Word Subsets | Leetcode 916

This video explains word subsets problem using the the most optimal hashmap approach.
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[@divyanshsharma1694]

Very nicely explained, thank you sir!

[@techdose4u]

Glad it was helpful!

[@divyasingh6771]

Thank you sir for the solution. I did it with hashmap . This is my code:-class Solution { public static MapgetfrequencyMap(String str) { Map frequencyMap=new HashMap(); for(char c:str.toCharArray()) { frequencyMap.put(c,frequencyMap.getOrDefault(c,0)+1); } return frequencyMap; } public List wordSubsets(String[] words1, String[] words2) { List result=new ArrayList(); Map subset=new HashMap(); for(String word:words2) { Map part=getfrequencyMap(word); for(Map.Entry entry:part.entrySet()) { char key=entry.getKey(); int count=entry.getValue(); subset.put(key,Math.max(subset.getOrDefault(key,0),count)); } } for(String Word:words1) { Map wordFreqMap=getfrequencyMap(Word); int sum=0; for(Map.Entry entry:subset.entrySet()) { char key=entry.getKey(); int count=entry.getValue(); if(wordFreqMap.getOrDefault(key, 0)

[@anaghabharadwaj3586]

Thank you sir , well explained.

[@techdose4u]

welcome :)

[@rakshitshandilya9934]

I think we can do word2 operations on a single map also

[@Anikait-h3d]

first week is always full of easy questions. Difficult question ana chalu hone hi wala hai

[@techdose4u]

hoping for harder ones :)

[@bhavukkalra1786]

what is the reasoning for the usage of the array to keep track of the number of instances of the characters? The same can be accomplished using maps right?

[@techdose4u]

array is the map :) array is faster on runtime while maps are slower

[@bhavukkalra1786]

@ hey thanks for taking the time out replying. Correct, I did went on and changed the map implementation for tracking the max character count in word2 and my solution was finally out of the TLE. Its crazy on how such a minor DS change can make or break a solution. Thanks for the video once again, It is due to your video that my streak is still alive hehe😎

[@Marvel9-j4d]

🎉🎉🎉🎉

[@techdose4u]

🎉