You’ve got it, just keep going! And be careful with the indexes… To simplify a bit I’ll assume that N and M are > k, so the complexity here is O(log k), which is O(log N + log M). Pseudo-code: For the demonstration you can use the loop invariant i + j = k, but … Read more
It looks like you are using Python 3.x. One of the important differences in Python 3.x is the way division is handled. When you do x / y, an integer is returned in Python 2.x because the decimal is truncated (floor division). However in 3.x, the / operator performs ‘true’ division, resulting in a float instead of an integer (e.g. 1 … Read more
You can use the localeCompare() method.
What you need is a way to look up some data given a key. With the key being an unsigned int, this gives you several possibilities. Of course, you could use a std::map: However, there’s other possibilities as well. For example, it’s quite likely that a hash map would be even faster than a binary … Read more
Here a more mathematical way of seeing it, though not really complicated. IMO much clearer as informal ones: The question is, how many times can you divide N by 2 until you have 1? This is essentially saying, do a binary search (half the elements) until you found it. In a formula this would be … Read more