Dada una array arr[] que consta de N coordenadas en el plano cartesiano , la tarea es encontrar el número de coordenadas, como (X, Y) , que satisface todas las siguientes condiciones:
- Todos los posibles arr[i][0] deben ser menores que X y arr[i][1] deben ser iguales a Y .
- Todos los arr[i][0] posibles deben ser mayores que X y arr[i][1] debe ser igual a Y .
- Todos los posibles arr[i][0] deben ser menores que Y y arr[i][1] deben ser iguales a X .
- Todos los arr[i][0] posibles deben ser mayores que Y y arr[i][1] debe ser igual a X .
Ejemplos:
Entrada: arr[][] = {{0, 0}, {0, 1}, {1, 0}, {0, -1}, {-1, 0}}
Salida: 1
Explicación: Solo existe una coordenada que satisface la condición dada, es decir (0, 0) , en base a las siguientes condiciones:
- arr[2] = {1, 0}: Dado que arr[2][0](= 1) > X(= 0) y arr[2][1](= 0) == Y(= 0), condición 1 está satisfecho.
- arr[4] = {-1, 0}: Dado que arr[4][0](= -1) < 0 y arr[4][1](= 0) == Y(= 0), la condición 2 es satisfecho.
- arr[1] = {0, 1}: Dado que arr[1][0](= 0) == X(= 0) y arr[1][1](= 1) > Y(= 0), condición 3 está satisfecho.
- arr[3] = {0, -1}: Dado que arr[3][0](= 0) == X(= 0) y arr[3][1](= -1) < Y(= 0) , se cumple la condición 4.
Por lo tanto, el número total de puntos es 1.
Entrada: arr[][] = {{1, 0}, {2, 0}, {1, 1}, {1, -1}}
Salida: 0
Enfoque: El problema dado se puede resolver considerando cada coordenada, digamos (arr[i][0], arr[i][1]) como las coordenadas resultantes, y si la coordenada (arr[i][0], arr [i][1]) satisface todas las condiciones dadas, luego cuente las coordenadas actuales. Después de verificar todas las coordenadas, imprima el valor del conteo total obtenido.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the number of
// coordinates from a given set
// that satisfies the given conditions
int centralPoints(int arr[][2], int N)
{
// Stores the count of central points
int count = 0;
// Store the count of
// each x and y coordinates
int c1, c2, c3, c4;
// Find all possible pairs
for (int i = 0; i < N; i++) {
// Initialize variables c1, c2,
// c3, c4 to define the status
// of conditions
c1 = 0, c2 = 0, c3 = 0;
c4 = 0;
// Stores value of each point
int x = arr[i][0];
int y = arr[i][1];
// Check the conditions for
// each point by generating
// all possible pairs
for (int j = 0; j < N; j++) {
// If arr[j][0] > x and
// arr[j][1] == y
if (arr[j][0] > x
&& arr[j][1] == y) {
c1 = 1;
}
// If arr[j][0] < x and
// arr[j][1] == y
if (arr[j][1] > y
&& arr[j][0] == x) {
c2 = 1;
}
// If arr[j][1] > y and
// arr[j][0] == x
if (arr[j][0] < x
&& arr[j][1] == y) {
c3 = 1;
}
// If arr[j][1] < y and
// arr[j][0] == x
if (arr[j][1] < y
&& arr[j][0] == x) {
c4 = 1;
}
}
// If all conditions satisfy
// then point is central point
if (c1 + c2 + c3 + c4 == 4) {
// Increment the count by 1
count++;
}
}
// Return the count
return count;
}
// Driver Code
int main()
{
int arr[4][2] = { { 1, 0 }, { 2, 0 },
{ 1, 1 }, { 1, -1 } };
int N = sizeof(arr) / sizeof(arr[0]);
cout << centralPoints(arr, N);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG {
// Function to count the number of
// coordinates from a given set
// that satisfies the given conditions
static int centralPoints(int arr[][], int N)
{
// Stores the count of central points
int count = 0;
// Store the count of
// each x and y coordinates
int c1, c2, c3, c4;
// Find all possible pairs
for (int i = 0; i < N; i++) {
// Initialize variables c1, c2,
// c3, c4 to define the status
// of conditions
c1 = 0;
c2 = 0;
c3 = 0;
c4 = 0;
// Stores value of each point
int x = arr[i][0];
int y = arr[i][1];
// Check the conditions for
// each point by generating
// all possible pairs
for (int j = 0; j < N; j++) {
// If arr[j][0] > x and
// arr[j][1] == y
if (arr[j][0] > x && arr[j][1] == y) {
c1 = 1;
}
// If arr[j][0] < x and
// arr[j][1] == y
if (arr[j][1] > y && arr[j][0] == x) {
c2 = 1;
}
// If arr[j][1] > y and
// arr[j][0] == x
if (arr[j][0] < x && arr[j][1] == y) {
c3 = 1;
}
// If arr[j][1] < y and
// arr[j][0] == x
if (arr[j][1] < y && arr[j][0] == x) {
c4 = 1;
}
}
// If all conditions satisfy
// then point is central point
if (c1 + c2 + c3 + c4 == 4) {
// Increment the count by 1
count++;
}
}
// Return the count
return count;
}
// Driver Code
public static void main(String[] args)
{
int arr[][]
= { { 1, 0 }, { 2, 0 }, { 1, 1 }, { 1, -1 } };
int N = arr.length;
System.out.println(centralPoints(arr, N));
}
}
// This code is contributed by Kingash.
