可以使用给定的矩形块形成的最大正方形

给定正整数的数组arr [] ，其中数组的每个元素代表矩形块的长度。任务是找到可以使用矩形块形成的正方形的最大长度。

例子：

Input: arr[] = {3, 2, 1, 5, 2, 4} 
Output: 3 
Explanation: 
Using rectangular block of length 3, 5 and 4, square of side length 3 can be constructed as shown below: 
 

Input: arr[] = {1, 2, 3} 
Output: 2 

方法：

1. 以降序对给定数组进行排序。
2. 初始化最大边长(例如maxLength )为0。
3. 遍历数组arr [] ，如果arr [i]> maxLength，则增加maxLength并检查此条件以进行下一次迭代。
4. 如果上述条件不满足，则中断循环并打印maxLength 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find maximum side
// length of square
void maxSide(int a[], int n)
{
    int sideLength = 0;
 
    // Sort array in asc order
    sort(a, a + n);
 
    // Traverse array in desc order
    for (int i = n - 1; i >= 0; i--) {
 
        if (a[i] > sideLength) {
            sideLength++;
        }
        else {
            break;
        }
    }
    cout << sideLength << endl;
}
 
// Driver Code
int main()
{
    int N = 6;
 
    // Given array arr[]
    int arr[] = { 3, 2, 1, 5, 2, 4 };
 
    // Function Call
    maxSide(arr, N);
    return 0;
}

Java

// Java program for the above approach
import java.util.Arrays;
 
class GFG{
     
// Function to find maximum side
// length of square
static void maxSide(int a[], int n)
{
    int sideLength = 0;
 
    // Sort array in asc order
    Arrays.sort(a);
 
    // Traverse array in desc order
    for(int i = n - 1; i >= 0; i--)
    {
       if (a[i] > sideLength)
       {
           sideLength++;
       }
       else
       {
           break;
       }
    }
    System.out.println(sideLength);
}
     
// Driver code
public static void main (String[] args)
{
    int N = 6;
 
    // Given array arr[]
    int arr[] = new int[]{ 3, 2, 1,
                           5, 2, 4 };
     
    // Function Call
    maxSide(arr, N);
}
}
 
// This code is contributed by Pratima Pandey

Python3

# Python3 program for the above approach
 
# Function to find maximum side
# length of square
def maxSide(a, n):
 
    sideLength = 0
 
    # Sort array in asc order
    a.sort
 
    # Traverse array in desc order
    for i in range(n - 1, -1, -1):
        if (a[i] > sideLength):
            sideLength += 1
        else:
            break
             
    print(sideLength)
     
# Driver code
N = 6
 
# Given array arr[]
arr = [ 3, 2, 1, 5, 2, 4 ]
 
# Function Call
maxSide(arr, N)
 
# This code is contributed by divyeshrabadiya07

C#

// C# program for the above approach
using System;
class GFG{
     
// Function to find maximum side
// length of square
static void maxSide(int []a, int n)
{
    int sideLength = 0;
 
    // Sort array in asc order
    Array.Sort(a);
 
    // Traverse array in desc order
    for(int i = n - 1; i >= 0; i--)
    {
        if (a[i] > sideLength)
        {
            sideLength++;
        }
        else
        {
            break;
        }
    }
    Console.Write(sideLength);
}
     
// Driver code
public static void Main()
{
    int N = 6;
 
    // Given array arr[]
    int []arr = new int[]{ 3, 2, 1,
                           5, 2, 4 };
     
    // Function Call
    maxSide(arr, N);
}
}
 
// This code is contributed by Code_Mech

Javascript

// Javascript program for the above approach
    // Function to find maximum side
    // length of square
    function maxSide( a, n) {
        let sideLength = 0;
 
        // Sort array in asc order
        a.sort();
 
        // Traverse array in desc order
        for ( i = n - 1; i >= 0; i--) {
            if (a[i] > sideLength) {
                sideLength++;
            } else {
                break;
            }
        }
        document.write(sideLength);
    }
 
    // Driver code
      
        let N = 6;
 
        // Given array arr
        let arr = [3, 2, 1, 5, 2, 4 ];
 
        // Function Call
        maxSide(arr, N);
     
 
// This code contributed by aashish1995

输出：

3

时间复杂度： O(N * log N)
辅助空间： O(1)