给定两个由N和M个整数组成的数组X []和Y [] ，使得有N条线平行于y轴， M条线平行于x轴。任务是找到在坐标平面上由这些线形成的正方形总数。
Each integer(say a) in the array X[] denotes lines having equation x = a, parallel to y-axis.
Each integer(say b) in the array Y[] denotes lines having equation y = b, parallel to x-axis.
例子：
Input: N = 3, M = 4, X[] = {1, 3, 7}, Y[] = {2, 4, 6, 1}
Output: 3
Explanation:
3 lines are parallel to y-axis for x = 1, x = 3 and x = 7.
4 lines are parallel to x-axis for y = 2, y = 4, y = 6 and y = 1.
From the above image, below are three possible squares formed:
1) square CDEF (x = 1, x = 3, y = 2, y = 4), side = 2 units.
2) square ABDC (x = 1, x = 3, y = 4, y = 6), side = 2 units.
3) square BGHF (x = 3, x = 7, y = 2, y = 6), side = 4 units.
Input: N = 5, M = 4, X[] = {1, 9, 2, 3, 7}, Y[] = {1, 2, 4, 6}
Output: 8
方法：请按照以下步骤解决问题：
- 在X []数组中找到所有对之间的距离，并将计数存储在Map中，例如M1 。
- 在Y []数组中找到所有对之间的距离，并将计数存储在Map M2中。
- 如果在M2中存在M1对的距离，则可以通过使用这两个对来形成平方。
- 因此，可以通过将存储在M1和M2中的所有距离计数相加来计算平方总数。
- 完成上述步骤后，打印平方总数。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to count all the possible
// squares with given lines parallel
// to both the X and Y axis
int numberOfSquares(int X[], int Y[],
int N, int M)
{
// Stores the count of all possible
// distances in X[] & Y[] respectively
unordered_map m1, m2;
int i, j, ans = 0;
// Find distance between all
// pairs in the array X[]
for (i = 0; i < N; i++) {
for (j = i + 1; j < N; j++) {
int dist = abs(X[i] - X[j]);
// Add the count to m1
m1[dist]++;
}
}
// Find distance between all
// pairs in the array Y[]
for (i = 0; i < M; i++) {
for (j = i + 1; j < M; j++) {
int dist = abs(Y[i] - Y[j]);
// Add the count to m2
m2[dist]++;
}
}
// Find sum of m1[i] * m2[i]
// for same distance
for (auto i = m1.begin();
i != m1.end(); i++) {
// Find current count in m2
if (m2.find(i->first)
!= m2.end()) {
// Add to the total count
ans += (i->second
* m2[i->first]);
}
}
// Return the final count
return ans;
}
// Driver Code
int main()
{
// Given lines
int X[] = { 1, 3, 7 };
int Y[] = { 2, 4, 6, 1 };
int N = sizeof(X) / sizeof(X[0]);
int M = sizeof(Y) / sizeof(Y[0]);
// Function Call
cout << numberOfSquares(X, Y, N, M);
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG{
// Function to count all the possible
// squares with given lines parallel
// to both the X and Y axis
static int numberOfSquares(int[] X, int[] Y, int N,
int M)
{
// Stores the count of all possible
// distances in X[] & Y[] respectively
HashMap m1 = new HashMap();
HashMap m2 = new HashMap();
int i, j, ans = 0;
// Find distance between all
// pairs in the array X[]
for(i = 0; i < N; i++)
{
for(j = i + 1; j < N; j++)
{
int dist = Math.abs(X[i] - X[j]);
// Add the count to m1
m1.put(dist, m1.getOrDefault(dist, 0) + 1);
}
}
// Find distance between all
// pairs in the array Y[]
for(i = 0; i < M; i++)
{
for(j = i + 1; j < M; j++)
{
int dist = Math.abs(Y[i] - Y[j]);
// Add the count to m2
m2.put(dist, m2.getOrDefault(dist, 0) + 1);
}
}
// Find sum of m1[i] * m2[i]
// for same distance
for(Map.Entry entry : m1.entrySet())
{
// Find current count in m2
if (m2.containsKey(entry.getKey()))
{
// Add to the total count
ans += (entry.getValue() *
m2.get(entry.getKey()));
}
}
// Return the final count
return ans;
}
// Driver Code
public static void main(String[] args)
{
// Given lines
int X[] = { 1, 3, 7 };
int Y[] = { 2, 4, 6, 1 };
int N = X.length;
int M = Y.length;
// Function call
System.out.println(numberOfSquares(X, Y, N, M));
}
}
// This code is contributed by akhilsaini Python3
# Python3 program for the above approach
# Function to count all the possible
# squares with given lines parallel
# to both the X and Y axis
def numberOfSquares(X, Y, N, M):
# Stores the count of all possible
# distances in X[] & Y[] respectively
m1 = {}
m2 = {}
ans = 0
# Find distance between all
# pairs in the array X[]
for i in range(0, N):
for j in range(i + 1, N):
dist = abs(X[i] - X[j])
# Add the count to m1
if dist in m1:
m1[dist] = m1[dist] + 1
else:
m1[dist] = 1
# Find distance between all
# pairs in the array Y[]
for i in range(0, M):
for j in range(i + 1, M):
dist = abs(Y[i] - Y[j])
# Add the count to m2
if dist in m2:
m2[dist] = m2[dist] + 1
else:
m2[dist] = 1
# Find sum of m1[i] * m2[i]
# for same distance
for key in m1:
# Find current count in m2
if key in m2:
# Add to the total count
ans = ans + (m1[key] * m2[key])
# Return the final count
return ans
# Driver Code
if __name__ == "__main__":
# Given lines
X = [ 1, 3, 7 ]
Y = [ 2, 4, 6, 1 ]
N = len(X)
M = len(Y)
# Function call
print(numberOfSquares(X, Y, N, M))
# This code is contributed by akhilsainiC#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to count all the possible
// squares with given lines parallel
// to both the X and Y axis
static int numberOfSquares(int[] X, int[] Y, int N,
int M)
{
// Stores the count of all possible
// distances in X[] & Y[] respectively
Dictionary m1 = new Dictionary();
Dictionary m2 = new Dictionary();
int i, j, ans = 0;
// Find distance between all
// pairs in the array X[]
for(i = 0; i < N; i++)
{
for(j = i + 1; j < N; j++)
{
int dist = Math.Abs(X[i] - X[j]);
// Add the count to m1
if (m1.ContainsKey(dist))
m1[dist]++;
else
m1.Add(dist, 1);
}
}
// Find distance between all
// pairs in the array Y[]
for(i = 0; i < M; i++)
{
for(j = i + 1; j < M; j++)
{
int dist = Math.Abs(Y[i] - Y[j]);
// Add the count to m2
if (m2.ContainsKey(dist))
m2[dist]++;
else
m2.Add(dist, 1);
}
}
// Find sum of m1[i] * m2[i]
// for same distance
foreach(KeyValuePair entry in m1)
{
// Find current count in m2
if (m2.ContainsKey(entry.Key))
{
// Add to the total count
ans += (entry.Value *
m2[entry.Key]);
}
}
// Return the final count
return ans;
}
// Driver Code
public static void Main()
{
// Given lines
int[] X = { 1, 3, 7 };
int[] Y = { 2, 4, 6, 1 };
int N = X.Length;
int M = Y.Length;
// Function call
Console.WriteLine(numberOfSquares(X, Y, N, M));
}
}
// This code is contributed by akhilsaini 输出：
3
时间复杂度： O(N 2 )
辅助空间： O(N)