Khanlou.com Website Review

about finding common algorithms and extracting them into the generic (and testable) functions. One thing in this vein that I like to look for is a few collection operations that cluster together that can be collected into a single operation. Often times, when combining these operations, you’ll also find a beneficial optimization that can be drawn out. The predicate block stays exactly the same and the result is the same, but it’s shorter, more semantic, and more performant, since it’s not allocating a new array and finding every single object in the array that pas*es the test — just the first. Another example is the count(where:) function that I helped add to Swift 5 Again: shorter, more semantic, and more performant. No arrays allocated, no extra work done. In a particular project of ours, we had one common pattern where we would do a sort followed by a prefix. This code, for example, needed     return images         .sorted(by: { $0.createdAt > $1.createdAt }) You could also imagine something like a leaderboard, where you want the top 5 users with the highest scores. I stared at this code until my eyes began to bleed. Something was fishy about it. I started thinking about the time complexity. If you code. The sort is O(nlogn), and prefixing is a very cheap operation, O(1). (Prefixing can be as slow as O(m), but arrays, which are what we are dealing with here, are random access collections and can prefix in constant time. Here’s what was bugging me: Getting the single smallest element of a sequence (using min()) requires only a single pas* over the all of its In our case, there are lots of images (n is about 55,000) and the amount we’re actually interested in small (m is 20). There should be For each element, we need to find the index of the first item that’s larger than it. Given our areInIncreasingOrder function, we can do this by pas*ing the element first, and each element of result second. If an index is found, that means that there’s a value less than element in our result array, and if we insert our new value at that