What are the fundamentals of calculating space complexity in loo

ghz 13hours ago ⋅ 6 views

What are the fundamentals of calculating space complexity in loops?

Imagine you loop n times, and every iteration you create a string of space n with scope only within that iteration (thus it is no longer accessible in the next iteration). I would look and say that I use O(n^2) space because for n iterations, I use n space.

However, logically, if every loop you destroy the previous iteration's string (of n space) and overwrite it with this iteration's string (of n space), throughout the entire loop, you would only be using O(n) space. I am confused about whether to confirm O(n) or O(n^2) space?

Take this example:

s = "hello"
for _ in range(len(s)):
    newString = s[:]
return newString

Answer

In the scenario you're describing, the space complexity depends on how memory is used during each iteration of the loop and what happens to the previous iterations' variables.

Let’s break down your example step by step:

s = "hello"
for _ in range(len(s)):
    newString = s[:]
return newString

What Happens in the Loop:

  1. Initial String (s): s = "hello" is a string of length 5.
  2. The Loop:
    • The loop runs n = len(s) times, so it will run 5 times in this case.
    • In each iteration, newString = s[:] creates a new string that is a copy of s (in other words, the entire string is copied each time, taking up new memory).
  3. Space Usage:
    • String Copy (newString): In each iteration, newString is a new string with the same content as s. Since strings in Python are immutable, this copy involves allocating memory for a new string object, which is O(n) where n is the length of the string.
    • Iteration Scope: The newString variable is overwritten each iteration, meaning that at the end of each iteration, the previous newString is discarded.

Space Complexity Analysis:

  • Per Iteration: Each iteration of the loop allocates memory for newString (which is a string of length n), but after the iteration ends, the previous value of newString is discarded.
  • Total Memory Usage:
    • Even though you're creating a new string in each iteration, once each iteration finishes, the memory used by newString from the previous iteration is no longer accessible. Therefore, at any given moment, you only need space to store one copy of the string, not all copies.

Thus, in total, the space complexity for the entire loop is O(n), because:

  • You are only ever storing one string of size n (the copy of s in each iteration).
  • The space for the string is overwritten each time, not accumulated.

Conclusion:

  • Space Complexity: O(n), where n is the length of the string.
  • Why: Although the string is copied in each iteration, you only need space for one copy of the string at any given time. Hence, the space usage does not accumulate, and it’s bounded by the size of the string in memory (O(n)).

To sum up: Even though you are creating new strings in each iteration, they are not all retained in memory at the same time, so the space complexity is O(n), not O(n^2).

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