Count and Say | Made Super Easy | Google | Explanation | Live Coding

[@tilakbhujade4438]

Very nice explanation. Pura ka pura logic samajh gaya. Apke logic se code maine khud hi solve kar diya. Thanks Sir!!! (That superbike sound in the BG tho...)

[@codestorywithMIK]

😇🙌🙏 Thanks a lot

[@aakarsh987]

Damn I saw so many explanations but couldn't understand anyone, you explained in a magical way, it was so amazing, thank you!!

[@Anikashukla1808]

whats the time and space complexity?

[@MounikaEMB]

Just started strings playlist.😊 As per question its hard to understand but after your explanation it is easy for me to understand... Thank you for making such a great videos for us🙏

[@codestorywithMIK]

So glad it helped Thank you 😇🙏

[@csengineer8819]

sir pheli baar hindi dekhi coding ke question ... pura 1-12th ka hindi medium school yaad agya

[@ArjunSaxena-wl3qs]

Amazing ! . . . . . . class Solution: def countAndSay(self, n: int) -> str: if n == 1 : return '1' def helper(n) : if n == 1 : return '1' say = helper(n - 1) # processing i,j = 0,1 res = '' count = 1 while j < len(say) : if say[i] == say[j] : count += 1 else: res += str(count) + say[i] count = 1 i = j j += 1 res += str(count) + say[i] return res return helper(n)

[@vineetjadhav1785]

Very Great Explaination ❤❤👋👋

[@souravjoshi2293]

Dude this became so easy. Thanks

[@sp_9467]

Very nice handwriting

[@akshatprakash126]

great explanation! hidden gem of a video

[@knight-z1x]

bdiya smjha ya apne

[@_PRANAYMATE]

Understood and code it on my own thanks bro

[@AjitKumar-jr7jo]

Today i sit in a OA of a company ,and it is the 1st question out of 3 , i could not understand this question, i totally got panicked and won't able to solve a single question.;)

[@codestorywithMIK]

Thank you for sharing but it’s ok, keep practising and soon you will be able to crack interviews. Would you mind sharing which company’s OA was it ?

[@AjitKumar-jr7jo]

@@codestorywithMIK it is on-campus company name is zessta.

[@DevOpskagyaan]

Such a fine explanation man. 🙌

[@leo71648]

One of the best explanation and as you said explanation on Leetcode is Pathetic, Keep it up

[@jagadeeshp1163]

class Solution: def rec(self,n): if n==1: return "1" temp=self.rec(n-1) count=1 ch=temp[0] ns="" for i in range(1,len(temp)): if temp[i]==ch: count+=1 elif ch!=temp[i]: ns+=str(count)+ch ch=temp[i] count=1 ns+=str(count)+ch return ns def countAndSay(self, n: int) -> str: return self.rec(n) Done✅

[@sayantanpoddar5428]

please came up with " Remove Outermost Parentheses" solution

[@RohitGupta-xf3zp]

nice bhaiya tgra logic build and inituiton

[@codestorywithMIK]

Thank you so much ❤️

[@kiranshelar2961]

Excellent explanation!

[@codestorywithMIK]

Glad it was helpful! ❤️❤️

[@Rajnish-pu8mj]

great teacher

[@pranavmaiya4386]

bro but this goes off to exponential complexity (2^n) , if interviewer expects us to optimize it then ?

[@_Deepak__Verma]

why length is say.length() instead of i

[@ShivamGupta-cx3hy]

Thank you sir 😊

[@codestorywithMIK]

Appreciate your comments ❤️

[@pranavmaiya4386]

what about the follow up question ?

[@mohdafzal4017]

Thanks alot.. feedback- you need to level up your thumbnail game

[@santoshbhatnagar2155]

what is the time complexity for this question?

[@TechVIP]

What is the time complexity for this?

[@mgnfy-view]

I coded the same solution with typescript but got a low score on time and space complexity.

[@jeethora586]

please explain time ans space complexity

[@B_Patidar88]

Bhaiya java me solve karvao pls

[@PakIslam2012]

why recursion, why not a for loop?

[@codestorywithMIK]

You can loop this too i guess. Recursion seems more easy to me . So i choose Recursion mostly

[@beinginnit]

Paypal asked this question in it's offcampus interview😅😅

[@allmovies2473]

Noise cancellation wala mic use✅Please✅

[@codestorywithMIK]

Sure ❤️

[@mayankkumar6628]

bhai tera explaination poora dekhne se pehle hi karliya if n ==1: return str(1) else : temp = str(self.countAndSay(n-1)) prev = temp[0] length=1 ans="" for i in range(1,(len(temp))): if prev == temp[i] : length+=1 else : ans =ans+str(length) ans +=prev prev = temp[i] length =1 ans +=str(length) ans +=prev return ans

[@carefree_ladka]

var countAndSay = function (n) { if (n === 1) return '1'; return countAndSay(n - 1) .match(/1+|2+|3+/g) .reduce((acc, nums) => acc += `${nums.length}${nums[0]}`, '') }; much simpler