NOTES.md

Approach

1. In the problem statement, we have to print elements from N to 1.
2. Let's understand the logic:

# let take n = 5
# First, print one element that is 5
# recursion: TRUST func will print n-1 elements from 4 to 1
# f(n) = 1 + f(n-1)

# n = 5
# f(5) = 1 + f(4) 
# f(4) = 1 + f(3) 
# f(3) = 1 + f(2) 
# f(2) = 1 + f(1) 
# f(1) = 1 + f(0)  base case: stop here because n = 0

def recursiveFunction(n:int):
    # base case: if n reaches 0, then return
    if n==0:
        return
    else:
        print(n) # print first element - 5 (for n = 5)
        
        recursiveFunction(n-1) # TRUST it will print n-1 elements from n-1 to 1

Time Complexity

The time complexity of this code is O(n), where n is the input value passed to the printNos function.

In the recursiveFunction, the function is called recursively with a decremented value x-1 until x reaches the base case x == 0. This results in exactly n recursive calls. Each recursive call involves constant time operations (appending to the ans list and comparisons), and the total number of recursive calls is directly proportional to n.

Therefore, the overall time complexity is linear in terms of n, making it O(n).

Space Complexity

The space complexity of this code is also O(n), where n is the input value passed to the printNos function.

In the recursive function recursiveFunction, there are n recursive calls placed on the call stack. Each call has its own set of local variables, including the x parameter and the ans list. Therefore, the space required on the call stack is proportional to n.

Additionally, since the code appends the elements to the ans list, it consumes additional space on the stack for the append operation. However, this space is constant relative to the input size and doesn't affect the overall space complexity significantly.