BS-13. Minimum days to make M bouquets | Binary Search

[@udaykulkarni5639]

5 mins into the video and I knew exactly how to code it up! Cheers man. This guy has some serious skills!

[@shashankvashishtha4454]

That's candidate master for ya.

[@anonymousanonymous7507]

gawd

[@AyushSharma-sd1ny]

Not a easy problem to understand directly by reading it.. Thanks you have made it very simple.

[@adityarajvermaa]

bro literally I coded it down myself...just came here to understand the problem coz it was too complex to understand. Now i can code medium range problems by myself....sooner or later I will be able to understand the problem as well as we code further

[@danielhaag756]

Ya I am also facing same problem, I am also not able to understand the problem

[@samit840]

@@danielhaag756 tell me one thing bro, i watched the video till 4-5 mins, and then i noticed that we would be doing it same way as we did in KOKO bananas, so is it fine that i watched the video till this???

[@Vivekkumar-zc7mz]

how much time you takes to reach at that level and what are the hacks u have learnt that helped u ??

[@pranavJha93]

there is one tip i will give u see there two types of cases either you are able to solve a problem either you are not when when you are not able to solve a problem first make notes of it after watching tutorial of it and then do the same question by yourself after 7 days if u are able to do it in by your own without copying that means u actually grabbed the concept of it and it will also boost your confidence @@Vivekkumar-zc7mz

[@ranjeet_sohanpal]

@@Vivekkumar-zc7mz there are no hacks

[@morty6456]

My strategy of following the striver sheet is that at first I try to solve the question myself and if I am unable to then I come to videos for understanding, After reading this question on leetcode, I was clueless even about the question itself 1:35 mins into video and I have solved it on my own. Is this magic or something? 😅 Thank you bhaiyaaaaaaaaaaa❤

[@syedfazil2998]

exact same thing but at 2:15

[@pulkitjain5159]

currently 259 questions kar liye hai apki sheet ke , dekhte hai aur kitne kar pata hu , quality of questions awesome hai bhaiya , monotnic stacks m aur sliding windows m toh kya tagde question daal rakhe hai every question improves thinking in other direction. Ab bas graphs , greedy , trees , heaps aur strings bach raha hai usme bhi umeed hai bahut shandar questions honge. karta hu bhiaya dekhte hai kab tak khatam hogi , jab binary search kar raha tha tab i really wanted ki yeh videos ajaye but jo khud grind kiya na woh alag hi satisfaction tha. kabhi bhulayega nahi yeh questions , jitna video dekh ke bhool jaate hai. ek video ke peeche itni mehnat hoti hai itni research hoti hai , samajh agya. Waise app thoda rest bhi karliya karo aur jyada soya karo , peace of mind jyada important hai. Thank you for free knowledge 😊.

[@krishnaramtripathi958]

nunnu touching comment

[@pulkitjain5159]

@@krishnaramtripathi958 thanks bro BTW completed the sheet :).

[@GauravKashyap-ei6mb]

​@@pulkitjain5159Bro Is it possible to connect with you? I have few doubts regarding dsa.

[@vaibhavvm4147]

@@pulkitjain5159 bro from where did u learn dsa

[@KrishnaKumar-b4m9p]

@@pulkitjain5159 hi are u placed now??

[@raZer.7_]

Little by little feeling confident in solving DSA problems. Thank you man!!

[@thegreekgoat98]

Saw the first 9 minutes of the video and able to code it by myself. Such is the clarity in his explanations. Thanks Striver.

[@thisismr900]

Thank u striver sir. Really an inspiration to me. U know what ,I feel NOW very confident in solving the questions. Bas logic samajh aa jaye to code to bas 5 min lagta hai. 2 days complete , now at 13th video of this playlist .

[@momentcoder]

Finally I completed the first medium problem by myself, without understanding the approach first from the video. 😀 Thanks for the guidance Raj bhaiya (Striver). 🙏

[@BroodWar4Ever]

I could not wrap my head around this problem until I saw your video, thank you!

[@indroneelgoswami5654]

Only 5 mins watching and solved the Question!! Striver you beauty!!!!!!!!!!!!!!!!!!!!!!!

[@SumitSingh-dc8pm]

Solved this today, 13 Dec 2023. Thank you man for making the problem a cakewalk.

[@Nexus-bg3]

I wasn't able to solve mid level binary search problems. Now i can easily. I can't thank you enough

[@JoyBoy_013]

It becomes easier to understand when Striver starts explaining

[@reshmah1497]

Coded myself in 3 minutes. Thanks for great teaching.