Python3
# Python3 program for the above approach # Function to count the number of # coordinates from a given set # that satisfies the given conditions def centralPoints(arr, N): # Stores the count of central points count = 0 # Find all possible pairs for i in range(N): # Initialize variables c1, c2, # c3, c4 to define the status # of conditions c1 = 0 c2 = 0 c3 = 0 c4 = 0 # Stores value of each point x = arr[i][0] y = arr[i][1] # Check the conditions for # each point by generating # all possible pairs for j in range(N): # If arr[j][0] > x and # arr[j][1] == y if (arr[j][0] > x and arr[j][1] == y): c1 = 1 # If arr[j][0] < x and # arr[j][1] == y if (arr[j][1] > y and arr[j][0] == x): c2 = 1 # If arr[j][1] > y and # arr[j][0] == x if (arr[j][0] < x and arr[j][1] == y): c3 = 1 # If arr[j][1] < y and # arr[j][0] == x if (arr[j][1] < y and arr[j][0] == x): c4 = 1 # If all conditions satisfy # then point is central point if (c1 + c2 + c3 + c4 == 4): # Increment the count by 1 count += 1 # Return the count return count # Driver Code if __name__ == '__main__': arr = [ [ 1, 0 ], [ 2, 0 ], [ 1, 1 ], [ 1, -1 ] ] N = len(arr) print(centralPoints(arr, N)); # This code is contributed by SURENDRA_GANGWAR
C#
// C# program for the above approach
using System;
class GFG{
// Function to count the number of
// coordinates from a given set
// that satisfies the given conditions
static int centralPoints(int[,] arr, int N)
{
// Stores the count of central points
int count = 0;
// Store the count of
// each x and y coordinates
int c1, c2, c3, c4;
// Find all possible pairs
for(int i = 0; i < N; i++)
{
// Initialize variables c1, c2,
// c3, c4 to define the status
// of conditions
c1 = 0;
c2 = 0;
c3 = 0;
c4 = 0;
// Stores value of each point
int x = arr[i, 0];
int y = arr[i, 1];
// Check the conditions for
// each point by generating
// all possible pairs
for(int j = 0; j < N; j++)
{
// If arr[j][0] > x and
// arr[j][1] == y
if (arr[j, 0] > x && arr[j, 1] == y)
{
c1 = 1;
}
// If arr[j][0] < x and
// arr[j][1] == y
if (arr[j, 1] > y && arr[j, 0] == x)
{
c2 = 1;
}
// If arr[j][1] > y and
// arr[j][0] == x
if (arr[j, 0] < x && arr[j, 1] == y)
{
c3 = 1;
}
// If arr[j][1] < y and
// arr[j][0] == x
if (arr[j, 1] < y && arr[j, 0] == x)
{
c4 = 1;
}
}
// If all conditions satisfy
// then point is central point
if (c1 + c2 + c3 + c4 == 4)
{
// Increment the count by 1
count++;
}
}
// Return the count
return count;
}
// Driver Code
public static void Main(string[] args)
{
int[,] arr = { { 1, 0 }, { 2, 0 },
{ 1, 1 }, { 1, -1 } };
int N = arr.GetLength(0);
Console.WriteLine(centralPoints(arr, N));
}
}
// This code is contributed by ukasp
Javascript
<script>
// Javascript program to implement the
// above approach
// Function to count the number of
// coordinates from a given set
// that satisfies the given conditions
function centralPoints(arr, N)
{
// Stores the count of central points
let count = 0;
// Store the count of
// each x and y coordinates
let c1, c2, c3, c4;
// Find all possible pairs
for (let i = 0; i < N; i++) {
// Initialize variables c1, c2,
// c3, c4 to define the status
// of conditions
c1 = 0;
c2 = 0;
c3 = 0;
c4 = 0;
// Stores value of each point
let x = arr[i][0];
let y = arr[i][1];
// Check the conditions for
// each point by generating
// all possible pairs
for (let j = 0; j < N; j++) {
// If arr[j][0] > x and
// arr[j][1] == y
if (arr[j][0] > x && arr[j][1] == y) {
c1 = 1;
}
// If arr[j][0] < x and
// arr[j][1] == y
if (arr[j][1] > y && arr[j][0] == x) {
c2 = 1;
}
// If arr[j][1] > y and
// arr[j][0] == x
if (arr[j][0] < x && arr[j][1] == y) {
c3 = 1;
}
// If arr[j][1] < y and
// arr[j][0] == x
if (arr[j][1] < y && arr[j][0] == x) {
c4 = 1;
}
}
// If all conditions satisfy
// then point is central point
if (c1 + c2 + c3 + c4 == 4) {
// Increment the count by 1
count++;
}
}
// Return the count
return count;
}
// Driver Code
let arr = [[ 1, 0 ], [ 2, 0 ], [ 1, 1 ], [ 1, -1 ]];
let N = arr.length;
document.write(centralPoints(arr, N));
// This code is contributed by splevel62.
</script>
Producción:
0
Complejidad Temporal: O(N 2 )
Espacio Auxiliar: O(1), ya que no se ha tomado ningún espacio extra.
Publicación traducida automáticamente
Artículo escrito por abhijeet010304 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA