Efficiently reverse the order of the words (not characters) in an array of characters

Question

Given an array of characters which forms a sentence of words, give an efficient algorithm to reverse the order of the words (not characters) in it.

Example input and output:

>>> reverse_words("this is a string")
'string a is this'

It should be O(N) time and O(1) space (split() and pushing on / popping off the stack are not allowed).

The puzzle is taken from here.

Answer 1

A solution in C/C++:

void swap(char* str, int i, int j){
    char t = str[i];
    str[i] = str[j];
    str[j] = t;
}

void reverse_string(char* str, int length){
    for(int i=0; i<length/2; i++){
        swap(str, i, length-i-1);
    }
}
void reverse_words(char* str){
    int l = strlen(str);
    //Reverse string
    reverse_string(str,strlen(str));
    int p=0;
    //Find word boundaries and reverse word by word
    for(int i=0; i<l; i++){
        if(str[i] == ' '){
            reverse_string(&str[p], i-p);
            p=i+1;
        }
    }
    //Finally reverse the last word.
    reverse_string(&str[p], l-p);
}

This should be O(n) in time and O(1) in space.

Edit: Cleaned it up a bit.

The first pass over the string is obviously O(n/2) = O(n). The second pass is O(n + combined length of all words / 2) = O(n + n/2) = O(n), which makes this an O(n) algorithm.

Answer 2

pushing a string onto a stack and then popping it off - is that still O(1)? essentially, that is the same as using split()...

Doesn't O(1) mean in-place? This task gets easy if we can just append strings and stuff, but that uses space...

EDIT: Thomas Watnedal is right. The following algorithm is O(n) in time and O(1) in space:

1. reverse string in-place (first iteration over string)
2. reverse each (reversed) word in-place (another two iterations over string)
   1. find first word boundary
   2. reverse inside this word boundary
   3. repeat for next word until finished

I guess we would need to prove that step 2 is really only O(2n)...