[@himangshusarma6384]

UNDERSTOOD ! Instead, I solved it by my own before watching the video... Thanks to you

[@akshaygupta811]

A small optimization: defining long longs and checking for the condition m*k>n is not necessary, once the binary search finishes, if m*k>n is true, then our Low pointer would point at n. Hence at last we can just simple write: if (low==n) return -1; else return low;

[@6mahine_mein_google]

It really isn't an optimization per se. Because if we calculate it beforehand we dont need to apply binary search at all , if(m*k>n) is true then it will straight away throw -1 whereas in your approach it would still take O(N*logN) , because it will go into binary search loop nevertheless.

[@dank7044]

how is it an optimization lol? it ulta increases the Tc.

[@23cash86]

at 3:53 I understood it is binary search, implemented on my own using binary search + sliding window ....Made some modification to sliding window as window cannot be overlapped to count consecutive k elements, got overflow error while checking n< m*k ... typecasted to n< (long) m*k ,finally passed all test cases after 40min..... Edit: after submiting i watched video and realised its easier to implement possible function , same complexity as sliding window but i could probably code you idea faster and it is simpler XD. Grateful for free knowledge from you.

[@aksharshah]

Striver always on fire🔥❤ excellently explained 😊

[@div9210]

I forgot to write noOfB += (cnt / k); again outside the loop in possible function, and it took me so much time to solve it myself, but i was able to do it after some time. So much thanks!

[@johndurai2226]

Thank you striver bro I am from the south india (Tamilnadu). You are the consistency person keep it up and post videos as well. These are helpfull for us. Please made videos like this consistency.Thank you

[@priyanshunautiyal1662]

you are a great striver love the way you approach problems ....... Clearly Understood

[@adamyagaur9638]

Minor optimization in TC: Instead of taking the complete range from lowest element to the highest element, just take unique sorted elements from the array for the binary search. This will reduce the log(max-min) complexity part to log(N), however the space complexity will change to O(N) from O(1)

[@Manishgupta200]

isPossible() function and taking long long as val = m*k to avoid overflow error is too tricky and helpful to understand the question clearly

[@krishnavamsiyerrapatruni5385]

Understood everything clearly and was able to code it myself. Thanks a lot Striver :)

[@danish15]

Thanks it wasn't an easy problem statement to understand but after your explanation I was able to solve it on my own. 🙏

[@joeljacob4685]

Wow !! I am able to understand the intuition behind these problems. Excellent Problem and awesome explanation.

[@ArijitsClasses]

Good evening sir, this is Arijit from Kolkata. Your videos are super amazing and helpful, getting confidence in attempting leetcode problems...could even solve few of them. Sir the purpose of this comment is, as you said in one of the TEDX videos that in your college days you used to earn through blogging...may you please suggest me some tips, how can I earn through blogging, and recommend some blogging websites from where I can earn too....

[@VinuthaTJ-u7o]

Best! Best! Bestest explanation ever!!

[@faizanahmed9304]

Thank you Striver bhaiya!

[@MJBZG]

could code it by myself before reading the explanation, thanks!

[@tanujaSangwan]

Completed the code without watching the video all by myself

[@nukuldevsinghkhichi]

Greato🎉

[@sauravchandra10]

Was able to code this on my own, thanks and understood clearly!

[@kaushiksen2190]

Thank you so much Dada, i was able to solve it on my own by taking help from the previous question intuition

[@cinime]

Understood! Amazing explanation as always, thank you very very much for your continuous effort!!

[@alessandrocamilleri1239]

Great explanation. TC should include an additional O(n) to find range.

[@elizabethr5161]

yes correct

[@akhileshagarwal5913]

What a Great Approach Man

[@Avi-rc5ff]

The explanation is so good

[@udatyadeb101]

thanks to you was able to do the problem on my own.

[@BiswajitDas-lk7pp]

Better than Paid Courses

[@divya1005]

Thank you so much for the super clear explanation 🙂

[@stephan24297]

thanks for the explanation definitely helped me to understand the problem!

[@shamanthhegde2820]

while solving the same problem in leetcode, i encountered a condition where the code might fail. for instance when there are enough flowers but they are not adjacent to make enough bouquets. so during that case our low pointer will go beyond the size of the array. In order to counter that , we have to check the position of low...if the position of low > maxDay then we have to return -1 as we were not able to satisfy number of bouquets.

[@ShravanKumar-wg9pv]

understood very well sir

[@statusupdate7007]

Understand bro great teaching 👍

[@abhishekkumargupta117]

wow! very well explained.

[@AtulKumar-c4x7l]

understood Thank you striver for such an amazing explanation...

[@dipingrover1970]

amazing explantion i was able to do it on my own. 😊

[@adityaroychowdhury3709]

understood. thank you very much for this series.

[@pushankarmakar1783]

why cant we do it in a way where the possible() function would return the bouquets we are able to form and then we deal with what to do in the binary search?

