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kandi X-RAY | Leetcode Summary
kandi X-RAY | Leetcode Summary
| # | title | c | c | python | go | js | java | difficulty | |---| ----- | -------- | ---------- | ---------- | ---------- | ---------- | ---------- | ---------- | |0001|[two sum] | [c] ./src/0001-two-sum/two_sum.c) | [c] ./src/0001-two-sum/two_sum.cpp) |[python] ./src/0001-two-sum/two_sum.py)|[go] ./src/0001-two-sum/two_sum.go)|| |easy| |0002|[add two numbers] | [c] ./src/0002-add-two-numbers/add_two_numbers.c) | [c] ./src/0002-add-two-numbers/add_two_numbers.cpp) |[python] ./src/0002-add-two-numbers/add_two_numbers.py)|[go] ./src/0002-add-two-numbers/add_two_numbers.go)|| |medium| |0003|[longest substring without repeating characters] | c | [c] ./src/0003-longest-substring-without-repeating-characters/0003.cpp) |[python] ./src/0003-longest-substring-without-repeating-characters/0003.py)|[go] ./src/0003-longest-substring-without-repeating-characters/0003.go)|| |medium| |0004|[median of two sorted arrays] | c | [c] ./src/0004-median-of-two-sorted-arrays/0004.cpp) |[python] ./src/0004-median-of-two-sorted-arrays/0004.py)||| |hard| |0005|[longest palindromic substring] | c | [c] ./src/0005-longest-palindromic-substring/0005.cpp) |[python] ./src/0005-longest-palindromic-substring/0005.py)|[go] ./src/0005-longest-palindromic-substring/0005.go)|| |medium| |0006|[zigzag conversion] | c | [c] ./src/0006-zigzag-conversion/0006.cpp) |[python] ./src/0006-zigzag-conversion/0006.py)|[go] ./src/0006-zigzag-conversion/0006.go)|| |medium| |0007|[reverse integer] | c | [c] ./src/0007-reverse-integer/0007.cpp) |[python] ./src/0007-reverse-integer/0007.py)|[go] ./src/0007-reverse-integer/0007.go)|| |easy| |0008|[string to integer (atoi)] | c | [c] ./src/0008-string-to-integer-(atoi)/0008.cpp) |[python] ./src/0008-string-to-integer-(atoi)/0008.py)|[go] ./src/0008-string-to-integer-(atoi)/0008.go)|| |medium| |0009|[palindrome number] | c | [c] ./src/0009-palindrome-number/0009.cpp) |[python] ./src/0009-palindrome-number/0009.py)|[go] ./src/0009-palindrome-number/0009.go)|| |easy| |0010|[regular expression matching] | c | [c] ./src/0010-regular-expression-matching/0010.cpp) |[python] ./src/0010-regular-expression-matching/0010.py)|[go] ./src/0010-regular-expression-matching/0010.go)|| |hard| |0011|[container with most water] | c | [c] ./src/0011-container-with-most-water/0011.cpp)|[python] ./src/0011-container-with-most-water/0011.py)|[go] ./src/0011-container-with-most-water/0011.go)|| |medium| |0012|[integer to roman] | c | [c] ./src/0012-integer-to-roman/0012.cpp) |[python] ./src/0012-integer-to-roman/0012.py)|[go] ./src/0012-integer-to-roman/0012.go)|| |medium| |0013|[roman to integer] | c | [c] ./src/0013-roman-to-integer/0013.cpp) |[python] ./src/0013-roman-to-integer/0013.py)|[go] ./src/0013-roman-to-integer/0013.go)|| |easy| |0014|[longest common prefix] | c | [c] ./src/0014-longest-common-prefix/0014.cpp) |[python] ./src/0014-longest-common-prefix/0014.py)|[go] ./src/0014-longest-common-prefix/0014.go)|| |easy| |0015|[3sum] | c | [c] ./src/0015-3sum/0015.cpp) |[python] ./src/0015-3sum/0015.py)||| |medium| |0016|[3sum closest] | c | [c] ./src/0016-3sum-closest/0016.cpp) |[python] ./src/0016-3sum-closest/0016.py)||| |medium| |0017|[letter combinations of a phone number] | c | [c] ./src/0017-letter-combinations-of-a-phone-number/0017.cpp) |[python] ./src/0017-letter-combinations-of-a-phone-number/0017.py)||| |medium| |0018|[4sum] | c | [c] ./src/0018-4sum/0018.cpp) |[python] ./src/0018-4sum/0018.py)||| |medium| |0019|[remove nth node from end of
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QUESTION
Here is the question:
...ANSWER
Answered 2021-Jun-15 at 16:06You 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.
QUESTION
I am writing code to answer the following LeetCode question:
Given the head of a linked list and an integer
Example 1 ...val
, remove all the nodes of the linked list that hasNode.val == val
, and return the new head
ANSWER
Answered 2021-Jun-15 at 10:52Some 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.skipper
only 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
prev
inskipper
is not the same variable as the other, outerprev
. So the assignmentprev=curr
inskipper
is quite uselessUnless the list starts with the searched value,
dummy.next
will always remainNone
, which is what the function returns. You should initialisedummy
so it links tohead
as its next node. In your updated code you took care of this, but it is done in an obscure way (in theelse
part).dummy
should 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 whileprev
is 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:
QUESTION
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:
...ANSWER
Answered 2021-Jun-14 at 19:33This will work:
QUESTION
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.
...ANSWER
Answered 2021-Jun-13 at 16:48Let'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
QUESTION
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.
ANSWER
Answered 2021-Jun-12 at 00:42I have ran it through gdb. After a bit of digging, the error is this:
QUESTION
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.
ANSWER
Answered 2021-Jan-13 at 12:43The main thing you want to check for overflow is this:
QUESTION
[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.
...ANSWER
Answered 2021-Jun-09 at 06:39A 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:
QUESTION
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.
...ANSWER
Answered 2021-Jun-07 at 07:00you have a typo in these code block:
QUESTION
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?
...ANSWER
Answered 2021-Jun-06 at 15:46The shadowing is defined by the C++ standard, and must work on all conforming compilers.
Your code prints garbage, because carry
in carry + 1
reads the new variable (which isn't initialized at that point yet, causing UB), not the old one.
QUESTION
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 char
datatype.
For example:
...ANSWER
Answered 2021-Jun-06 at 03:23Assuming your system is using ASCII or UTF-8 or something like that, 'a' == 97
.
So, this is equivalent to chars[97]++
.
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