451. Sort Characters By Frequency | 3 Ways | Sorting | Hash Table | Bucket Sort

[@6mahine_mein_google]

Hats off to your consistency bro!! . Whenever i dont feel doing leetcode , ur video pops out in my feed outta nowhere

[@dhruvrawatt9]

same for me

[@adityashekharsingh2669]

Love your explaining skills and hand movements ....helped me to get the patterns

[@siddarthreddykoppera5930]

In bucket sort I think time complexity is O(2N) , because we iterate over N characters and 0 to N frequencies so total N+N 2N... as you said N+k here k is always N so 2N however it's O(N) at the end

[@anonymous10906]

he did not consider the time taken by append function

[@halfthing2925]

The last approach was awesome 🔥

[@mainakseal5027]

isnt the last solution o(2*n) which is not better than o(n + klogk)

[@aidennav7336]

nice explanation

[@harshal8781]

🤩

[@learningmaths786]

good aryan

[@Vaidehi_IIT]

u r grt

[@Algorithmswithsubham]

why this is wrong ans string frequencySort(string s) { mapm; string ans=""; for(auto it:s)m[it]++; for(auto it :m){ char key=it.first; int val=it.second; while(val--){ ans+=key; } } return ans; }

[@deepakbhallavi1612]

because, It is storing character in sorted order. Example:- string s = "abbbc" Then map will store 'a' -> 1 'b' -> 2 'c' -> 1 So we need to store data of map in additional data structure, which will store based on the frequencies of characters 1 -> 'a' 2 -> 'b' 1 -> 'c' And then sort the data structure and All set 😉

[@6mahine_mein_google]

class Solution { public: string frequencySort(string s) { vector freq(26, 0); int n = s.length(); for (char c : s) { freq[c - 'a']++; } vector bucket(n + 1); for (int i = 0; i < 26; i++) { char ch = 'a' + i; int f = freq[i]; if (f > 0) { bucket[f].push_back(ch); } } cout

[@agambedi4142]

I don't want to look bad but if you just use Hindi as well that'll be really good bro.

[@sahilshrivastava6455]

bucket sort space complexity would be n^2

[@kgjr.6224]

Using this code the Generation of the ans String can be done in O(1) (not including the time complexity of mapping ie O(b)) class Solution { public String frequencySort(String s) { if (s == null || s.isEmpty()) return s; int[] freq=new int[128]; for(int i=0;i

[@manoor0858]

no lol repeat will take max amount of times as well

[@kgjr.6224]

​@@manoor0858Yes but this can easily be replaced by a string builder. Also the internal methods are faster than the normal lopping. So in the actual test it performs better then expected

[@manoor0858]

@@kgjr.6224 internal methods logic would be the same to insert elements how do u think the internal system insert n numbers , obv one by one ->o(n)

[@kgjr.6224]

@@manoor0858 the logic is kind of same but those are well optimized and and when you like write a sort method and use the inbuilt method the inbuilt method performs better. Also try running this code it will result in i guess 90 to 100 % efficient.

[@kgjr.6224]

@@manoor0858 because if you go with normal concatenation of string so it takes o(n,) it self so if the process repeat the concat will take around O(n^2) so it's better to use .repeat() or a stringbuilder

[@ChristianoSts]

I failed my interview with this question