reverse in O(log n) time

The basic framework for divide and conquer algorithm can be used for any operations like reverse, checking if array is sorted etc. The possible implementation of reversing a string/vector is given as follows which takes O(n) time where n is size of collection string/vector.

template<class BidirIt>
void reverse(BidirIt first, BidirIt last)
 while ((first != last) && (first != --last)) {
 std::iter_swap(first++, last);

Better algorithm for this can be given as follows:

#include <iostream>
#include <algorithm>
#include <string>

template <typename T>
void reverse_rotate(T first, T last){
 if((first == last) || std::next(first) == last) return;
 T middle = first;
 std::advance(middle, std::distance(first, last)/2);
 reverse_rotate(first, middle);
 reverse_rotate(middle, last);
 std::rotate(first, middle, last);

int main(){
 std::string s = "This is super cool";
 reverse_rotate(s.begin(), s.end());
 std::cout << s << '\n';

Author: Saurabh Purnaye

VP - Low Latency Developer @jpmchase... Linux, C++, Python, Ruby. pursuing certificate in #QuantFinance and Passed CFA L1

One thought on “reverse in O(log n) time”

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: