# leetcode | 😏 LeetCode solutions in any programming language | Learning library

## kandi X-RAY | leetcode Summary

## kandi X-RAY | leetcode Summary

😏 LeetCode solutions in any programming language | Multiple programming languages to achieve LeetCode, "Sword Pointer Offer (2nd Edition)", "Programmer Interview Classic (6th Edition)" problem solutions

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### Top functions reviewed by kandi - BETA

- Returns all cells in a corner order .
- merge sort .
- Removes the specified value .
- Serialize a TreeNode into a string .
- Get news feed for a given user
- Request chunk ID .
- shell sort .
- Print number .
- Determine if the specified value is in the counter .
- Move the counter at the specified position in the player counter

## leetcode Key Features

## leetcode Examples and Code Snippets

## Community Discussions

Trending Discussions on leetcode

QUESTION

I'm learning Rust and have been going through leetcode problems. One of them includes merging two linked lists, whose nodes are optional. I want to write a while loop that would go on until at least 1 node becomes `None`

, and I was trying to use the `while let`

loop for that.

However, it looks like the `while let`

syntax supports only one optional, e.g.:

ANSWER

Answered 2022-Apr-10 at 08:22You can pattern match with `Option::zip`

:

QUESTION

I can use the normal F2 rename variable functionality in regular python files in vscode. But not when editing python in a jupyter notebook.

When I press F2 on a variable in a jupyter notebook in vscode I get the familiar change variable window but when I press enter the variable is not changed and I get this error message:

No result. No result.

Is there a way to get the F2 change variable functionality to work in jupyter notebooks?

Here's my system info:

jupyter module version

...ANSWER

Answered 2022-Jan-17 at 02:49Notice that you put up a bug report in GitHub and see this issue: Renaming variables didn't work, the programmer replied:

Some language features are currently not supported in notebooks, but we are making plans now to hopefully bring more of those online soon.

So please wait for this feature.

QUESTION

I am solving a problem on LeetCode:

Given an unsorted array of integers

`nums`

, return the length of the longest consecutive elements sequence. You must write an algorithm that runs in`O(n)`

time. So for`nums`

=`[100,4,200,1,3,2]`

, the output is`4`

.

The Union Find solution to solve this is as below:

...ANSWER

Answered 2022-Mar-14 at 07:33They are right. A properly implemented Union Find with *path compression* and *union by rank* has linear run time complexity as a whole, while any individual operation has an amortized constant run time complexity. The exact complexity of `m`

operations of any type is `O(m * alpha(n))`

where `alpha`

is the inverse Ackerman function. For any possible `n`

in the physical world, the inverse Ackerman function doesn't exceed 4. Thus, we can state that individual operations are constant and algorithm as a whole linear.

The key part for path compression in your code is here:

QUESTION

I am using a functional programming style to solve the Leetcode easy question, Count the Number of Consistent Strings. The premise of this question is simple: count the amount of values for which the predicate of "all values are in another set" holds.

I have two approaches, one which I am fairly certain behaves as I want it to, and the other which I am less sure about. Both produce the correct output, but ideally they would stop evaluating other elements after the output is in a final state.

...ANSWER

Answered 2022-Jan-12 at 09:30The actual term you’re asking for is *short-circuiting*

Further, some operations are deemed

short-circuitingoperations. An intermediate operation is short-circuiting if, when presented with infinite input, it may produce a finite stream as a result. A terminal operation is short-circuiting if, when presented with infinite input, it may terminate in finite time. Having a short-circuiting operation in the pipeline is a necessary, but not sufficient, condition for the processing of an infinite stream to terminate normally in finite time.

The term “lazy” only applies to intermediate operations and means that they only perform work when being requested by the terminal operation. This is always the case, so when you don’t chain a terminal operation, no intermediate operation will ever process any element.

Finding out whether a terminal operation is short-circuiting, is rather easy. Go to the `Stream`

API documentation and check whether the particular terminal operation’s documentation contains the sentence

This is a short-circuiting terminal operation.

`allMatch`

has it, `reduce`

has not.

