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    <p>You don't really "add" the complexities. Sorting is, as you say, O(n * log n), and searching is O(log n). If you were to do "normal math" on them, then it would be (n+1)*log n, which is still n*log n.</p> <p>When you're performing multiple steps like that, you typically take the highest complexity and call it that. After all, when n is sufficiently large, n*log n dwarfs log n.</p> <p>Think of it this way: when n is 1,000,000, n*log n is 20 million. log n is 20. So what's the difference between 20,000,000 and 20,000,020? The (log n) term is irrelevant. So (n log n) + (log n) is, for all intents and purposes, equal to (n log n). Even when n is 100, log n is 7. The (log n) term just won't make a difference when n is even moderately large.</p> <p>In your particular case, if you only need to search the list one time, then sequential search is the way to go. If you need to search it multiple times, then you have to weigh the cost of m searches O(m * n) against the cost of sorting and then searching. If you're interested in the minimum time and you know how many times you'll be searching the list, then you'd use sequential search if (m*n) is less than (n * log n). Otherwise use the sort and then binary search.</p> <p>But that's not the only consideration. Binary search on a sorted list gives you very quick response time, whereas linear search can take a very long time for a single item. If you can afford to sort the list during program startup then that's probably the best way to go because items will be found (or not found) much faster once the program is operating. Sorting the list gives you better response time. It's better to pay the price of sorting during startup than to experience very unpredictable response times during operation. Or to find out that you need to do more searches than you thought. . .</p>
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    1. USJim Mischel
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    1. COThanks for the answer!
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      1. USVikram
 

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