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1. Top results related to time limit exceeded leetcode

   • Leetcode: Time limit exceeded, Longest palindromic substring

     Top Answer

     Answered Feb 18, 2022 · 0 votes

     The algorithm loses a lot of time by working from the outside inwards. Realise that a palindrome has a "center" (on, or between neighboring indexes), and your algorithm will often look for palindromes using the same center but with decreasing sizes. You could reduce work by working from the inside out, i.e. select all possible centers, and see what the largest palindrome is that has that center. This way you will only do one scan per center. So, iterate over all possible centers (both on indexes, and between two indexes), and see which is the largest palindrome you can find this way.

     You can then further optimise by starting with centers in the middle of the string (since they can potentially offer the largest strings), and then "fan" your center away from the middle of the string, until you come too close to an outer end of the string, making it impossible to improve on the longest palindrome that you had already found.

     Finally, avoid creating new strings, and just work with indexes (left, right, ...etc), and only really create a string when you already have identified the start/end index of the largest palindrome.

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     1/5

   • "Time Limit Exceeded" on LeetCode's three-sum problem?

     Top Answer

     Answered Oct 27, 2018 · 2 votes

     This has worked for me, uses a few optimizations for a lot of repeated elements. We store the count of the appearances of each element and then only iterate over each different element. The rest is similar to what you have already done

     from collections import Counterimport itertools-class Solution: def threeSum(self, nums): counts = Counter(nums) numSet = list(set(nums)) result = set() for idx, num1 in enumerate(numSet): for idx2, num2 in enumerate(itertools.islice(numSet, idx, None), start=idx): num3 = (num1 + num2) * -1 if num3 in counts: d = Counter([num1, num2, num3]) if not any(d[n] > counts[n] for n in d): result.add(tuple(sorted([num1, num2, num3]))) return list(result)

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     2/5

   • "Time Limit Exceeded" on LeetCode's Longest Palindromic Subsequence question

     Top Answer

     Answered Sep 27, 2018 · 2 votes

     String slicing in Python is O(n) (n being the length of the slice) while java's substring is O(1) as it merely creates a view on the same underlying char[]. You can take the slices out of the equation, however, by simply operating on the same string with two moving indexes. Moreover, you can move indexes past blocks of identical letters when first and last are not the same:

     @functools.lru_cache(maxsize=None)def longest_palindromic_subsequence(s, start=None, end=None): if start is None: start = 0 if end is None: end = len(s) - 1 if end < start: return 0 if end == start: return 1 if s[start] == s[end]: return 2 + longest_palindromic_subsequence(s, start+1, end-1) # you can move indexes until you meet a different letter! start_ = start end_ = end while s[start_] == s[start]: start_ += 1 while s[end_] == s[end]: end_ -= 1 return max( longest_palindromic_subsequence(s, start, end_), longest_palindromic_subsequence(s, start_, end))

     Memoizaton should help significantly. Take input "abcde". In the return max(...) part, eventually two recursive calls will be made for "bcd", and even more calls for the further embedded substrings.

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     3/5

   • leetcode, passes all test cases but time limit exceeded

     Top Answer

     Answered May 12, 2022 · 1 votes

     You can cache a lot of data. For instance:

     or[i..j] == vec[i] | or[i+1..j].-

     So you can readily create a 2d array of all or-values fairly quickly.

     So my algorithm would be:

     for (int index = 0; index < n; ++index) { for (int innerIndex = 0; innerIndex < n; ++innerIndex) { if (or[index][innerIndex] == maxOr) { ++count; } else if (or[index][innerIndex] > maxOr) { maxOr = or[index][innerIndex]; count = 1; } }}

     • Build that 2d array
     • Traverse the array to find maxOr and you can get the count at the same time. for (int index = 0; index < n; ++index) { for (int innerIndex = 0; innerIndex < n; ++innerIndex) { if (or[index][innerIndex] == maxOr) { ++count; } else if (or[index][innerIndex] > maxOr) { maxOr = or[index][innerIndex]; count = 1; } }}

     This works because when you encounter a new maxOr to continue, clearly this is the first instance of it, so you can reset count.

