Number of Students Unable to Eat Lunch - Leetcode 1700 - Python

[@loflog]

With the inflation these days, no one’s eating lunch

[@amongus-rq2hb]

Finished on my own and came for the explanation. Thanks

[@AhmedMarzookisabeast]

Solution looks good I did it thinking of queues in mind and did a .pop() and .append() as I came from the Neetcode queue section seems a bit overengineered now. class Solution: def countStudents(self, students: list[int], sandwiches: list[int]) -> int: students_left = len(students) while students_left > 0: if students[0] == sandwiches[0] and students_left: students.pop(0) sandwiches.pop(0) students_left = len(students) else: student = students.pop(0) students.append(student) students_left -= 1 return len(students)

[@souljarohill8795]

same

[@xavier7769]

class Solution: def countStudents(self, students: List[int], sandwiches: List[int]) -> int: i = 0 # initialize pointer to first student while i < len(students) and len(students) != 0: # break if we have not found a student to pair with the first sandwich or all students are fed (no students left) if students[i] == sandwiches[0]: # if i-th student matches first sandwich students.pop(i) # removes the i-th student and first sandwich as a pair sandwiches.pop(0) i=0 # then resets pointer to first student else: i+=1 # otherwise if student sandwich pair not found, point to next student return len(students) # return students who have not eaten O(n^2) but still idk how to work with queues

[@gabrieldavi9174]

I was really overthinking this problem.

[@licokr]

I initially approached with a dequeue (the same approach the problem tells) and the time complexity was O(n ^ 2). I thought there would be a better solution and there is. I focused on the question too much... Thank you so much👍 This is the code I wrote after watching your solution. I used cnt to calc the remaining students cnt = Counter(students) for s in sandwiches: if cnt[s] == 0: break cnt[s] -= 1 return cnt[0] + cnt[1] ps: either cnt[0] or cnt[1] is not zero if there are students unable to eat lunch.

[@CodingResoures]

how you did with the queue? how to handle the deadlock situation?

[@abdelrhmantarek3937]

@@CodingResoures class Solution { public: int countStudents(vector& students, vector& sandwiches) { std::queue StudentQueue; for(const auto& it : students) { StudentQueue.push(it); } int sandPointer = 0; int counter = 0; while(!StudentQueue.empty() && sandPointer != sandwiches.size()) { if(StudentQueue.front() == sandwiches[sandPointer]) { sandPointer++; StudentQueue.pop(); counter = 0; } else { counter++; if(counter == StudentQueue.size()) break; int rotatedValue = StudentQueue.front(); StudentQueue.pop(); StudentQueue.push(rotatedValue); } } return StudentQueue.size(); } };

[@Shanoro]

You have this video as a solution in your Pro course in the section Queues. This video does not solve anything with a queue.

[@servantofthelord8147]

Any issue with just simulating the problem with deques here? : from collections import deque class Solution: def countStudents(self, students: List[int], sandwiches: List[int]) -> int: sandwiches, students = deque(sandwiches),deque(students) skips = 0 while skips < len(students): if students[0] == sandwiches[0]: skips = 0 students.popleft() sandwiches.popleft() else: skips+=1 students.append(students.popleft()) return len(students)

[@spsc07]

I did a simulation using deque in cpp class Solution { public: int countStudents(vector& st, vector& sd) { deque q(st.begin(),st.end()); int i=0,cnt=0; while(!q.empty()) { int curr=q.front(); q.pop_front(); if(curr!=sd[i]) { ++cnt; q.push_back(curr); } else { ++i; cnt=0; } if(cnt==q.size()) break; } return q.size(); } }; Edit1: Did your method as well class Solution { public: int countStudents(vector& st, vector& sd) { int ans=st.size(); vectorhashh(2); hashh[0]=count(st.begin(),st.end(),0); hashh[1]=ans-hashh[0]; for(int i:sd) { if(hashh[i]>0) { --ans; --hashh[i]; } else return ans; } return 0; } };

[@juanmacias5922]

I was so close to get this problem using the Python Counter(), but didn't realize the order of the sandwich mattered lol so I also just did the simulation with deques using Python: def countStudents(students, sandwiches): students_deque = deque(students) sandwiches_deque = deque(sandwiches) i = 0 while (students_deque or sandwiches_deque) and i < len(students_deque): if students_deque[0] == sandwiches_deque[0]: students_deque.popleft() sandwiches_deque.popleft() i = 0 else: students_deque.append(students_deque.popleft()) i += 1 return len(students_deque)

[@doc9448]

You need to add this to the DSA course now. There is no link in the Queue's section even though you have the video here (I arrived here by chance via google). (Edit: never mind. I see why you maybe didn't add this video. Because you didn't use a Queue, no?) Nice to see you're still doing leetcodes though. This is the first recent leetcode video I've seen. I was worried I'd eventually run out of content.

[@keyulpatel4181]

True @neetCode

[@bohdanzaichenko]

I don't get it, he didn't use Queue here but it's still linked as a suggested problem for queue topic in his DSA beginner course

[@brianvega9231]

@@bohdanzaichenko you can solve it using queue

[@nehushtant]

@@brianvega9231 Yeah but even then if he's gonna include it in the queue section, might as well write code for it. We are paying for this course after all

[@GeekSlider]

@@bohdanzaichenko exactly werird

[@JoshPeterson]

This is in the DSA course, but the video isn't linked

[@1vader]

You don't even need the "res" variable. You can just return "cnt[1 - s]" (and 0 if we reach outside the loop). In the else branch, cnt[s] is zero. cnt[1 - s] is the count of the other preference which is obviously all that can remain. And if we reach outside the loop, all the sandwiches were eaten.

[@s8x.]

neet how did u learn everything? what’s resources did u use?

[@adama7752]

Life lessons and hashmaps

[@s8x.]

@@adama7752 neet

[@spsc07]

@@adama7752 lmao

[@sohampatil5670]

@@adama7752 bruh

[@Timjstewart]

Nice solution to what is, sadly and literally, a first world problem.

[@rithikgandhi3685]

Wow. I so overengineered this problem

[@krateskim4169]

Awesome explanation

[@satyamjha68]

Solved it!

[@get_out_it]

It’s a great solution

[@jenwans3055]

Do you have simple problems and sol. for blokes like me?

[@adityamwagh]

This is an easy tagged problem, but I agree it’s a bit on the harder end of easy questions. You can try some other easy questions on LeetCode or GeeksForGeeks.

[@varunpalsingh3822]

thank you neet :-)

[@staywithmeforever]

we could do like return cnt[0]+cnt[1]

[@shepmax555]

this solution is acceptable, but not scalable in case you start offering more than 2 sandwiches. need to keep this in mind as well

[@ShreyBhardwaj-mr5tl]

How are you implementing the rule that if the student does not get what he wants i.e the top of the stack, then he moves to the end of the queue? Can anyone pls explain?

[@seanmorris6185]

What drawing pad do you use for writing your thoughts out on screen?

[@infoknow3278]

3d paint

[@neks2081]

mouse

[@DBagg-zz4ip]

oh I just did this one but messed around popping from the student list

[@lethnisoff]

man this is 300IQ solution

[@innazhogova3621]

The question makes no sense. Why is the video & the correct solution approaching both inputs like they're queues? Didn't it say that sandwiches is a stack? There's definitely a mistake in the description.

[@goatking443]

you are so evil for including this as exercise in queue part of the course