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

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:

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// C++ implementation of the approach
#include <bits/stdc++.h>
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[0]);
    cout << countPoints(n, points);
}
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// 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
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# Python3 implementation of the approach
from sys import maxsize as INT_MAX
  
MX = 2001
OFF = 1000
  
# Represents a point in 2-D space
class point:
    def __init__(self, x, y):
        self.x = x
        self.y = y
  
# Function to return the count of
# required points
def countPoints(n: int, points: list) -> int:
  
    # Initialize minimum values to infinity
    minx = [INT_MAX] * MX
    miny = [INT_MAX] * MX
  
    # Initialize maximum values to zero
    maxx = [0] * MX
    maxy = [0] * MX
    x, y = 0, 0
    for i in range(n):
  
        # 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)
  
    count = 0
    for i in range(n):
        x = points[i].x
        y = points[i].y
  
        # Check if condition is satisfied
        # for X coordinate
        if (x > minx[y] and x < maxx[y]):
  
            # Check if condition is satisfied
            # for Y coordinate
            if (y > miny[x] and y < maxy[x]):
                count += 1
  
    return count
  
# Driver Code
if __name__ == "__main__":
  
    points = [point(0, 0),
              point(0, 1),
              point(1, 0),
              point(0, -1),
              point(-1, 0)]
    n = len(points)
  
    print(countPoints(n, points))
  
# This code is contributed by
# sanjeev2552
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// 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
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Output:
1

Time Complexity: O(N)

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