Grumpy Bookstore Owner | Simplest Thought Process | Leetcode 1052 | codestorywithMIK

Whatsapp Community Link : www.whatsapp.c...
This is the 23rd Video of our Playlist "Sliding Window : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve a good and standard Sliding Window Problem : Grumpy Bookstore Owner | Simplest Thought Process | Leetcode 1052 | codestorywithMIK
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Grumpy Bookstore Owner | Simplest Thought Process | Leetcode 1052 | codestorywithMIK
Company Tags : will update soon
My solutions on Github(C++ & JAVA) : github.com/MAZ...
Leetcode Link : leetcode.com/p...
My DP Concepts Playlist : • Roadmap for DP | How t...
My Graph Concepts Playlist : • Graph Concepts & Qns -...
My Recursion Concepts Playlist : • Introduction | Recursi...
My GitHub Repo for interview preparation : github.com/MAZ...
Instagram : / codestorywithmik
Facebook : / 100090524295846
Twitter : / cswithmik
Subscribe to my channel : / @codestorywithmik
╔═╦╗╔╦╗╔═╦═╦╦╦╦╗╔═╗
║╚╣║║║╚╣╚╣╔╣╔╣║╚╣═╣
╠╗║╚╝║║╠╗║╚╣║║║║║═╣
╚═╩══╩═╩═╩═╩╝╚╩═╩═╝
Summary :
The approach used in the provided code to solve the problem of maximizing satisfied customers involves the following steps:
Initial Calculation:
Calculate the initial number of unsatisfied customers within the first minutes window. This is done by multiplying the number of customers by the grumpy state for the first minutes elements and summing the results.
Sliding Window Technique:
Utilize a sliding window of size minutes to determine the maximum number of unsatisfied customers that can be turned into satisfied customers by leveraging the special technique. As the window slides one position to the right:
Include the next element's contribution to unsatisfied customers if it falls within the grumpy period.
Exclude the first element of the previous window.
Update the maximum number of unsatisfied customers (maxUnsat) that can be converted to satisfied customers within any window of minutes.
Total Satisfied Customers Calculation:
Calculate the total number of satisfied customers by adding:
The maximum number of potentially converted unsatisfied customers (maxUnsat).
The customers who are already satisfied (i.e., where grumpy[i] is 0).
By combining these steps, the algorithm efficiently finds the maximum possible number of satisfied customers by utilizing a sliding window technique to optimize the period during which the special technique can be applied.
✨ Timelines✨
00:00 - Introduction
#coding #helpajobseeker #easyrecipes #leetcode #leetcodequestionandanswers #leetcodesolution #leetcodedailychallenge #leetcodequestions #leetcodechallenge #hindi #india #coding #helpajobseeker #easyrecipes #leetcode #leetcodequestionandanswers #leetcodesolution #leetcodedailychallenge#leetcodequestions #leetcodechallenge #hindi #india #hindiexplanation #hindiexplained #easyexplaination #interview#interviewtips #interviewpreparation #interview_ds_algo #hinglish #github #design #data #google #video #instagram #facebook #leetcode #computerscience #leetcodesolutions #leetcodequestionandanswers #code #learning #dsalgo #dsa #newyear2024

[@aniketnama8390]

Solution by the approach taught by The legend(MIK sir) in his previous videos of sliding window class Solution { public: int maxSatisfied(vector& customers, vector& grumpy, int minutes) { int n = customers.size(); int i = 0; int j = 0; int maxunsatcost = 0; int currunsatcost = 0; int ans = 0; for(int i=0;i

[@monikavaid5083]

Hi Mike, today I followed your approach of writing brute force first then writing the story for optimized solution and then the code finally. I was able to solve this in first attempt. Thanks a lot for building this power in your solutions. I can now get the intuition from the questions and I am also able to solve medium level questions now on leetcode. god bless you MIK 😇 you are doing a great job!

[@adritaadi8027]

you are so underrated bro!! thanks for making my leetcode journey easier

[@davidmiller814]

Thought of sliding window on the first minute of thinking, all thanks to you sir

[@ActorSidhantBhat]

Amazing work sir app bhaut badiya samjhate ho varnah sabh hindi-urdu bolne wale toh angreez banne kai chakkar mai hai

[@justanuhere]

did this on my own but it took 305ms. thank you so much for your solution.

