Given N points in a 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:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define MX 2001#define OFF 1000// Represents a point in 2-D spacestruct point { int x, y;};// Function to return the count of// required pointsint 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 codeint 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);} |
Java
// Java implementation of the approachimport java.util.*;class GFG {static int MX = 2001;static int OFF = 1000;// Represents a point in 2-D spacestatic class point { int x, y; public point(int x, int y) { this.x = x; this.y = y; } };// Function to return the count of// required pointsstatic 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 codepublic 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 |
Python3
# Python3 implementation of the approachfrom sys import maxsize as INT_MAXMX = 2001OFF = 1000# Represents a point in 2-D spaceclass point: def __init__(self, x, y): self.x = x self.y = y# Function to return the count of# required pointsdef 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 Codeif __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 |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GFG {static int MX = 2001;static int OFF = 1000;// Represents a point in 2-D spacepublic 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 pointsstatic 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 codepublic 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 |
Javascript
<script>// Javascript implementation of the approachvar MX = 2001;var OFF = 1000;// Function to return the count of// required pointsfunction countPoints( n, points){ var minx = Array(MX).fill(1000000000); var miny = Array(MX).fill(1000000000); // Initialize maximum values to zero var maxx = Array(MX).fill(0); var maxy = Array(MX).fill(0); var x, y; for (var i = 0; i < n; i++) { // Add offset to deal with negative // values points[i][0] += OFF; points[i][1] += OFF; x = points[i][0]; y = points[i][1]; // 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); } var count = 0; for (var i = 0; i < n; i++) { x = points[i][0]; y = points[i][1]; // 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 codevar points = [ [ 0, 0 ], [ 0, 1 ], [ 1, 0 ], [ 0, -1 ], [ -1, 0 ] ];var n = points.length;document.write( countPoints(n, points));// This code is contributed by famously.</script> |
Output
1
Complexity Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(N) because it is using auxiliary space for array minx and miny
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