Recursively reverse a list in place

Question

For the below code, even though reverse(s) returns the reversed string, the string itself is not reversed.

s = ["h", "e", "l", "l", "o"]
def reverse(s):
    if len(s) == 0:
        return
    if len(s) == 1:
        return s;
    temp = s[0]
    s = reverse(s[1:])
    s.append(temp)
    return s

print(reverse(s))
print(s)

Output

['o', 'l', 'l', 'e', 'h'] #reverse(s)
['h', 'e', 'l', 'l', 'o'] #s

As per this, Python: How do I pass a string by reference?. I understand i need to set s= reverse(s) and then the list gets reversed and this works alright.

But why is s getting reversed for this one?

s = ["h", "e", "l", "l", "o"]
def reverse(s, index):
    if index >= len(s) or len(s) == 0:
        return
    reverse(s, index + 1)
    s.insert((len(s) - 1 - index), s.pop())


reverse(s, 0)
print(s)

Output

['o', 'l', 'l', 'e', 'h']

While debugging I have found out that for the first piece of code when the handle comes back to the print statement, s becomes the original list, but for the second piece of code, s remains reversed. I am not sure what is happening here Any help is appreciated!

score 3 · Answer 1 · edited Jul 09 '19 at 14:48

3

The first method never changes the original s argument, as it slices it (using s[0] and s[1:]), thus making copies (and when you're calling s.append(temp), s is already a copy)

The second method keeps changing the original s argument (with s.pop() and s.insert())

If you want to avoid that the second method changes the original argument, create a wrapper method that creates a copy beforehand (which also avoids showing the auxiliary index argument)

def reverse(s):
    def internal_reverse(s, index):
        if index >= len(s) or len(s) == 0:
            return
        internal_reverse(s, index + 1)
        s.insert((len(s) - 1 - index), s.pop())

    s2 = s[:] # s[:] is a way to copy the input
    internal_reverse(s2, 0)  
    return s2

edited Jul 09 '19 at 14:48

bruno desthuilliers

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answered Jul 09 '19 at 14:38

Jean-François Fabre

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Thats right! But i thought "s.append(temp)" will actually edit the list by inserting temp to it? – pks Jul 09 '19 at 14:41
Sorry i did not get that. Has already changes as in? Could you please elaborate – pks Jul 09 '19 at 14:44
1

@pks the call to `internal_reverse` passes a copy of `s` (the `s[:]` part) – bruno desthuilliers Jul 09 '19 at 14:46
ah just saw your edited answer. So slice creates a copy and at the point its already sliced. Appending happens to the copy and not the original list. Thanks a lot :) – pks Jul 09 '19 at 14:47
1

@pks you may want to read this: https://nedbatchelder.com/text/names.html – bruno desthuilliers Jul 09 '19 at 14:49
thanks @brunodesthuilliers the return part was missing. too sure of myself => no test/test with mental interpretor => mistake. – Jean-François Fabre Jul 09 '19 at 14:52

Recursively reverse a list in place

1 Answers1