Count of Subarrays which Contain the Length of that Subarray

Given an array A[] of length N, the task is to count the number of subarrays of A[] that contain the length of that subarray.

Examples:

Input: A = {10, 11, 1}, N = 3
Output: 1
Explanation: Only the subarray {1}, with a length 1, contains its own length.

Input: A = [1, 2, 3, 4, 5], N = 5
Output: 9
Explanation: The subarrays {1}, {1, 2}, {2, 3}, {1, 2, 3}, {2, 3, 4}, {3, 4, 5},  
{1, 2, 3, 4}, {2, 3, 4, 5}, {1, 2, 3, 4, 5} contain their own length.

Approach: Follow the below idea to solve the problem:

First, form each and every subarray of A. Then, check if the length of the subarray is present in that subarray.

Follow the steps mentioned below to implement the idea:

• Iterate over the array from i = 0 to N:
  • Iterate in a nested loop from j = i to N:
  • The subarray created is from i to j.
  • Traverse the subarray and check if the length is present in the subarray.
  • If present, then increment the count.
• The final count is the required answer.

Below is the implementation for the above approach:

// C++ code to implement the approach
#include <iostream>
using namespace std;

// Function to find the count of the subarrays
// that contain their own length
int findCount(int arr[], int N)
{
    int counts = 0;

    // Forming all possible subarrays
    for (int i = 0; i < N; i++) {
        for (int j = i + 1; j < N + 1; j++) {

            // Checking if the length is pre