We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
This solution, like many others, claims to be O(n) because there is only one 'for' loop involved.
But that idea is totally flawed because, the access in dictionary (or STL maps in C++)is O(log(n)).
This means the overall worst case complexity when all the elements are distinct, is O(nlog(n)).
Cookie support is required to access HackerRank
Seems like cookies are disabled on this browser, please enable them to open this website
Count Triplets
You are viewing a single comment's thread. Return to all comments →
This solution, like many others, claims to be O(n) because there is only one 'for' loop involved. But that idea is totally flawed because, the access in dictionary (or STL maps in C++)is O(log(n)). This means the overall worst case complexity when all the elements are distinct, is O(nlog(n)).