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

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Problem Link: leetcode.com/problems/number-...
0:00 - Read the problem
0:19 - Drawing Explanation
6:03 - Coding Explanation
leetcode 1700
#neetcode #leetcode #python

[@loflog]

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

[@amongus-rq2hb]

Finished on my own and came for the explanation. Thanks

[@gabrieldavi9174]

I was really overthinking this problem.

[@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

[@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

[@Timjstewart]

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

[@JoshPeterson]

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

[@satyamjha68]

Solved it!

[@krateskim4169]

Awesome explanation

[@varunpalsingh3822]

thank you neet :-)

[@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?

[@get_out_it]

It’s a great solution

[@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)

[@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)

[@rithikgandhi3685]

Wow. I so overengineered this problem

[@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.

[@staywithmeforever]

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

[@DBagg-zz4ip]

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

[@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

[@seanmorris6185]

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

[@infoknow3278]

3d paint

[@neks2081]

mouse

[@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.

[@miaowcat1]

I'm just here for the comment section