Given an integer N which denotes the length of an array, the task is to count the number of subarray and subsequence possible with the given length of the array.
Examples:
Input: N = 5
Output:
Count of subarray = 15
Count of subsequence = 32Input: N = 3
Output:
Count of subarray = 6
Count of subsequence = 8
Approach: The key observation fact for the count of the subarray is the number of ends position possible for each index elements of the array can be (N – i), Therefore the count of the subarray for an array of size N can be:
Count of Subarrays = (N) * (N + 1)  2
The key observation fact for the count of the subsequence possible is each element of the array can be included in a subsequence or not. Therefore, the choice for each element is 2.
Count of subsequences = 2^{N}
Below is the implementation of the above approach:
// C++ implementation to count // the subarray and subsequence of // given length of the array #include <bits/stdc++.h> using namespace std;
// Function to count the subarray // for the given array int countSubarray( int n){
return ((n)*(n + 1))/2;
} // Function to count the subsequence // for the given array length int countSubsequence( int n){
return pow (2, n);
} // Driver Code int main()
{ int n = 5;
cout << (countSubarray(n)) << endl;
cout << (countSubsequence(n)) << endl;
return 0;
} // This code is contributed by mohit kumar 29 
// Java implementation to count // the subarray and subsequence of // given length of the array class GFG{
// Function to count the subarray // for the given array static int countSubarray( int n){
return ((n)*(n + 1 ))/ 2 ;
} // Function to count the subsequence // for the given array length static int countSubsequence( int n){
return ( int ) Math.pow( 2 , n);
} // Driver Code public static void main(String[] args)
{ int n = 5 ;
System.out.print((countSubarray(n)) + "\n" );
System.out.print((countSubsequence(n)) + "\n" );
} } // This code is contributed by Princi Singh 
# Python implementation to count # the subarray and subsequence of # given length of the array # Function to count the subarray # for the given array def countSubarray(n):
return ((n) * (n + 1 )) / / 2
# Function to count the subsequence # for the given array length def countSubsequence(n):
return ( 2 * * n)
# Driver Code if __name__ = = "__main__" :
n = 5
print (countSubarray(n))
print (countSubsequence(n))

// C# implementation to count // the subarray and subsequence of // given length of the array using System;
class GFG{
// Function to count the subarray // for the given array static int countSubarray( int n){
return ((n)*(n + 1))/2;
} // Function to count the subsequence // for the given array length static int countSubsequence( int n){
return ( int ) Math.Pow(2, n);
} // Driver Code public static void Main(String[] args)
{ int n = 5;
Console.Write((countSubarray(n)) + "\n" );
Console.Write((countSubsequence(n)) + "\n" );
} } // This code is contributed by RajputJi 
15 32
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