[@amanasrani6405]

Amazing Explanation 🙌❤

[@Bigg_boss_trolls]

it becomes easy when striver explains💀

[@dayashankarlakhotia4943]

Good problem very well explained. Thanks.

[@RaunitJaiswal-s9v]

Done

[@AK-nj1je]

Done and Dusted !!!

[@ritikkurta3363]

happy birthday bhai bhai

[@cenacr007]

the opposite polarity is very useful.

[@prakhargarg4166]

Best intuition

[@AnmolGupta-oj4lm]

Understood Very Well!

[@culeforever5408]

understood 😇

[@ishangujarathi10]

Understood completely, tysmm striverrrrrr

[@deepakff7498]

17:33 AND 24:20

[@googleit2490]

Understood, coded it on my own, thanks!!

[@vedantrajanikantlatthe735]

you are great sir

[@manavsingh5919]

thank you striver understood everything

[@vaibhav5269]

Hi striver At 20:20 , you directly ignored less than 10 days, but what if bloom days array was something like this [7,7,7,7,13,11,12,7,7,7] In this case 2 bouquets are possible for minimum of 7 days, but we have ignored 7 days in binary search.

[@vivekprasannan9258]

Bro, in your case 10 days is a possible ans, so according to binary search the ans will store 10 and try to search for a min day than 10 (right half is ignored). Hope it helps

[@srija0608]

Understood 👍

[@Ina2003-q4s]

Love from Mysore ❤

[@oyeesharme]

understood bhaiya

[@AyushGupta-kb9iv]

class Solution { public: bool f(vector &a, int m, int k, int midDay) { int cnt=0, noOfBookey=0; for(auto x: a) { if(midDay>=x) cnt++; else { noOfBookey += (cnt/k); cnt=0; } } noOfBookey += (cnt/k); return noOfBookey >=m; } int minDays(vector& a, int m, int k) { if(1ll*m*k > a.size()) return -1; int lo=1e9, hi=-1e9, mid, ans=-1; for(auto x: a) { lo=min(lo, x); hi=max(hi, x); } while(lo>1; if(f(a, m, k, mid)) { ans=mid; hi=mid-1; } else lo=mid+1; } return ans; } };

[@CodeMode9313]

Habibi tussi mast bideo banate ho

Thank you sir

[@AbhishekKumar-nz9dn]

UNDERSTOOD 🥰

[@smoothgaming7822]

Understood sir!

[@DeadPoolx1712]

UNDERSTOOD;

[@Learnprogramming-q7f]

Thank you bhaiya

[@kaushiksen2190]

Keep going thanks alot

[@bharatmundra9568]

amazing sir

[@YourCodeVerse]

Understood✅🔥🔥

[@SYCOA12CHAITANYAASOLE]

Understood !! 😎😎

[@srikanthkumar6842]

Thank you

[@Anony.musk01]

Understood!

[@TON-108]

Understood!

[@md.imrankhan6912]

Legendary boss

[@stormragemc4375]

if i may once you've found out 12 (mid) is possible, why do you still need to eliminate the ranges and continue looking when it's safe to assume that since you've ordered the list the first number you found possible would be the smallest possible and u got ur answer already?

[@mohammedansil2302]

he is a fire❤‍🔥❤‍🔥❤‍🔥❤‍🔥❤‍🔥

[@lethalgaming7087]

Java Code : public class Solution { public static int roseGarden(int[] arr, int k, int m) { int n=arr.length,low=1,high=0,ans=0; if(k*m>n) return -1; for(int i:arr) if(i>high) high=i; while(low>>1; if(isPossible(arr,mid,k,m)) { ans=mid; high=mid-1; } else low=mid+1; } return ans; } private static boolean isPossible(int[] arr, int mid, int k, int m) { int adjl=0,bc=0; for(int i:arr) { if(mid>=i) adjl++; else adjl=0; if(adjl==k) { bc++; adjl=0; } if(bc==m) return true; } return false; } }

[@senseiAree]

Understood ❤

[@sahilsinghtomar]

nice one

[@sakshinaruka3440]

Understood :) Thank you so much

[@per.seus._]

UNDERSTOOD

[@arujain1690]

Can someone please help me understand what is polarity concept striver was talking about?

[@Bharat_Rider]

Initially low was on the not possible side and high on the possible side. This is polarity. Now as the binary search ends polarity changes i.e. low will be on possible and high on not possible.

[@darkpheonix6592]

simply binary search is converting o(n2) solution to o(nlogn) solution

[@rohithdarlanka]

understood

[@raghavkansal3765]

"understood"

[@ShamsiaJim]

Can you kindly upload the String Videos on Striver's A2Z DSA Course? (!)

[@RaviKumar-sn6tu]

understood!!

[@AnandSharma-ei1fv]

understoood

[@humanity7880]

understood!

[@soumiyamuthuraj3516]

Awesome