[@PratyushBirole]

You are a legend.. Keep teaching like this. Also one more request. Can you please make an excel sheet for all the important DSA questions that you are teaching us ?I would be great. also we can track our progress

[@bunnypubg3475]

Pehle 6 min mei hi Moot dia mere approach par

[@gui-codes]

🤣🤣🤣 are bhai

[@25-cse-csmohitkumarmandal59]

Bs bhaii🤣🤣🤣

[@vanshgambhir7805]

Itna sach nhi bolna tha bhai 😂😂

[@DevOpskagyaan]

😂😂😂

[@Polly10189]

Hahah... Ak dam se approach badal di, example badal diya... Waqt badal diya...jazzbat badal diye😂😂😂😂

[@gouravshaw6939]

Here is the code -> class Solution { public: int maxSatisfied(vector& customers, vector& grumpy, int minutes) { // we will use sliding window approach here // 1. maxUnsatCust findout // 2. add this to totalSat; int n = customers.size(); int maxUnsatCust = 0; int curUnsatCust = 0; for(int i=0; i O(n) { curUnsatCust += grumpy[i]*customers[i]; } maxUnsatCust=curUnsatCust; // sliding window approach below int i=0; int j=minutes; while(jO(n) curUnsatCust += customers[j]*grumpy[j]; // adding the new element in the window curUnsatCust -= customers[i]*grumpy[i]; // removing the old element in the window i++; j++; maxUnsatCust=max(maxUnsatCust,curUnsatCust); } int totalSat=maxUnsatCust; for(int i=0;i O(n) { if(grumpy[i]==0) totalSat+=customers[i]; } return totalSat; } }; // tc-> O(n) // sc -> O(1)

[@gouravshaw6939]

wonderful explanation thanks!! I was able to do this on my own after watching the explanation!!

[@musaddikkhan9720]

Wowww ! and i solved this problem using DP 🤐 and i come here to watch optimized solution and w t h this is the sliding window problem so now i am going to solve it using sliding window 💝💝

[@musaddikkhan9720]

and i Solved that problem using sliding window also SatisfyCustomer + maximumUnsatisfyCustomer in that minutes window

[@musaddikkhan9720]

@@utkpal First pada Love Babbar then From Priyansh agarwal and abhi iss channel se

[@bombblaster6945]

just love your explaination...❤Crystal Clear

[@nishantkshirsagar2150]

I solved this by calculating which subarray will cause the maximum increase in sum rather than finding the subarray with maximum sum but it runs in O(n*m) complexity. Accept ho gaya class Solution: def maxSatisfied(self, customers: List[int], grumpy: List[int], minutes: int) -> int: n = len(customers) all_sum = 0 secret_sum_increase = -1 for i in range(n-minutes+1): j = i+minutes temp = 0 for k in range(i,j): if grumpy[k] == 1: temp += customers[k] secret_sum_increase = max(secret_sum_increase,temp) # O(n * minutes) for a in range(len(customers)): if grumpy[a] == 0: all_sum+= customers[a] #O(n) return all_sum+secret_sum_increase

[@pranavpranjal2717]

Solved this one on my own today!

[@tharunkumar8133]

Excellent explanation!! Appreciate your daily efforts!!

[@vishalvishwakarma7621]

I don't believe this is same solution that i have implemented 😮

[@THOSHI-cn6hg]

similar to find longest subarray of 1s by filpping k 0s

[@imPriyansh77]

Crystal clear explanation!!! Thanks bhaiya...

[@thecuriouskid5828]

Thank you sir! Suggestion/Request: Can you also start providing weekly leetcode contest solutions?

[@imPriyansh77]

6:17 bhaiya maine yahi socha tha lekin phir galti mil gyi iss approach mai 🥲🥲🥲🥲

[@lostcyrus8578]

kya he explanation dete ho bhaiya 😍😍. Segment tree ki new video jaldi lao na please waiting for it.

