# Find the number of points that have atleast 1 point above, below, left or right of it

Given N points in 2 dimensional plane. A point is said to be above another point if the X coordinates of both points are same and the Y coordinate of the first point is greater than the Y coordinate of the second point. Similarly, we define below, left and right. The task is to count the number of points that have atleast one point above, below, left or right of it.

Examples:

Input: arr[] = {{0, 0}, {0, 1}, {1, 0}, {0, -1}, {-1, 0}}
Output: 1
The only point which satisfies the condition is the point (0, 0).

Input: arr[] = {{0, 0}, {1, 0}, {0, -2}, {5, 0}}
Output: 0

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

Approach: For every X coordinate, find 2 values, the minimum and maximum Y coordinate among all points that have this X coordinate. Do the same thing for every Y coordinate. Now, for a point to satisfy the constraints, its Y coordinate must lie in between the 2 calculated values for that X coordinate. Check the same thing for its X coordinate.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define MX 2001 ` `#define OFF 1000 ` ` `  `// Represents a point in 2-D space ` `struct` `point { ` `    ``int` `x, y; ` `}; ` ` `  `// Function to return the count of ` `// required points ` `int` `countPoints(``int` `n, ``struct` `point points[]) ` `{ ` `    ``int` `minx[MX]; ` `    ``int` `miny[MX]; ` ` `  `    ``// Initialize minimum values to infinity ` `    ``fill(minx, minx + MX, INT_MAX); ` `    ``fill(miny, miny + MX, INT_MAX); ` ` `  `    ``// Initialize maximum values to zero ` `    ``int` `maxx[MX] = { 0 }; ` `    ``int` `maxy[MX] = { 0 }; ` `    ``int` `x, y; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Add offset to deal with negative  ` `        ``// values ` `        ``points[i].x += OFF; ` `        ``points[i].y += OFF; ` `        ``x = points[i].x; ` `        ``y = points[i].y; ` ` `  `        ``// Update the minimum and maximum ` `        ``// values ` `        ``minx[y] = min(minx[y], x); ` `        ``maxx[y] = max(maxx[y], x); ` `        ``miny[x] = min(miny[x], y); ` `        ``maxy[x] = max(maxy[x], y); ` `    ``} ` ` `  `    ``int` `count = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``x = points[i].x; ` `        ``y = points[i].y; ` ` `  `        ``// Check if condition is satisfied  ` `        ``// for X coordinate ` `        ``if` `(x > minx[y] && x < maxx[y]) ` ` `  `            ``// Check if condition is satisfied ` `            ``// for Y coordinate ` `            ``if` `(y > miny[x] && y < maxy[x]) ` `                ``count++; ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``struct` `point points[] = { { 0, 0 }, ` `                              ``{ 0, 1 }, ` `                              ``{ 1, 0 }, ` `                              ``{ 0, -1 }, ` `                              ``{ -1, 0 } }; ` `    ``int` `n = ``sizeof``(points) / ``sizeof``(points); ` `    ``cout << countPoints(n, points); ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` `static` `int` `MX = ``2001``; ` `static` `int` `OFF = ``1000``; ` ` `  `// Represents a point in 2-D space ` `static` `class` `point  ` `{ ` `    ``int` `x, y; ` ` `  `    ``public` `point(``int` `x, ``int` `y)  ` `    ``{ ` `        ``this``.x = x; ` `        ``this``.y = y; ` `    ``} ` `     `  `}; ` ` `  `// Function to return the count of ` `// required points ` `static` `int` `countPoints(``int` `n, point points[]) ` `{ ` `    ``int` `[]minx = ``new` `int``[MX]; ` `    ``int` `[]miny = ``new` `int``[MX]; ` ` `  `    ``// Initialize minimum values to infinity ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``minx[i]=Integer.MAX_VALUE; ` `        ``miny[i]=Integer.MAX_VALUE; ` `    ``} ` ` `  `    ``// Initialize maximum values to zero ` `    ``int` `[]maxx = ``new` `int``[MX]; ` `    ``int` `[]maxy = ``new` `int``[MX]; ` ` `  `    ``int` `x, y; ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` ` `  `        ``// Add offset to deal with negative  ` `        ``// values ` `        ``points[i].x += OFF; ` `        ``points[i].y += OFF; ` `        ``x = points[i].x; ` `        ``y = points[i].y; ` ` `  `        ``// Update the minimum and maximum ` `        ``// values ` `        ``minx[y] = Math.min(minx[y], x); ` `        ``maxx[y] = Math.max(maxx[y], x); ` `        ``miny[x] = Math.min(miny[x], y); ` `        ``maxy[x] = Math.max(maxy[x], y); ` `    ``} ` ` `  `    ``int` `count = ``0``; ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `        ``x = points[i].x; ` `        ``y = points[i].y; ` ` `  `        ``// Check if condition is satisfied  ` `        ``// for X coordinate ` `        ``if` `(x > minx[y] && x < maxx[y]) ` ` `  `            ``// Check if condition is satisfied ` `            ``// for Y coordinate ` `            ``if` `(y > miny[x] && y < maxy[x]) ` `                ``count++; ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``point points[] = {``new` `point(``0``, ``0``), ` `                      ``new` `point(``0``, ``1``), ` `                      ``new` `point(``1``, ``0``), ` `                      ``new` `point(``0``, -``1``), ` `                      ``new` `point(-``1``, ``0``)}; ` `    ``int` `n = points.length; ` `    ``System.out.println(countPoints(n, points)); ` `} ` `}  ` ` `  `// This code is contributed by PrinciRaj1992 `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG  ` `{ ` `static` `int` `MX = 2001; ` `static` `int` `OFF = 1000; ` ` `  `// Represents a point in 2-D space ` `public` `class` `point  ` `{ ` `    ``public` `int` `x, y; ` ` `  `    ``public` `point(``int` `x, ``int` `y)  ` `    ``{ ` `        ``this``.x = x; ` `        ``this``.y = y; ` `    ``} ` `}; ` ` `  `// Function to return the count of ` `// required points ` `static` `int` `countPoints(``int` `n, point []points) ` `{ ` `    ``int` `[]minx = ``new` `int``[MX]; ` `    ``int` `[]miny = ``new` `int``[MX]; ` ` `  `    ``// Initialize minimum values to infinity ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``minx[i]=``int``.MaxValue; ` `        ``miny[i]=``int``.MaxValue; ` `    ``} ` ` `  `    ``// Initialize maximum values to zero ` `    ``int` `[]maxx = ``new` `int``[MX]; ` `    ``int` `[]maxy = ``new` `int``[MX]; ` ` `  `    ``int` `x, y; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` ` `  `        ``// Add offset to deal with negative  ` `        ``// values ` `        ``points[i].x += OFF; ` `        ``points[i].y += OFF; ` `        ``x = points[i].x; ` `        ``y = points[i].y; ` ` `  `        ``// Update the minimum and maximum ` `        ``// values ` `        ``minx[y] = Math.Min(minx[y], x); ` `        ``maxx[y] = Math.Max(maxx[y], x); ` `        ``miny[x] = Math.Min(miny[x], y); ` `        ``maxy[x] = Math.Max(maxy[x], y); ` `    ``} ` ` `  `    ``int` `count = 0; ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``x = points[i].x; ` `        ``y = points[i].y; ` ` `  `        ``// Check if condition is satisfied  ` `        ``// for X coordinate ` `        ``if` `(x > minx[y] && x < maxx[y]) ` ` `  `            ``// Check if condition is satisfied ` `            ``// for Y coordinate ` `            ``if` `(y > miny[x] && y < maxy[x]) ` `                ``count++; ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``point []points = {``new` `point(0, 0), ` `                      ``new` `point(0, 1), ` `                      ``new` `point(1, 0), ` `                      ``new` `point(0, -1), ` `                      ``new` `point(-1, 0)}; ` `    ``int` `n = points.Length; ` `    ``Console.WriteLine(countPoints(n, points)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```1
```

Time Complexity: O(N)

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Improved By : princiraj1992, 29AjayKumar