Count of possible subarrays and subsequences using given length of Array

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 = 32
Input: 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 Sub-arrays = (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 = 2N

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 Rajput-Ji 

<script> // JavaScript implementation to count // the subarray and subsequence of // given length of the array // Function to count the subarray // for the given array function countSubarray(n){  return ((n)*(n + 1))/2; }   // Function to count the subsequence // for the given array length function countSubsequence(n){  return Math.pow(2, n); } // Driver Code  let n = 5;  document.write((countSubarray(n)) +"<br/>");  document.write((countSubsequence(n)) +"\n"); </script> 

15 32

Time Complexity: O(log n)

Auxiliary Space: O(1)

_mridul_bhardwaj_

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