[@2Deadly_]

Another way to doing this class Solution { public: int maxSatisfied(vector& customers, vector& grumpy, int minutes) { int n= customers.size(); int ans=0; for(int i=0;i

[@25-cse-csmohitkumarmandal59]

Can't believe,it is solved by me without watching your video 😍😍😍🤯🤯❤️❤️🎉🎉🎉

[@krishangbose6584]

well taught👏👏👏👏

[@abhisheksingh3795]

Just wow❤‍🔥❤‍🔥

[@Abhay14]

great explaination bhaiya

[@rushabhladdha4694]

Holy shit man. I was making it so complex

[@Sumit-wy4zp]

Please upload the next lecture on segment trees. Also, sir, could you provide some LeetCode questions that we can practice?

[@motives748]

Waiting for your video

[@varunpalsingh3822]

Today I solved this on my own, using sliding window algorithm 😊

[@aggarwalsachin4854]

ab tu nakuri lega aur naukar bnega🤣🤣

[@varunpalsingh3822]

​@@aggarwalsachin4854 शिक्षित बनो, संघर्ष करो

[@varunpalsingh3822]

@@aggarwalsachin4854 शिक्षित बन, संघर्ष कर, अपने पिताजी के पैसों पर ज्यादा मत उछल

[@mukulkumar5133]

@@aggarwalsachin4854 abe yr tu bhi wahi krega kuch din bad

[@RohitKumar-dz8dh]

Thanks 😊

[@viholvishwassinh1709]

maine socha tha ki mai pahle customer array ke hisab se customer aur grumpty ko sort kardunga taki muje customers ke hisab se sorted array mil jaye.and then mai last se traverse karta hua aaung and jaha pe muje grumpy mil jae vaha se mai vo power use karlunga and mere is approach se 1st example mai ans 18 aa raha tha to koi bata sakta hai ki kyu wrong tha?

[@IITians]

Power consecutive minutes me use honge...sort karne se minutes consecutive nahi rahenge

[@nishantkshirsagar2150]

minutes consecutive he to sort karne se order kharab ho raha he

[@viholvishwassinh1709]

@@nishantkshirsagar2150 matlab is question Mai order matter karta hai?

[@viholvishwassinh1709]

@@IITians pls thoda sa explanation de sakte ho?

[@nishantkshirsagar2150]

@@viholvishwassinh1709 consecutive minutes tak wo apna secret power use kar sakt ahe agar tum use sort karoge to conseuctive minutes konse he ye kaisse pata chalega?

[@Anonymousyr]

subscriber yesterday 52.6K and today 52.9K 👍

[@sarveshjoshi6913]

solved by myself today ❤

[@dayashankarlakhotia4943]

Good explanation 🎉❤

[@siddhantgoyal5103]

why in the starting we are finding the unsatisfied customers in the first minutes window using a seperate for loop?? Rather than this we could have kept i and j at 0 and applied sliding window by moving j till minutes and if size of window becomes minutes increasing both i and j .

[@anshtanwar1813]

Feedback: All you need is to improve the thumbnails and add some intro/outro. May your channel grow to 100k soon

[@codestorywithMIK]

This means a lot. Thank you for the feedback. Feel free to provide some suggestions on what changes should I make in the Thumbnail. Would love to hear. Also in intro and outro, what should I add ❤️

[@anshtanwar1813]

​@@codestorywithMIK 1) Your thumbnail have different components like different text, your signature image :). But all things seem to be placed/pasted in little unorganized fashion. Matlab sab milakar ek single componet nahi lagte. 2) For example you can see thumbails of other channels like anuj bhaiya, Aryan mittal, take you forward. most of them use their personal brand, image (Pattern 1- More attractive and catchier) 3) But if you dont like these types of thumbnails you can take inspiration from 3 blue one brown channel, Reducible channel. (Pattern 2- simplicity at its best Maybe you could try more on brand building and stuff. Intro toh aap jo motivation dete ho vohi best hai, thoda background music ke sath testing kar sakte ho I am just a college student with some experience in designing but as your regular viewer just wanted to tell you that I am attracted to a 1) catchy and professional thumbnail 2) the views on the video and 3) if the person is familiar to me or is famous. I know content is the power. But these small things can increase subs and views and eventually take the content to a more audience. 💖✌

[@codestorywithMIK]

Noted ❤️❤️

[@NiranjanJangirBCH]

Thanks

[@AmanKumar-qz4jz]

sir aapne bola tha *longest valid parantheses* pe video layenge weekend pe......please bana dena

[@DevOpskagyaan]

DSA legend 🫡

[@Chandakshaythakur]

Bhaiya Please make a playlist on rabin karp and rolling hash problems

[@divsworld5925]

Bhaiya please please make a video on leetcode 2659. Make array empty please explain the intuition of this problem . No one has covered it properly i wasnt able to understand the intuition after searching for all resources. I beleive your story to code formula will be able to do it . Please please please bhaiya.

