TLE | Discord Bot for Competitive Programming | Bot library
kandi X-RAY | TLE Summary
kandi X-RAY | TLE Summary
TLE is a Discord bot centered around Competitive Programming.
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Top functions reviewed by kandi - BETA
- Compute the statistics for the submissions
- Filters submitted submissions
- Return a dictionary of matching tags
- Check if the name matches any of the given markers
- Histogram
- Solve the given list of submissions
- Return a list of nice sub types
- Calculate ratings
- Plot the plot
- Resolve the given arguments
- Gimme
- Plot a curve
- List submissions
- Manage submissions
- Perform a vote
- Profile a ladderist
- Prints a pretty pretty listing of members
- Display the status of a problem
- Rating rating
- Get the rating for a guild
- Generate ratings for a match
- Define a challenge
- Generate rating plot
- Scatter
- Determine whether the game is complete
- Rate ratings
TLE Key Features
TLE Examples and Code Snippets
Community Discussions
Trending Discussions on TLE
QUESTION
Forgive me if I come off as a bit novice, I started teaching myself how to code just a few weeks ago.
I've been working through a problem set at cses.fi. I came across the "collecting numbers" problem which essentially asks us: given an array, such as [4, 2, 1, 5, 3], how many times would we have to "collect" the numbers such that the numbers are organized in increasing order from 1 to n.
In this particular case we would need 3 rounds, as we iterate once and collect "1", a second iteration gives us "2, 3", and finally a third will give us "4, 5".
I am programming in Python and came across the solution eventually, though I was getting a TLE error when submitting it. I made a number of tweaks and discovered that .method() was slowing my code considerably.
TLE code:
...ANSWER
Answered 2022-Mar-12 at 07:20Yes it will be much slower, especially as the size of 'n' grows.
In the TLE Code, what you're saying is:
- find the index of 'i' by searching through the array for 'i'. Remember that the python program cannot assume that the list is ordered or anything, so it will have to do a manual search for the first occurrence of 'i'
- this is fine when 'n' is small, because for example during the first iteration, .index() will first look for i=0 and return quickly because the first number in the list happens to be 0.
- but think about as n grows, you're basically asking .index() to traverse the list +1 more spot each time
- this for-loop is what is known as an O(n^2) algorithm, because for each 'i' within 'n', the worst case is searching through n items to find the index to return. The worse case happens at the last iteration of the for-loop.
Your second implementation does not require traversing the list n-times like your first implementation does.
Of course, like I mentioned, the first implementation will not have to traverse the entire list every time, but it gets worse as the loop goes on and as 'i' grows.
QUESTION
Question is We have to find if there is any sub array whose sum is zero.Return true if possible and false if not. Time contraint:3 seconds
This is the code using unordered set getting accepted .
...ANSWER
Answered 2022-Mar-10 at 13:02This line
QUESTION
Bob is a construction worker who does mathematics for increasing his efficiency. He is working on a site and has n buckets of cement-lined up with different characters (a – z) marked upon them. He has a strict command from the senior that he cannot change the order of the buckets.
Before starting his work, he has been given a string s of length n in which the character at position i (1 <= i <= n) gives us the mark on the i'th bucket. He can only carry one bucket at a time and bring it back to the base site. In each round, he has a criterion on which bucket to pick up. He will take the bucket with the smallest character marked upon it (a
Constraints
1 < t,m < 10^5
The sum of n over all test cases does not exceed 10^6
SAMPLE INPUT
2
badce
SAMPLE OUTPUT
7
Explanation
- badce - Firstly Bob takes the second basket with mark 'a' and adds 2 to the cost.
- bdce - Then he takes the first basket with the mark 'b' and adds 1 to the cost.
- dce - Then he takes the second basket with the mark 'c' and adds 2 to the cost.
- de - Again he takes the first basket with the mark 'd' and adds 1 to the cost.
- e - Again he takes the first basket with the mark 'e' and adds 1 to the cost.
The total cost becomes 7 units.
I have tried to code in Python but giving TLE for some cases. Here is my approach-->
...ANSWER
Answered 2022-Feb-28 at 20:15This runs in O(n). For every char, check how many previous chars will be transported later.
QUESTION
Question:
Given an array arr[] with N integers.
What is the maximum number of items that can be chosen from the array so that their GCD is greater than 1?
Example:
...ANSWER
Answered 2021-Dec-15 at 03:04I think you need to search among all possible prime numbers to find out which prime number can divide most element in the array.
Code:
QUESTION
I would like to ask why this link list will not run the result. After running, it is TLE. I want the head to be an indicator, and the head list can be returned without modifying the head.
