency) C2)` `      ``int` `sameYPairs` `        ``= (yFrequency * (yFrequency - ``1``)) / ``2``;` `      ``yAns += sameYPairs;` `    ``}`   `    ``// find pairs with same (Xi, Yi)` `    ``for` `(Map.Entry, Integer> XYPair : XY.entrySet())` `    ``{` `      ``int` `xyFrequency = XYPair.getValue();`   `      ``// 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` `  ``public` `static` `void` `main(String[] args)` `  ``{` `    ``int``[][] arr` `      ``= { { ``1``, ``2` `}, { ``1``, ``2` `}, { ``4``, ``3` `}, { ``1``, ``3` `} };` `    ``int` `n = arr.length;`   `    ``System.out.println(findManhattanEuclidPair(arr, n));` `  ``}` `}`   `// This code is contributed by phasing17`

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`

C#

 `// C# implementation of the above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG` `{` `  `  `  ``// Function to return the number of non coincident` `  ``// pairs of points with manhattan distance` `  ``// equal to euclidean distance` `  ``static` `int` `findManhattanEuclidPair(``int``[, ] arr, ``int` `n)` `  ``{` `    `  `    ``// To store frequency of all distinct Xi` `    ``Dictionary<``int``, ``int``> X = ``new` `Dictionary<``int``, ``int``>();`   `    ``// To store Frequency of all distinct Yi` `    ``Dictionary<``int``, ``int``> Y = ``new` `Dictionary<``int``, ``int``>();`   `    ``// To store Frequency of all distinct` `    ``// points (Xi, Yi);` `    ``var` `XY = ``new` `Dictionary<(``int` `first, ``int` `second), ``int``>();`   `    ``for` `(``int` `i = 0; i < n; i++) {` `      ``int` `xi = arr[i, 0];` `      ``int` `yi = arr[i, 1];`   `      ``// Hash xi coordinate` `      ``if` `(!X.ContainsKey(xi))` `        ``X[xi] = 0;` `      ``X[xi]++;`   `      ``// Hash yi coordinate` `      ``if` `(!Y.ContainsKey(yi))` `        ``Y[yi] = 0;` `      ``Y[yi]++;`   `      ``// Hash the point (xi, yi)` `      ``if` `(!XY.ContainsKey((xi, yi)))` `        ``XY[(xi, yi)] = 0;` `      ``XY[(xi, yi)]++;` `    ``}`   `    ``int` `xAns = 0, yAns = 0, xyAns = 0;`   `    ``// find pairs with same Xi` `    ``foreach` `(``var` `xCoordinatePair ``in` `X) {` `      ``int` `xFrequency = xCoordinatePair.Value;`   `      ``// calculate ((xFrequency) C2)` `      ``int` `sameXPairs` `        ``= (xFrequency * (xFrequency - 1)) / 2;` `      ``xAns += sameXPairs;` `    ``}`   `    ``// find pairs with same Yi` `    ``foreach` `(``var` `yCoordinatePair ``in` `Y) {` `      ``int` `yFrequency = yCoordinatePair.Value;`   `      ``// calculate ((yFrequency) C2)` `      ``int` `sameYPairs` `        ``= (yFrequency * (yFrequency - 1)) / 2;` `      ``yAns += sameYPairs;` `    ``}`   `    ``// find pairs with same (Xi, Yi)` `    ``foreach` `(``var` `XYPair ``in` `XY) {` `      ``int` `xyFrequency = XYPair.Value;`   `      ``// 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` `  ``public` `static` `void` `Main(``string``[] args)` `  ``{` `    ``int``[, ] arr` `      ``= { { 1, 2 }, { 1, 2 }, { 4, 3 }, { 1, 3 } };` `    ``int` `n = arr.GetLength(0);`   `    ``Console.WriteLine(findManhattanEuclidPair(arr, n));` `  ``}` `}`   `// This code is contributed by pphasing17`

Javascript

 ``

Output

`3`

Complexity Analysis:

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

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