kandi X-RAY | LeetCode Summary
kandi X-RAY | LeetCode Summary
LeetCode Algorithm Answers
Top functions reviewed by kandi - BETA
- Gets max price
- Restores the given range of raw text
- Computes the maximum allowable price for the given price
- Puts the current value of the current context
- Convenience method
- Compares two strings
- Compute distance between two words
- Add two numbers
- Add two numbers
- Rearrange the matrix
- Returns the longest common prefix of the given strings
- Sorted sum
- Get next permutation
- Approximates the given cardinality of the given value
- Calculate intersection
- Convert a string to an integer
- Returns the kthth least kth kth
- Return the nth odd number
- Multiply two strings
- Convert a string to an integer
- Solves the linear equations
- Convert roman string to integer
- Decode string
- Parses the given number of parameters
- Compute min window width and t
- Get zigzag tree order
LeetCode Key Features
LeetCode Examples and Code Snippets
Trending Discussions on LeetCode
Here is the question:...
ANSWERAnswered 2021-Jun-15 at 16:06
You have to assign the number of elements of the array to return (
2 in this case) to what the argument
returnSize points at (
*returnSize) to tell the judge system how large the array you returned is.
I am writing code to answer the following LeetCode question:
Given the head of a linked list and an integerExample 1 ...
val, remove all the nodes of the linked list that has
Node.val == val, and return the new head
ANSWERAnswered 2021-Jun-15 at 10:52
Some issues in your code (as it was first posted):
return skipper(prev,curr)is going to exit the loop, but there might be more nodes to be removed further down the list.
skipperonly takes care of a sub sequence consisting of the same value, but it will not look beyond that. The list is not necessarily sorted, so the occurrences of the value are not necessarily grouped together.
Be aware that the variable
skipperis not the same variable as the other, outer
prev. So the assignment
skipperis quite useless
Unless the list starts with the searched value,
dummy.nextwill always remain
None, which is what the function returns. You should initialise
dummyso it links to
headas its next node. In your updated code you took care of this, but it is done in an obscure way (in the
dummyshould just be initialised as the head of the whole list, so it is like any other node.
In your updated code, there some other problems as well:
while prev.next:risks to be an infinite loop, because it continues while
previs not the very last node, but it also doesn't move forward in that list if its value is not equal to the searched value.
I would suggest doing this without the
skipper function. Your main loop can just deal with the cases where
current.val == val, one by one.
Here is the corrected code:
I have two lists of strings and i want to check if the words within one list contain the strings within the other list throughout
Here is an example of what i mean:...
ANSWERAnswered 2021-Jun-14 at 19:33
This will work:
So I was solving this LeetCode question - https://leetcode.com/problems/palindrome-partitioning-ii/ and have come up with the following most naive brute force recursive solution. Now, I know how to memoize this solution and work my way up to best possible with Dynamic Programming. But in order to find the time/space complexities of further solutions, I want to see how much worse this solution was and I have looked up in multiple places but haven't been able to find a concrete T/S complexity answer....
ANSWERAnswered 2021-Jun-13 at 16:48
Let's take a look at a worst case scenario, i.e. the palindrome check will not allow us to have an early out.
For writing down the recurrence relation, let's say n = end - start, so that n is the length of the sequence to be processed. I'll assume the indexed array accesses are constant time.
is_palindrome will check for palindromity in O(end - start) = O(n) steps.
dfs_helper for a subsequence of length n, calls
is_palindrome once and then has 2n recursive calls of lengths 0 through n - 1, each being called two times, plus the usual constant overhead that I will leave out for simplicity.
So, we have
I am coding a solution program for leetcode. I have encountered memory read errors and segmentation fault errors which I am unable to resolve. I have tried tinkering with the code to try and remove the errors.
For example: in the
numIslands function's last lines, I had tried declaring
coordinates as an array rather than a pointer but the results did not change.
*(last + 1) = *(coordinates + 1); does not seem to work at all in the
numIslands function. I am totally blank about what to do next.
ANSWERAnswered 2021-Jun-12 at 00:42
I have ran it through gdb. After a bit of digging, the error is this:
I'm writing this code for more than 3 hours already..
I gave up about the overflow thing and tried to google and look it up on stackoverflow.
I did not find any solution besides the one that I wrote in my code as you can see in
lines 27-28 (where it returns 0).
But this condition also does not work.
ANSWERAnswered 2021-Jan-13 at 12:43
The main thing you want to check for overflow is this:
[link to the question] (https://leetcode.com/problems/process-tasks-using-servers/). The problem statement is clearly written in the provided link. I've highlighted a difference between my result and the expected result.
I'm not able to figure out which edge case or condition I've missed?
If the problem statement is still unclear then please do let me know.
Update: The updated solution is in the comment section....
ANSWERAnswered 2021-Jun-09 at 06:39
A point to remember: Servers present inside runningServers heap have higher priority than the servers that are present in the pool serverId waiting for some task.
Now, assume that we have 3 servers s1, s2, s3 having priority order s1 > s2 > s3, and 4 tasks t1, t2, t3, t4 with arrival time in increasing order. Also, assume that when we were about to allocate a server for t4, then by that time s2 and s3 got free.
Now, ideally, we should choose server s2 for task t4 but the above-given algorithm is going to choose s3 for t4 because it's maintaining a min-heap property on the finish time of the servers. So it's important to first free all the servers that have finish time <= currTime, and then choose a new server for t4 from the pool.
Now, what would happen if no server can be chosen from runningServer and we don't even have any server available in the pool serverId?
solution: We'll have to wait for some time for one of the servers to get free. let's say a server is going to get free at time = x ms, and it is also possible that there are multiple servers that are going to get free at time = x ms.
So, the final solution for the problem statement is:
I've been trying to attempt some leetcode questions as a complete beginner, and a section requires me to convert/return the thousands, hundreds, tens and units place of a number.
This is the code in python....
ANSWERAnswered 2021-Jun-07 at 07:00
you have a typo in these code block:
I ran this code on Leetcode.com but it prints random numbers. It works on my local machine, however. Anybody know if variable shadowing is supposed to work across all compilers?...
ANSWERAnswered 2021-Jun-06 at 15:46
The shadowing is defined by the C++ standard, and must work on all conforming compilers.
Your code prints garbage, because
carry + 1 reads the new variable (which isn't initialized at that point yet, causing UB), not the old one.
I'm working my way through the leetcode problems in C++ for practice.
I'm at problem number 3. I don't quite understand why you can access vector using
ANSWERAnswered 2021-Jun-06 at 03:23
Assuming your system is using ASCII or UTF-8 or something like that,
'a' == 97.
So, this is equivalent to
No vulnerabilities reported
You can use LeetCode like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the LeetCode component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .
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