给定两个阵列ARR1 []以及N表示的形式的N个坐标(ARR1 [I]，0)和M的正整数，表示的形式的M个坐标(ARR2 [J]，1)，其中1≤I≤ÑARR2 []和1≤j≤M 。任务是找到可以使用这些坐标形成的最大矩形区域。如果无法形成矩形，则打印0 。
例子：
Input: arr1[] = {1, 2, 4}, arr2[] = {1, 3, 4}
Output: 3
Explanation: The largest rectangle possible is {(1, 0), (1, 1), (4, 0), (4, 1)}.
Therefore, area of rectangle = length * breadth = (4 – 1)*(1 – 0) = 3
Input: arr1[] = {1, 3, 5}, arr2[] = {4, 2, 10}
Output: 0
Explanation: No rectangle can be formed. Therefore, the answer is zero.
天真的方法：最简单的方法是遍历给定的数组arr1 [] ，对于每个i ，使用变量j遍历arr2 []中的点。如果arr1 [i]等于arr2 [j] ，则将值arr1 [i]存储在单独的数组中，例如ans [] 。找到所有这些值之后，对ans []数组进行排序并将最大面积打印为ans [L] – ans [0] ，其中L是ans []的最后一个元素的索引，因为矩形的宽度将始终为1 。
时间复杂度： O(N * M)
辅助空间： O(M + N)
高效的方法：想法是使用排序算法和两指针技术。请注意，矩形的宽度将始终为1 。因此，可以通过最大长度来找到最大面积。请按照以下步骤解决问题：
- 对给定数组arr1 []和arr2 []进行排序。
- 初始化变量，开始和结束0来存储长度的开始和结束点。另外，初始化变量i和j分别遍历数组arr1 []和arr2 [] 。
- 当i小于N且j小于M时，请检查以下条件：
- 如果arr1 [i]与arr2 [j]相同且start为0 ，则将start更新为start = arr1 [i] 。否则，更新结束作为端= ARR1 [I]然后通过递增1 i和j。
- 如果ARR1 [i]是比ARR2 [j]时，通过1个增量Ĵ更大。否则，将i增加1 。
- 如果开始或结束为0 ，则打印0 。否则，请打印(结束–开始)作为最大可能区域。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the maximum possible
// area of a rectangle
int largestArea(int arr1[], int n,
int arr2[], int m)
{
// Initialize variables
int end = 0, start = 0, i = 0, j = 0;
// Sort array arr1[]
sort(arr1, arr1 + n);
// Sort array arr2[]
sort(arr2, arr2 + m);
// Traverse arr1[] and arr2[]
while (i < n and j < m) {
// If arr1[i] is same as arr2[j]
if (arr1[i] == arr2[j]) {
// If no starting point
// is found yet
if (start == 0)
start = arr1[i];
else
// Update maximum end
end = arr1[i];
i++;
j++;
}
// If arr[i] > arr2[j]
else if (arr1[i] > arr2[j])
j++;
else
i++;
}
// If no rectangle is found
if (end == 0 or start == 0)
return 0;
else
// Return the area
return (end - start);
}
// Driver Code
int main()
{
// Given point
int arr1[] = { 1, 2, 4 };
// Given length
int N = sizeof(arr1) / sizeof(arr1[0]);
// Given points
int arr2[] = { 1, 3, 4 };
// Given length
int M = sizeof(arr2) / sizeof(arr2[0]);
// Function Call
cout << largestArea(arr1, N, arr2, M);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the maximum possible
// area of a rectangle
static int largestArea(int arr1[], int n,
int arr2[], int m)
{
// Initialize variables
int end = 0, start = 0, i = 0, j = 0;
// Sort array arr1[]
Arrays.sort(arr1);
// Sort array arr2[]
Arrays.sort(arr1);
// Traverse arr1[] and arr2[]
while (i < n && j < m)
{
// If arr1[i] is same as arr2[j]
if (arr1[i] == arr2[j])
{
// If no starting point
// is found yet
if (start == 0)
start = arr1[i];
else
// Update maximum end
end = arr1[i];
i++;
j++;
}
// If arr[i] > arr2[j]
else if (arr1[i] > arr2[j])
j++;
else
i++;
}
// If no rectangle is found
if (end == 0 || start == 0)
return 0;
else
// Return the area
return (end - start);
}
// Driver Code
public static void main(String args[])
{
// Given point
int arr1[] = { 1, 2, 4 };
// Given length
int N = arr1.length;
// Given points
int arr2[] = { 1, 3, 4 };
// Given length
int M = arr2.length;
// Function Call
System.out.println(largestArea(arr1, N,
arr2, M));
}
}
// This code is contributed by bolliranadheerPython3
# Python3 program for the above approach
# Function to find the maximum possible
# area of a rectangle
def largestArea(arr1, n, arr2, m):
# Initialize variables
end = 0
start = 0
i = 0
j = 0
# Sort array arr1[]
arr1.sort(reverse = False)
# Sort array arr2[]
arr2.sort(reverse = False)
# Traverse arr1[] and arr2[]
while (i < n and j < m):
# If arr1[i] is same as arr2[j]
if (arr1[i] == arr2[j]):
# If no starting point
# is found yet
if (start == 0):
start = arr1[i]
else:
# Update maximum end
end = arr1[i]
i += 1
j += 1
# If arr[i] > arr2[j]
elif (arr1[i] > arr2[j]):
j += 1
else:
i += 1
# If no rectangle is found
if (end == 0 or start == 0):
return 0
else:
# Return the area
return (end - start)
# Driver Code
if __name__ == '__main__':
# Given point
arr1 = [ 1, 2, 4 ]
# Given length
N = len(arr1)
# Given points
arr2 = [ 1, 3, 4 ]
# Given length
M = len(arr2)
# Function Call
print(largestArea(arr1, N, arr2, M))
# This code is contributed by ipg2016107C#
// C# program for the above approach
using System;
class GFG{
// Function to find the maximum possible
// area of a rectangle
static int largestArea(int[] arr1, int n,
int[] arr2, int m)
{
// Initialize variables
int end = 0, start = 0, i = 0, j = 0;
// Sort array arr1[]
Array.Sort(arr1);
// Sort array arr2[]
Array.Sort(arr2);
// Traverse arr1[] and arr2[]
while (i < n && j < m)
{
// If arr1[i] is same as arr2[j]
if (arr1[i] == arr2[j])
{
// If no starting point
// is found yet
if (start == 0)
start = arr1[i];
else
// Update maximum end
end = arr1[i];
i++;
j++;
}
// If arr[i] > arr2[j]
else if (arr1[i] > arr2[j])
j++;
else
i++;
}
// If no rectangle is found
if (end == 0 || start == 0)
return 0;
else
// Return the area
return (end - start);
}
// Driver code
static void Main()
{
// Given point
int[] arr1 = { 1, 2, 4 };
// Given length
int N = arr1.Length;
// Given points
int[] arr2 = { 1, 3, 4 };
// Given length
int M = arr2.Length;
// Function Call
Console.WriteLine(largestArea(arr1, N,
arr2, M));
}
}
// This code is contributed by divyeshrabadiya07Javascript
输出：
3时间复杂度： O(N * log N + M * log M)
辅助空间： O(M + N)