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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Article Tags :

• Mathematical
• Python
• Competitive Programming
• C Language
• DSA
• Arrays
• subsequence
• subarray

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Practice Tags :

• Arrays
• Mathematical
• python