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     4/5

   • Getting time limit exceeded in leetcode

     Top Answer

     Answered Feb 03, 2022 · 4 votes

     This is a problem with your time complexity. In the leetcode question, it asks for:

     You must implement a solution with a linear runtime complexity and use only constant extra space.

     Your answer has time complexity n^2 because of your nested for loops. Try to implement your solution only using one for loop to obtain linear runtime complexity.

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     5/5

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2. stackoverflow.com › questions › 57931883How to tackle time limit exceeded error in leetcode

   stackoverflow.com › questions › 57931883

   Sep 14, 2019 · while l1[j] == l2[j] and l2[j] == l3[j]: tmp += l1[j] candidate.append(tmp) print(max(candidate)) Error Message. Time Limit Exceeded means they ran your code on a bigger input, and the code is slow. You need to optimize your code to reduce runtime complexity.

3. hatchjs.com › leetcode-time-limit-exceededLeetCode Time Limit Exceeded: Causes and Solutions

   hatchjs.com › leetcode-time-limit-exceeded

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   In this tutorial, we will discuss the causes and solutions to the time limit exceeded (TLE) error in LeetCode. We will also provide some tips on how to avoid TLE errors in your own code.

4. support.leetcode.com › hc › en-usWhat is the difference between Time Limit Exceeded (TLE) and...

   support.leetcode.com › hc › en-us

   • Cached

   If your solution is judged Time Limit Exceeded, it could be one of the following reasons: Your code has an underlying infinite loop. Your algorithm is too slow and has a high time complexity. The data structure you returned is in an invalid state. For example, a linked list that contains a cycle.

5. www.geeksforgeeks.org › overcome-time-limit-exceedtleHow to overcome Time Limit Exceed (TLE)? - GeeksforGeeks

   www.geeksforgeeks.org › overcome-time-limit-exceedtle

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   Sep 12, 2023 · Overcome Time Limit Errors. Change methods of Input-Output: You must choose proper input-output functions and data structure that would help you in optimization. In C++, do not use cin/cout – use scanf and printf instead. In Java, do not use a Scanner – use a BufferedReader instead.

6. stackoverflow.com › questions › 71724672How do you handle time limit exceeded errors in leetcode?

   stackoverflow.com › questions › 71724672

   Apr 3, 2022 · Did 5-7 fixes to your code, main errors were in addAtIndex() and deleteAtIndex(), other functions contained small errors or typos. Some errors resulted in infinite loop, that's why there was Time Exceeded error on LeetCode server. Corrected code snippet is below.

7. leetcode.com › discuss › general-discussionWorks on "Run Code'" but does not work on 'Submit' showing 'Time ...

   leetcode.com › discuss › general-discussion

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   Works on "Run Code'" but does not work on 'Submit' showing 'Time limit exceeded' - LeetCode Discuss. Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.

8. stackoverflow.com › questions › 70977391python - Getting time limit exceeded in leetcode - Stack Overflow

   stackoverflow.com › questions › 70977391

   Feb 3, 2022 · In the leetcode question, it asks for: You must implement a solution with a linear runtime complexity and use only constant extra space. Your answer has time complexity n^2 because of your nested for loops. Try to implement your solution only using one for loop to obtain linear runtime complexity.

9. www.youtube.com › watchLec 4.15: Time Limit Exceeded | Leetcode Problem Solved | how to...

   www.youtube.com › watch

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   Related Tags Time limit exceeded time limit exceeded in leetcode tle how to avoid tle overcome tle avoid Time limit exceeded iterative and recursive solution of factorial of number basic maths ...

10. leetcode.com › i-am-getting-time-limit-exceeded-as-an-error-what-does-that-mean- LeetCode

    leetcode.com › i-am-getting-time-limit-exceeded-as-an-error-what-does-that-mean

    Can you solve this real interview question? - Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.

11. leetcode.com › solutions › 30634Decode Ways - LeetCode

    leetcode.com › solutions › 30634

    Can you solve this real interview question? Decode Ways - Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.