# Count of subarrays which forms a permutation from given Array elements

Given an array A[] consisting of integers [1, N], the task is to count the total number of subarrays of all possible lengths x (1 ≤ x ≤ N), consisting of a permutation of integers [1, x] from the given array.
Examples:

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

Naive Approach:
Follow the steps below to solve the problem:

• The simplest approach to solve the problem is to generate all possible subarrays.
• For each subarray, check if it is a permutation of elements in the range [1, length of subarray].
• For every such subarray found, increase count. Finally, print the count.

Time Complexity: O(N3
Auxiliary Space: O(1)

Efficient Approach:
To optimize the above approach, follow the steps below:

• For every element from i = [1, N], check the maximum and minimum index, at which the elements of the permutation [1, i] is present.
• If the difference between the maximum and minimum index is equal to i, then it means there is a valid contiguous permutation for i.
• For every such permutation, increase count. Finally, print the count.

Below is the implementation of the above approach:

## 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 `

Output:

```4
```

Time Complexity: O(N)
Auxiliary Space: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up A BTech sophomore in Information Technology major, having a keen interest in computer science and programming, particularly in algorithms design and analysis, data structures, Python and C++

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : chitranayal, GauravRajput1