LeetCode 1657. Determine if Two Strings Are Close - Interview Prep Ep 100

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Algorithm:
Based on the definition of the two given operations, we can infer these:
1. the order of each of the letter in the two words doesn't matter any more;
2. we need to make sure all unique characters that appear in word1 does also appear in word2.
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[@siddharthgupta848]

keep producing these leetcode videos daily, it is really so valuable.Thanks

[@varunpalsingh3822]

you way of explanation is good

[@FisherCoder]

Glad it was helpful!

[@kevinmanuel6711]

This is brilliant!

[@NikolayMishin]

cool explanation, thank you!

[@FisherCoder]

Glad it was helpful!

[@Meenereem]

Thank you!

[@FisherCoder]

You're welcome!

[@amintalukder7189]

I did not understand why count1 and count2 should be equal here ? they sometimes can not be equal, is not it ?

[@mhmd5ld81]

I think because u r mapping characters by their frequency for example if I got 3 "a" in word1 and I got 3 "z" in word2, 'a' is at the first of the list word1 and z is at the bottom of the list word2 so we want to perform the 2 operations and in order to do that we need to check the equality of frequent chars so we need to sort each list and comparing them and yes the two count lists could not be equal and that misses with the 2 operations so we return false

[@abdulmziya748]

Is there a way to solve this at linear time if we use a hashmap?