从给定的数组元素形成排列的子数组的计数

给定一个由整数[1, N ]组成的数组A[] ，任务是计算所有可能长度x ( 1 ≤ x ≤ N ) 的子数组的总数，由整数[1, x]的排列组成给定的数组。

例子：

Input: A[] = {3, 1, 2, 5, 4}  Output: 4 
Explanation: 
Subarrays forming a permutation are {1}, {1, 2}, {3, 1, 2} and {3, 1, 2, 5, 4}.
 

Input: A[] = {4, 5, 1, 3, 2, 6} Output: 4 
Explanation: 
Subarrays forming a permutation are {1}, {1, 3, 2}, {4, 5, 1, 3, 2} and {4, 5, 1, 3, 2, 6}.  

天真的方法：
请按照以下步骤解决问题：

• 解决问题的最简单方法是生成所有可能的子数组。
• 对于每个子数组，检查它是否是[1, length of subarray]范围内元素的排列。
• 对于找到的每个这样的子数组，增加count 。最后，打印计数。

时间复杂度： O(N ³ )
辅助空间： O(1)

有效的方法：
要优化上述方法，请按照以下步骤操作：

• 对于i = [1, N] 中的每个元素，检查最大和最小 index ，其中存在排列[1, i]的元素。
• 如果最大和最小索引之间的差等于i ，则意味着i存在有效的连续排列。
• 对于每个这样的排列，增加计数。最后，打印计数。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function returns the required count
int PermuteTheArray(int A[], int n)
{
 
    int arr[n];
 
    // Store the indices of the
    // elements present in A[].
    for (int i = 0; i < n; i++) {
        arr[A[i] - 1] = i;
    }
 
    // Store the maximum and
    // minimum index of the
    // elements from 1 to i.
    int mini = n, maxi = 0;
    int count = 0;
 
    for (int i = 0; i < n; i++) {
 
        // Update maxi and mini, to
        // store minimum and maximum
        // index for permutation
        // of elements from 1 to i+1
        mini = min(mini, arr[i]);
        maxi = max(maxi, arr[i]);
 
        // If difference between maxi
        // and mini is equal to i
        if (maxi - mini == i)
 
            // Increase count
            count++;
    }
 
    // Return final count
    return count;
}
 
// Driver Code
int main()
{
 
    int A[] = { 4, 5, 1, 3, 2, 6 };
    cout << PermuteTheArray(A, 6);
 
    return 0;
}

Java

// Java program to implement
// the above approach
class GFG{
 
// Function returns the required count
static int PermuteTheArray(int A[], int n)
{
    int []arr = new int[n];
 
    // Store the indices of the
    // elements present in A[].
    for(int i = 0; i < n; i++)
    {
        arr[A[i] - 1] = i;
    }
 
    // Store the maximum and
    // minimum index of the
    // elements from 1 to i.
    int mini = n, maxi = 0;
    int count = 0;
 
    for(int i = 0; i < n; i++)
    {
 
        // Update maxi and mini, to
        // store minimum and maximum
        // index for permutation
        // of elements from 1 to i+1
        mini = Math.min(mini, arr[i]);
        maxi = Math.max(maxi, arr[i]);
 
        // If difference between maxi
        // and mini is equal to i
        if (maxi - mini == i)
 
            // Increase count
            count++;
    }
 
    // Return final count
    return count;
}
 
// Driver Code
public static void main(String[] args)
{
    int A[] = { 4, 5, 1, 3, 2, 6 };
     
    System.out.print(PermuteTheArray(A, 6));
}
}
 
// This code is contributed by gauravrajput1

Python3

# Python3 program to implement
# the above approach
 
# Function returns the required count
def PermuteTheArray(A, n):
 
    arr = [0] * n
 
    # Store the indices of the
    # elements present in A[].
    for i in range(n):
        arr[A[i] - 1] = i
 
    # Store the maximum and
    # minimum index of the
    # elements from 1 to i.
    mini = n
    maxi = 0
    count = 0
 
    for i in range(n):
 
        # Update maxi and mini, to
        # store minimum and maximum
        # index for permutation
        # of elements from 1 to i+1
        mini = min(mini, arr[i])
        maxi = max(maxi, arr[i])
 
        # If difference between maxi
        # and mini is equal to i
        if (maxi - mini == i):
 
            # Increase count
            count += 1
 
    # Return final count
    return count
 
# Driver Code
if __name__ == "__main__":
 
    A = [ 4, 5, 1, 3, 2, 6 ]
     
    print(PermuteTheArray(A, 6))
 
# This code is contributed by chitranayal

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function returns the required count
static int PermuteTheArray(int []A, int n)
{
    int []arr = new int[n];
 
    // Store the indices of the
    // elements present in []A.
    for(int i = 0; i < n; i++)
    {
        arr[A[i] - 1] = i;
    }
 
    // Store the maximum and
    // minimum index of the
    // elements from 1 to i.
    int mini = n, maxi = 0;
    int count = 0;
 
    for(int i = 0; i < n; i++)
    {
 
        // Update maxi and mini, to
        // store minimum and maximum
        // index for permutation
        // of elements from 1 to i+1
        mini = Math.Min(mini, arr[i]);
        maxi = Math.Max(maxi, arr[i]);
 
        // If difference between maxi
        // and mini is equal to i
        if (maxi - mini == i)
 
            // Increase count
            count++;
    }
 
    // Return final count
    return count;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []A = { 4, 5, 1, 3, 2, 6 };
     
    Console.Write(PermuteTheArray(A, 6));
}
}
 
// This code is contributed by gauravrajput1

Javascript

 
// Javascript Program to implement
// the above approach
 
// Function returns the required count
function PermuteTheArray(A, n)
{
 
    var arr = Array(n);
 
    // Store the indices of the
    // elements present in A[].
    for (var i = 0; i < n; i++) {
        arr[A[i] - 1] = i;
    }
 
    // Store the maximum and
    // minimum index of the
    // elements from 1 to i.
    var mini = n, maxi = 0;
    var count = 0;
 
    for (var i = 0; i < n; i++) {
 
        // Update maxi and mini, to
        // store minimum and maximum
        // index for permutation
        // of elements from 1 to i+1
        mini = Math.min(mini, arr[i]);
        maxi = Math.max(maxi, arr[i]);
 
        // If difference between maxi
        // and mini is equal to i
        if (maxi - mini == i)
 
            // Increase count
            count++;
    }
 
    // Return final count
    return count;
}
 
// Driver Code
var A = [4, 5, 1, 3, 2, 6];
document.write( PermuteTheArray(A, 6));
 

输出：

4

时间复杂度： O(N)
辅助空间： O(N)

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