大小为K的子数组的计数，它是数字从1到K的排列

给定一个由不同整数组成的数组arr ，任务是查找大小为i的子数组的数量，该子数组具有从1到i的所有元素，换句话说，子数组是元素从1到i的任何排列，其中1 <= i <= N。
例子：

Input: arr[] = {2, 3, 1, 5, 4} 
Output: 3 
Explanation: 
we have {1}, {2, 3, 1} and {2, 3, 1, 5, 4} subarray for i=1, i=3, i=5 respectively. 
Permutation of size 4 and size 2 can’t be made because 5 and 3 are in the way respectively.

Input: arr[] = {1, 3, 5, 4, 2} 
Output: 2 
Explanation: 
we have {1} and {1, 3, 5, 4, 2} subarray for i=1 and i=5 respectively.

幼稚的方法是从每个索引开始，尝试找到每个大小(i)的子数组，并检查是否存在从1到i的所有元素。

时间复杂度：O(N ² )

通过检查是否有可能为i的每个值(从1到N)创建大小为i的子数组，从而提供一种有效的方法。
我们知道大小为K的每个子数组必须是从1到K的所有元素的排列，知道我们可以按顺序查看从1到N的数字的索引，并在每一步中计算最小值和最大值的索引。

• 如果maximum_ind – minimum_ind + 1 = K ，则我们有一个大小为K的排列，否则就没有。
• 在每个步骤中更新minimum_ind和maximum_ind的值。

时间复杂度： O(n)
插图：

Given Arr = {2, 3, 1, 5, 4}, let’s start with min_ind = INF and max_ind = -1

1. index of 1 is 2, so min_ind = min(min_ind, 2) = 2 and max_ind = max(max_ind, 2) = 2, 
   2-2+1 = 1 so we have a permutation of size 1
2. index of 2 is 0, so min_ind = min(min_ind, 0) = 0 and max_ind = max(max_ind, 0) = 2, 
   2-0+1 = 3 so we don’t have a permutation of size 2
3. index of 3 is 1, so min_ind = min(min_ind, 1) = 0 and max_ind = max(max_ind, 1) = 2, 
   2-0+1 = 3 so we have a permutation of size 3
4. index of 4 is 4, so min_ind = min(min_ind, 4) = 0 and max_ind = max(max_ind, 4) = 4, 
   4-0+1 = 5 so we don’t have a permutation of size 4
5. index of 5 is 3, so min_ind = min(min_ind, 3) = 0 and max_ind = max(max_ind, 4) = 4, 
   4-0+1 = 5 so we have a permutation of size 5

So answer is 3

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
  
#include 
#include 
#include 
using namespace std;
  
int find_permutations(vector& arr)
{
    int cnt = 0;
    int max_ind = -1, min_ind = 10000000;
    int n = arr.size();
    unordered_map index_of;
  
    // Save index of numbers of the array
    for (int i = 0; i < n; i++) {
        index_of[arr[i]] = i + 1;
    }
  
    for (int i = 1; i <= n; i++) {
  
        // Update min and max index
        // with the current index
        // and check if it's a valid permutation
        max_ind = max(max_ind, index_of[i]);
        min_ind = min(min_ind, index_of[i]);
        if (max_ind - min_ind + 1 == i)
            cnt++;
    }
  
    return cnt;
}
  
// Driver code
int main()
{
    vector nums;
    nums.push_back(2);
    nums.push_back(3);
    nums.push_back(1);
    nums.push_back(5);
    nums.push_back(4);
  
    cout << find_permutations(nums);
  
    return 0;
}

Java

// Java program to implement 
// the above approach 
import java.util.*;
  
class GFG{
      
public static int find_permutations(
    Vector arr) 
{ 
    int cnt = 0; 
    int max_ind = -1, min_ind = 10000000; 
    int n = arr.size(); 
      
    HashMap index_of = new HashMap<>(); 
              
    // Save index of numbers of the array 
    for(int i = 0; i < n; i++) 
    { 
        index_of.put(arr.get(i), i + 1);
    } 
  
    for(int i = 1; i <= n; i++) 
    { 
          
        // Update min and max index with 
        // the current index and check 
        // if it's a valid permutation 
        max_ind = Math.max(max_ind, index_of.get(i)); 
        min_ind = Math.min(min_ind, index_of.get(i)); 
          
        if (max_ind - min_ind + 1 == i) 
            cnt++; 
    } 
    return cnt; 
} 
  
// Driver Code
public static void main(String[] args)
{
    Vector nums = new Vector(); 
    nums.add(2); 
    nums.add(3); 
    nums.add(1); 
    nums.add(5); 
    nums.add(4); 
      
    System.out.print(find_permutations(nums));
}
}
  
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to implement
# the above approach
def find_permutations(arr):
      
    cnt = 0
    max_ind = -1
    min_ind = 10000000;
      
    n = len(arr)
    index_of = {}
  
    # Save index of numbers of the array
    for i in range(n):
        index_of[arr[i]] = i + 1
  
    for i in range(1, n + 1):
  
        # Update min and max index with the
        # current index and check if it's a
        # valid permutation
        max_ind = max(max_ind, index_of[i])
        min_ind = min(min_ind, index_of[i])
          
        if (max_ind - min_ind + 1 == i):
            cnt += 1
              
    return cnt
  
# Driver code
if __name__ == "__main__":
  
    nums = []
    nums.append(2)
    nums.append(3)
    nums.append(1)
    nums.append(5)
    nums.append(4)
  
    print(find_permutations(nums))
  
# This code is contributed by chitranayal

C#

// C# program to implement
// the above approach
using System; 
using System.Collections; 
using System.Collections.Generic; 
  
class GFG{
      
static int find_permutations(ArrayList arr)
{
    int cnt = 0;
    int max_ind = -1, min_ind = 10000000;
    int n = arr.Count;
      
    Dictionary index_of = new Dictionary(); 
              
    // Save index of numbers of the array
    for(int i = 0; i < n; i++) 
    {
        index_of[(int)arr[i]] = i + 1;
    }
  
    for(int i = 1; i <= n; i++)
    {
          
        // Update min and max index with 
        // the current index and check 
        // if it's a valid permutation
        max_ind = Math.Max(max_ind, index_of[i]);
        min_ind = Math.Min(min_ind, index_of[i]);
          
        if (max_ind - min_ind + 1 == i)
            cnt++;
    }
    return cnt;
}
  
// Driver Code
public static void Main(string[] args)
{
    ArrayList nums = new ArrayList(); 
  
    nums.Add(2);
    nums.Add(3);
    nums.Add(1);
    nums.Add(5);
    nums.Add(4);
  
    Console.Write(find_permutations(nums));
}
}
  
// This code is contributed by rutvik_56

输出：

3