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    <p>Answer to this question totally depend on the size of unique words, If the unique word count is small then you can use any string->number mapping data-structure (e.g. Trie tree) to count word frequency. Complexity will be <code>n log(m)</code> (m is the length of individual words), easy to implement. But the way problem was described, most probably unique word count is big enough to be able to store in the memory. In that case this following approach can be used:</p> <p>1 TB data means there are about <code>1.0*10^12</code> bytes of data in the input file. 1 byte is one character and lets say on average a single word has 4 characters then we have about <code>2.5*10^11</code> words. We will divide this word list in to <code>50k</code> different word lists. So, every time we will read about <code>5m</code> unread words from the input file, sort this <code>5m</code> word list and write this sorted list in a file. We will use <code>50k</code> array of numbers (lets call it <code>Parray</code>) to store the starting location of all sorted list in the file (initially <code>Parray</code> will have numbers like: 0, 5m+1, 10m+1 etc). Now read the top <code>50k</code> words from all list, put them in a min heap, you will get the smallest word on top of the heap. After getting the current smallest word (lets call it <code>cur_small</code>) from all sorted list read out that word from each list (after this operation your <code>Parray</code> will point to the next smallest word in each list). Here you will get the count of <code>cur_small</code> - so take the decision based on <code>K</code> then remove all entries of <code>cur_small</code> from the heap and lastly add a new word from each list to the heap from where at-least one word was <code>cur_small</code>. Continue this process until you read out all the sorted lists. Over all complexity is <code>n log(n)</code> </p>
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    1. POHow to print frequency k words from more than 1TB of data?
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    1. USsowrov
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    1. COWhile overall complexity is n * log(n) (because you are essentially just doing a huge sort) the practical performance will be dominated by I/O (because of the non-computational overhead --- writing and reading all these segment files). In an interview I highly recommend maintaining a clear awareness of the difference between complexity analysis and practical performance implications. A very elegant academic solution may be out performed by a hack that avoids some real world capacity limitations. On the other hand, consider the impliciations of SSDs, filesystems tuned for them.
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      1. USJim Dennis
    2. COTuning and tweaking can be done any time as long as you have a proper algorithm to perform the actual job. So while selecting an algorithm I think all you have to be concern with is the complexity because, a `n log n` algo will always out perform a `n^2` one.
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    3. COAny "Top K" algorithm can be reduced to sorting --- so n * log(n) should be considered the upper bound of any feasible approach. The question then become a matter of whether you can improve on sorting for any given specific case -- such as when K is significantly smaller than the number of distinct/countable "words" in the data set.
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      1. USJim Dennis
 

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