...ANSWER
Answered 2022-Jan-30 at 15:50The function has undefined behavior because the pointer head and correspondingly the pointer node are not initialized
QUESTION
I have an ESP32 T-CALL with an integrated GSM-unit and used this as a base in Arduino IDE.
My code makes a server-call and processes the result. This code reads from the buffer and outputs it to a String. It uses a manually generated POST header, sent serially. However, I need to remove the HTTP header, leaving only the JSON.
...ANSWER
Answered 2022-Jan-30 at 02:01This is not a direct answer on how to use the regex, however, if you want to skip the headers and get the payload, other than using regex, or a httpclient library that I suggested in the comment, it is not difficult to do that without using any library.
To skip the header and get the payload, you need to modify your code to find the end of the header.
QUESTION
I'm trying to solve a competitive programming task which I cannot find out effective solution to tackle the last 5 test cases with.
You are given with t (t >= 1 && t <= 5) inquiries each of which consists of n numbers num (num >= 1 && num <= 1000000). The sequence that represents each inquiry isn’t sorted and there can be repeating numbers in it. The sum of all ns is less than 1000000. Let's call the GCD of all numbers in a single inquiry x. The task is to find out what is the minimum number of removals that have to be made in order to maximize x. Time limit is 0.7 s.
Consider that this (8, 2, 6, 4, 10, 12) is the inquiry that I’m given. GCD (8, 2, 6, 4, 10, 12) = 2 but If I remove 2, 6 and 10 GCD (8, 4, 12) = 4. I increased initial GCD x from 2 to 4 with 3 removals namely 2, 6 and 10. The answer of this inquiry is 3.
The best ideas I've come up with so far are the following ones:
...ANSWER
Answered 2022-Jan-19 at 14:39Here is a rather fast solution.
The first step consists in calculating the global gcd and divide all numbers by it.
The global GCD is now equal to 1. The problem is now to find the minimum number of removals such that this GC is no longer equal to 1.
For that, we perform the prime decomposition of each number, and find the most frequent prime in them.
The result is the array size minus the number of times this prime is present.
The complexity is dominated by the prime decomposition: O(n sqrt(m)), where m is the maximum data value.
QUESTION
I'm new at Haskell and Alex. I'm trying to make tokens of operators in Lexer.x
here is an example of my code
...ANSWER
Answered 2021-Dec-25 at 16:34You can use string literals to prevent escaping all characters, so you can use:
QUESTION
Why is initializing the diagonal in a 2d matrix using an external loop much faster than doing it in a list comprehension? I was using the list comprehension and I was getting time limit exceeded in Leetcode in a problem involving dynamic programming. I thought my algorithm was wrong. Switching to an external loop solved it in half the time and my solution got accepted. Here is sample of my code with timing. The first one is 6 times slower than the second approach on my machine.
...ANSWER
Answered 2021-Nov-09 at 05:13The two methods are not equivalent. In the list comprehension you perform 1,000,000 comparisons (if i==j) while in the second one you don't have any comparisons at all.
Also [False]*1000 is a built-in shortcut and probably executes faster than a for loop.
Note that the time complexity for both methods is O(n^2), but that doesn't mean that one cannot be faster than the other.
QUESTION
Here's a problem from Leetcode that I've spent nearly 2 days on.
https://leetcode.com/problems/sliding-window-maximum/
I have been able to pass 59/61 test cases but am getting TLE on the 60th. Could you please suggest if I could tweak my program somewhere to pass all the test cases ?
The approach I've used is as follows: slide is a list that consists of all the elements in the slide at a given time. As the slide slides one place to the right ,the element at the front is popped off, we'd have to check the max element each time. But the time can be saved if we check whether the max element really has been popped off. In case it is not the max element that has been popped off , we just have to compare the max element with the newly added element to the window and add that to the answer. However in case it is the max element indeed , we'd have to find the max element again. With the above approach I could pass 59/61 cases.
Then I thought:
It is possible that the max element could be present at more than one place in the slide. So I initially created a set from all the elements currently in the slide and then found the max element , which might help to save time But that didn't help either. Here's the code that I've tried:
...ANSWER
Answered 2021-Oct-31 at 11:37Let's use a deque (double-ended queue), the structure which pops from/ pushes to either side with the same O(1) performance.
It's more handy to store in the deque indexes instead of elements since both are used during an array parsing. Once you realized that, it becomes natural to approach this in different que way.
[Note] The similar idea has been discussed in many forum and credit should be shared to the "community".
Algorithm
It will be involved these steps:
Process the first k elements separately to initiate the deque. Iterate over the array. At each step :
Clean the deque:
Keep only the indexes of elements from the current sliding window. Remove indexes of all elements smaller than the current one, since they will not be the maximum ones.
Append the current element to the deque. Append deque[0] to the output. Return the output array.
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