# Count of subarrays of size K which is a permutation of numbers from 1 to K

• Last Updated : 31 May, 2021

Given an array arr of distinct integers, the task is to find the count of sub-arrays of size i having all elements from 1 to i, in other words, the sub-array is any permutation of elements from 1 to i, with 1 < = i <= N.

Examples:

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

A Naive approach is to start from each index and try to find the subarray of every size(i) and check whether all elements from 1 to i are present.
Time complexity: O(N2)

An Efficient approach can be given by checking if it is possible to create a subarray of size i for every value of i from 1 to N.
As we know, every subarray of size K must be a permutation of all elements from 1 to K, knowing that we can look at the index of the numbers from 1 to N in order and calculate the index of the minimum and maximum values at every step.

• If maximum_ind – minimum_ind + 1 = K, then we have a permutation of size K, else not.
• Update the value of minimum_ind and maximum_ind at every step.

Time complexity: O(n)
Illustration:

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

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach` `#include ``#include ``#include ``using` `namespace` `std;` `int` `find_permutations(vector<``int``>& arr)``{``    ``int` `cnt = 0;``    ``int` `max_ind = -1, min_ind = 10000000;``    ``int` `n = arr.size();``    ``unordered_map<``int``, ``int``> 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<``int``> 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<``int``,``               ``int``> index_of = ``new` `Dictionary<``int``,``                                              ``int``>();``            ` `    ``// 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`

## Javascript

 ``

Output:

`3`

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