[@AbhijeetMuneshwar]

Respected MIK Sir, May I know what motivates you to deliver such a high quality content with lot of dedication, efforts and with total selflessness for free? 🙇🙏

[@rishabhpanesar9621]

i guess you're our indian @NeetcodeIO

[@doomhead9109]

one suggestion sir if possible please avoid telling the hint in the start of the video like you mentioned sliding window in the starting of this video. This thing make our mind to focus on one direction only.

[@rohitaggarwal8776]

Can’t we just find maximum sub array sum starting with 1?

[@mailatkaushal]

bruteforce got accepted in this question

[@dhairyachauhan6622]

this is how i did it class Solution { public: int maxSatisfied(vector& customers, vector& grumpy, int minutes) { int n = customers.size(); int j = 0; int i = 0; int sum = 0; int maxSum = 0; int start = -1; int end = -1; while(j < n){ sum += (grumpy[j] == 1)?customers[j]:0; while((j - i + 1) > minutes){ sum -= (grumpy[i] == 1)?customers[i]:0; i++; } if(j-i+1 == minutes && maxSum < sum){ maxSum = sum; start = i; end = j; } j++; } int ans = 0; for(int i = 0;i= start && i

[@gauravbanerjee2898]

Thanks a lot bhaiya ❤❤

[@techyou6163]

class Solution { public: int maxSatisfied(vector& customers, vector& grumpy, int minutes) { int n=customers.size(); int sum=0; for(int i=0;i

[@chitranshjain9714]

Did my self🎉

[@TUSHARKHANDELWAL-i8s]

again did this on my own , bhaiya kal vali video ke comment mai apne similar type ke q ki list di thi wo ab dikh nhi rhi comments mai ? dlt to nhi krdi?

[@nishantkshirsagar2150]

parso vali video ke niche he

[@atharvsoni8961]

Segment Tree video 4, when??

[@yashkumar7461]

class Solution { public: int maxSatisfied(vector& customers, vector& grumpy, int minutes) { int n = customers.size(); int unsatisfied = 0, maxUnsatisfied = 0; int i = 0; for(int j=0; j minutes) { if(grumpy[i] == 1) { unsatisfied -= customers[i]; } i++; } maxUnsatisfied = max(maxUnsatisfied , unsatisfied); } // Calculate total satisfied customers int totalSatisfied = 0; for(int i=0; i

[@knowledgepedia6976]

Bahiya 12.06 me logic click ho gaya ...3ms me solve ho gaya .....but khud se kyu ho nahi ho pata😰😰😰😰😰😰

[@ShardulPrabhu]

Sir meh sirf problem statement samjhne ata hu par first page pr uska type samjh jata hai fir mazha nahi ata solve krne aap please timestamp dal sakte ho kya taki direct mein problem statement pr ja saku first slide skip kr k

[@AbhishekJain-yf7nl]

Ese agr video dekh kr smjoge to kabhi question pdne ki ability improve nhi hogi.. Khud se try kro.. 1-2 baar pd kar. ..

[@gui-codes]

first try reading the Qns and try understanding from the examples on your own. Try 1-2 times. If your really get stuck, then get help. Waise mai most of the times video 1 minutes k baad se start karta hu so that start me topic ka pata na chal sake.

[@aggarwalsachin4854]

customer ko nhi, girlfriend ko satisfy kro

[@Mainak908]

Size increase karke

[@22gauravkumar70]

Please segment tree continue kijiye

[@mohdaqibkhan2135]

Can someone explain what am i doing wrong. I used max heap for this ques Intuition : shopkeeper wants to be non grumpy when he has more customers in his shop better name for cnt= minute_Used_To_Make_not_grumpy class Solution { public: int maxSatisfied(vector& customers, vector& grumpy, int minutes) { int n=customers.size(); priority_queuepq; for(int i=0;i0 && cnt=minutes) { if(pq.top().second==0) satisfied=pq.top().first; } pq.pop(); } return satisfied; } };

[@banadipmudi3186]

thanks