Is this time complexity actually O(n^2)?

问题

I am working on a problem out of CTCI.

The third problem of chapter 1 has you take a string such as

'Mr John Smith '

and asks you to replace the intermediary spaces with %20:

'Mr%20John%20Smith'

The author offers this solution in Python, calling it O(n):

def urlify(string, length):
    '''function replaces single spaces with %20 and removes trailing spaces'''
    counter = 0
    output = ''
    for char in string:
        counter += 1
        if counter > length:
            return output
        elif char == ' ':
            output = output + '%20'
        elif char != ' ':
            output = output + char
    return output

My question:

I understand that this is O(n) in terms of scanning through the actual string from left to right. But aren't strings in Python immutable? If I have a string and I add another string to it with the + operator, doesn't it allocate the necessary space, copy over the original, and then copy over the appending string?

If I have a collection of n strings each of length 1, then that takes:

1 + 2 + 3 + 4 + 5 + ... + n = n(n+1)/2

or O(n^2) time, yes? Or am I mistaken in how Python handles appending?

Alternatively, if you'd be willing to teach me how to fish: How would I go about finding this out for myself? I've been unsuccessful in my attempts to Google an official source. I found https://wiki.python.org/moin/TimeComplexity but this doesn't have anything on strings.

回答1:

In CPython, the standard implementation of Python, there's an implementation detail that makes this usually O(n), implemented in the code the bytecode evaluation loop calls for + or += with two string operands. If Python detects that the left argument has no other references, it calls realloc to attempt to avoid a copy by resizing the string in place. This is not something you should ever rely on, because it's an implementation detail and because if realloc ends up needing to move the string frequently, performance degrades to O(n^2) anyway.

Without the weird implementation detail, the algorithm is O(n^2) due to the quadratic amount of copying involved. Code like this would only make sense in a language with mutable strings, like C++, and even in C++ you'd want to use +=.

回答2:

The author relies on an optimization that happens to be here, but is not explicitly dependable. strA = strB + strC is typically O(n), making the function O(n^2). However, it is pretty easy to make sure it the whole process is O(n), use an array:

output = []
    # ... loop thing
    output.append('%20')
    # ...
    output.append(char)
# ...
return ''.join(output)

In a nutshell, the append operation is amortized O(1), (although you can make it strong O(1) by pre-allocating the array to the right size), making the loop O(n).

And then the join is also O(n), but that's okay because it is outside the loop.

回答3:

I found this snippet of text on Python Speed > Use the best algorithms and fastest tools:

String concatenation is best done with ''.join(seq) which is an O(n) process. In contrast, using the '+' or '+=' operators can result in an O(n^2) process because new strings may be built for each intermediate step. The CPython 2.4 interpreter mitigates this issue somewhat; however, ''.join(seq) remains the best practice

回答4:

For future visitors: Since it is a CTCI question, any reference to learning urllib package is not required here, specifically as per OP and the book, this question is about Arrays and Strings.

Here's a more complete solution, inspired from @njzk2's pseudo:

text = 'Mr John Smith'#13 
special_str = '%20'
def URLify(text, text_len, special_str):
    url = [] 
    for i in range(text_len): # O(n)
        if text[i] == ' ': # n-s
            url.append(special_str) # append() is O(1)
        else:
            url.append(text[i]) # O(1)

    print(url)
    return ''.join(url) #O(n)


print(URLify(text, 13, '%20'))

来源：https://stackoverflow.com/questions/34008010/is-this-time-complexity-actually-on2