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>
Output:
15 32
Time Complexity: O(log n)
Auxiliary Space: O(1)
