Determine if Two Strings Are Close | Live Coding with Explanation | Leetcode #1657

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AlgorithmsMadeEasy
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The reason why I like this channel the most is because of the explanation of the algorithms with some visuals. But nowadays instead of explaining the algorithm, the code is being explained which is not good.

bharathkv
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time complexity with the sorted approach is O(n) if and only if you're using number sort. Not sure about default java sort, but its probably a hybrid with an n log n rutime

abelsimon
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Isn't the time complexity of the final solution O(nlog(n)) since we are sorting the arrays ?
A slight optimization is we can also have this check at beginning
if(word1.length() != word2.length())
return false;
Nice solution and explanation guys

akashselvakumar
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We can use priority queue and map to solve this problem

only_for_funr
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How does comparing the frequencies help? for eg : "cabbba" and "abbccc" are close though the frequencies of individual characters are not same ...

PensivePortraits
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I think Array.sort() function is nlog(n), wdyt?

minarezkalla
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Background music is distracting and annoying.

MrPanthershah
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It's the solution. Where is the explanation?

vipulm