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

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: 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.

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

So answer is 3

Below is the implementation of the above approach:

// C++ program to implement
// the above approach

#include <iostream>
#include <unordered_map>
#include <vector>
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 program to implement 
// the above approach 
import java.util.*;

class GFG{
    
public static int find_permutations(
    Vector<Integer> arr) 
{ 
    int cnt = 0; 
    int max_ind = -1, min_ind = 10000000; 
    int n = arr.size(); 
    
    HashMap<Integer, 
            Integer> 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<Integer> nums = new Vector<Integer>(); 
    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 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# 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

<script>
// Javascript implementation
function find_permutations(arr)
{
    var cnt = 0;
    var max_ind = -1, min_ind = 10000000;
    var n = arr.length;
    var index_of = new Map();
  
    // Save index of numbers of the array
    for (var i = 0; i < n; i++) {
        index_of.set(arr[i], i + 1);
    }
  
    for (var 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;
}
 
var nums = [];
nums.push(2);
nums.push(3);
nums.push(1);
nums.push(5);
nums.push(4);

document.write(find_permutations(nums));
// This code contributed by shubhamsingh10
</script>

3