dunjen_master's blog

By dunjen_master, history, 9 months ago, In English

I am attaching a picture about the question. Kindly refer to the picture.

 
 
 
 
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9 months ago, # |
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Sample test cases?

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9 months ago, # |
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You didn't tell us whether the test is finished or not. We shouldn't help you if the test is not finished yet.

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9 months ago, # |
Rev. 4   Vote: I like it +3 Vote: I do not like it

Reverse all strings, so task becomes about prefix instead of suffix. Create segment tree, where each node have a trie that contains all strings from this node's corresponding segment (each string will be in $$$log~n$$$ tries, so memory will be $$$23 \cdot n~log~n$$$). With trie it's easy to count necessary sum of all string nested in this trie in $$$O(23)$$$, so overall query's complexity will be $$$O(23 \cdot log~n)$$$.

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    9 months ago, # ^ |
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    You can do it with a single trie

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      9 months ago, # ^ |
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      can you please explain

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        9 months ago, # ^ |
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        Basically what you and lemelisk said in the comment below, I'm not sure if it's better, just seems slightly easier to implement

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9 months ago, # |
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create a trie with reversed strings and store indices of strings in each node in sorted order then query for reversed string, you can find out for each each prefix of reversed string how many strings from given range match this prefix

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    9 months ago, # ^ |
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    For updates you will need to store indices in BSTs that supports query "how many elements is smaller than given value". But that a nice alternative to segment tree's solution and it even has lower memory consumption complexity.

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9 months ago, # |
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I solved it using map and policy based data structure My solution