给定一个由笛卡尔平面中的N 个坐标组成的数组arr[] ,任务是找到满足以下所有条件的坐标数,例如(X, Y) :
- 所有可能的arr[i][0]必须小于X并且arr[i][1]必须等于Y 。
- 所有可能的arr[i][0]必须大于X并且arr[i][1]必须等于Y 。
- 所有可能的arr[i][0]必须小于Y并且arr[i][1]必须等于X 。
- 所有可能的arr[i][0]必须大于Y并且arr[i][1]必须等于X 。
例子:
Input: arr[][] = {{0, 0}, {0, 1}, {1, 0}, {0, -1}, {-1, 0}}
Output: 1
Explanation: There exists only one coordinate that satisfy the given condition, i.e. (0, 0), based on the following conditions:
- arr[2] = {1, 0}: Since arr[2][0](= 1) > X(= 0) and arr[2][1](= 0) == Y(= 0), condition 1 is satisfied.
- arr[4] = {-1, 0}: Since arr[4][0](= -1) < 0 and arr[4][1](= 0) == Y(= 0), condition 2 is satisfied.
- arr[1] = {0, 1}: Since arr[1][0](= 0) == X(= 0) and arr[1][1](= 1) > Y(= 0), condition 3 is satisfied.
- arr[3] = {0, -1}: Since arr[3][0](= 0) == X(= 0) and arr[3][1](= -1) < Y(= 0), condition 4 is satisfied.
Therefore, the total number of points is 1.
Input: arr[][] = {{1, 0}, {2, 0}, {1, 1}, {1, -1}}
Output: 0
方法:给定的问题可以通过考虑每个坐标来解决,比如(arr[i][0], arr[i][1])作为结果坐标,如果坐标(arr[i][0], arr [i][1])满足所有给定条件,然后计算当前坐标。检查所有坐标后,打印获得的总计数值。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
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
输出:
0
时间复杂度: O(N 2 )
辅助空间: O(1)
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