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How efficient is the find() function on the std::map class? Does it iterate through all the elements looking for the key such that it's O(n), or is it in a balanced tree, or does it use a hash function or what?
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Time Complexity for Searching element : The time complexity for searching elements in std::map is O(log n).
find(): Searches for a particular element and returns the iterator pointing to the element if the element is found otherwise it will return the iterator returned by end(). Its time complexity is O(logN) where N is the size of the set.
end() if the key is not found. Show activity on this post. std::map operator[] inserts the default constructed value type in to the map if the key provided for the lookup doesn't exist. So you will get an empty string as the result of the lookup.
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Log(n) It is based on a red black tree.
Edit: n is of course the number of members in the map.
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std::map and std::set are implemented by compiler vendors using highly balanced binary search trees (e.g. red-black tree, AVL tree).
As correctly pointed out by David, find would take O(log n) time, where n is the number of elements in the container.
But that's with primitive data types like int, long, char, double etc., not with strings.
If std:string, lets say of size 'm', is used as key, traversing the height of the balanced binary search tree will require log n comparisons of the given key with an entry of the tree.
When std::string is the key of the std::map or std::set, find and insert operations will cost O(m log n), where m is the length of given string that needs to be found.
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