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

C++Server Side ProgrammingProgramming

In this problem, we are given N points that lie in a 2D plane. Our task is to find the number of points that have at least 1 point above, below, left or right of it.

We need to count all the points that have at least 1 point which satisfies any of the below conditions.

Point above it − The point will have the same X coordinate and the Y coordinate is one more than its current value.

Point below it − The point will have the same X coordinate and the Y coordinate is one less than its current value.

Point left of it − The point will have the same Y coordinate and the X coordinate is one less than its current value.

Point right of it − The point will have the same Y coordinate and the X coordinate is one more than its current value.

Let’s take an example to understand the problem,

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

Solution Approach

To solve the problem, we need to take each point form the plane and find the maximum and minimum value of X and Y coordinates its neighbours points can have for a valid count. And if any coordinate exists that has the same X coordinate and Y’s value lies in the range. We will increase point count. We will store the count in a variable and return it.

Example

Let’s take an example to understand the problem

#include <bits/stdc++.h>
using namespace std;
#define MX 2001
#define OFF 1000
struct point {
   int x, y;
};
int findPointCount(int n, struct point points[]){
   int minX[MX];
   int minY[MX];
   int maxX[MX] = { 0 };
   int maxY[MX] = { 0 };
   int xCoor, yCoor;
   fill(minX, minX + MX, INT_MAX);
   fill(minY, minY + MX, INT_MAX);
   for (int i = 0; i < n; i++) {
      points[i].x += OFF;
      points[i].y += OFF;
      xCoor = points[i].x;
      yCoor = points[i].y;
      minX[yCoor] = min(minX[yCoor], xCoor);
      maxX[yCoor] = max(maxX[yCoor], xCoor);
      minY[xCoor] = min(minY[xCoor], yCoor);
      maxY[xCoor] = max(maxY[xCoor], yCoor);
   }
   int pointCount = 0;
   for (int i = 0; i < n; i++) {
      xCoor = points[i].x;
      yCoor = points[i].y;
      if (xCoor > minX[yCoor] && xCoor < maxX[yCoor])
         if (yCoor > minY[xCoor] && yCoor < maxY[xCoor])
            pointCount++;
   }
   return pointCount;
}
int main(){
   struct point points[] = {{1, 1}, {1, 0}, {0, 1}, {1, 2}, {2, 1}};
   int n = sizeof(points) / sizeof(points[0]);
   cout<<"The number of points that have atleast one point above, below, left, right is "<<findPointCount(n, points);
}

Output

The number of points that have atleast one point above, below, left, right is 1
raja
Updated on 24-Jan-2022 13:24:34

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