Akamai Interview Question for Java Developers


Country: India
Interview Type: In-Person




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2
of 2 vote

Use a hash map (hash_map<string,list<string>). The key for the hash map will be the sorted form of the scanned word.If the key doesnt exists in the hash map then create a list and the append the word to the list . This list serves as the value.If the key exists then append the scanned word for the list

- avikodak December 15, 2012 | Flag Reply
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0
of 0 votes

So this is O(n* w log w) time complexity? (w is the size of the max word)

- Kamy December 16, 2012 | Flag
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0
of 0 votes

yes

- avikodak December 16, 2012 | Flag
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1
of 1 vote

Can u explain this in detail?? I could not get the solution.

- john April 21, 2013 | Flag Reply
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1
of 1 vote

Can u explain this in detail?? I could not get the solution.

- john April 21, 2013 | Flag Reply
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0
of 0 vote

I think avikodak's solution would be a good choice if would be important to not use any additional space.

But since the important thing is minimum time complexity .. we can do it in O(n * w) where word is the length of the word : we would use the same logic as in avikodak's algorithm, but in stead of using a n log n sorting alg. I would recommend counting alg which we'll involve additional space , but we'll decrease the time complexity.

- Kamy December 16, 2012 | Flag Reply


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