An app I'm building professionally includes rudimentary text messaging. It allows users to send from the phone, and it polls for new messages (I know, yikes. Push notifications are a future task) with variable frequency for new messages from other users.

Part of its functionality is of course keeping messages in date-sorted order, and each message has a timestamp on it. I came up with two sorting implementations and I was curious as to which one would be more efficient.

Implementation 1:

for (Message* message in messages) {
    [self.backingMessages addObject:message];
NSSortDescriptor* sortingDescriptor = [NSSortDescriptor sortDescriptorWithKey:@"fullTimestamp" ascending:YES];
[self.backingMessages sortUsingDescriptors:@[sortingDescriptor]];

Implementation 2:

for (Message* message in messages) {
    NSInteger insertionPoint = [self.backingMessages indexOfObject:message inSortedRange:NSMakeRange(0, self.backingMessages.count) options:NSBinarySearchingInsertionIndex usingComparator:^NSComparisonResult(Message*  _Nonnull obj1, Message*  _Nonnull obj2) {
        return [obj1.fullTimestamp compare:obj2.fullTimestamp];
    self.backingMessages[insertionPoint] = message;

Some observations:

  • The new messages being added are known to be sorted
  • The messages being added to are initially sorted, and so will also be sorted every time new messages are added.

Doing some back-of-the-napkin analysis...

n = length of existing message array  
m = length of new message array  
implementation 1: O(m) + O((n+m)log(n+m))  
implementation 2: O(mlog(n+m)) + O(m) <- for the shifting that I assume happens upon insertion

My gut says that implementation 2 would be faster by a hair, depending on how the shifting works. All of this being said, I'm assuming that Apple ends up doing some crazy polymorphic implementation for sorting depending on the data, so implementation 1 might end up being faster. I'm probably overthinking this since the amount of data I'm expecting is so small, but I'm still curious about what might theoretically be faster.


1 Answer 1


If both new and old messages are sorted it's basically a merge of two sorted arrays, similar to the merge step in merge sort. It can be done O(n + m) time. I'd suggest you to look into how that's implemented.

  • \$\begingroup\$ Good catch, I'll review the code and see if that observation fits with the data I'm getting back. \$\endgroup\$
    – Chris
    Commented Aug 5, 2016 at 22:15

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