This does not mean that such optimizations based on logic or algebra are impossible. But the responsibility lies at the JVM’s optimizer which might do the same for loops. However, this requires inlining of all involved methods to be sure that this conditions always applies and there are no side effect which must be retained. This behavioral compatibility implies that even if the processing gets optimized away, a `peek(System.out::println)`

would keep printing all elements as if they were processed. In practice, you should not expect such optimizations, as the Stream implementation code is too complex for the optimizer.

QUESTION

I'm currently studying for a technical interview and I'm just going through the leetcode grind. I came across a question that is apparently asked pretty frequently by the company I'm about to interview at so I attempted it. I couldn't quite get it so I looked to the solution and came across this solution.

...ANSWER

Answered 2021-Dec-19 at 00:30This happens because `prev`

's reference is being pointed by `res`

when you do `res=[prev]`

, basically the address where the actual `prev`

array is stored is pointed, as the prev updates, it also show changes in res.

QUESTION

I am doing the following problem on leetcode: https://leetcode.com/problems/contains-duplicate/

Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.

The solution I came up to the problem is the following:

...ANSWER

Answered 2021-Dec-18 at 21:29Will

`std::sort`

always compare equal values or sometimes it can skip comparing them and therefore duplicate values will not be found?

Yes, some equal value elements will always be compared if duplicates do exist.

Let us assume the opposite: initial array of elements *{e}* for sorting contains a subset of elements having the same value and a valid sorting algorithm does not call comparison `operator <`

for any pair of the elements from the subset.

Then we construct same sized array of tuples *{e,k}*, with the first tuple value from the initial array and arbitrary selected second tuple value *k*, and apply the same sorting algorithm using the lexicographic comparison operator for the tuples. The order of tuples after sorting can deviate from the order of sorted elements *{e}* only for same value elements, where in the case of array of tuples it will depend on second tuple value *k*.

Since we assumed that the sorting algorithm does not compare any pair of same value elements, then it will not compare the tuples with the same first tuple value, so the algorithm will be unable to sort them properly. This contradicts our assumptions and proves that some equal value elements (if they exist in the array) will always be compared during sorting.

QUESTION

I'm trying to do leet puzzle https://leetcode.com/problems/max-area-of-island/, requiring labelling connected (by sides, not corners) components.

How can I transform something like

...ANSWER

Answered 2021-Dec-16 at 12:26We can start off by enumerating the ones. We do the by applying the function `⍸`

(where, but since all are 1s, it is equivalent to 1,2,3,…) `@`

**at** the subset masked by `⊢`

the bits themselves, i.e. `⍸@⊢`

:

QUESTION

For an array of n integers, there are C(n,2)= n(n−1) / 2 pairs of integers. Thus, we may check all n(n−1) / 2 pairs and see if there is any pair with duplicates.

I was poking around a LeetCode question and the answer for one of the algorithms included the above formula in the question explanation.

What is the point of the `C(n, 2)`

nomenclature on the left hand side of the equation? Is this a known/named standard that I can read and interpret, or is this some more general information that must/should be ascertained from context? I understand the math on the right, but I don't have any preconceived notions that adds any detail to my understanding from the function on the left.

What is the 2 doing?

...ANSWER

Answered 2021-Dec-13 at 06:30It's called binomial coefficient, or "nCk" or "n Choose k".

Here *n* is the size of the set, and *k* = 2 is the number of elements to select, so that e.g. sets `{3, 6}`

and `{6,3}`

taken are considered equal.

AFAIK, the standard notation in combinatorics is as shown above and spelled "n choose k", where as `C(...)`

is non-standard requiring clarification when first introduced.

QUESTION

I am trying to solve the Leet Code challenge 14. Longest Common Prefix:

Write a function to find the longest common prefix string amongst an array of strings.

If there is no common prefix, return an empty string

Example 1: ...`""`

.

ANSWER

Answered 2021-Aug-09 at 00:42this:

QUESTION

I am grinding LeetCode these days and I encountered the challenge 162. Find Peak Element:

A peak element is an element that is strictly greater than its neighbors.

Given an integer array

`nums`

, find a peak element, and return its index. If the array contains multiple peaks, return the index toany of the peaks.You may imagine that

`nums[-1] = nums[n] = -∞`

.You must write an algorithm that runs in

Constraints:`O(log n)`

time.

`1 <= nums.length <= 1000`

`-231 <= nums[i] <= 231 - 1`

`nums[i] != nums[i + 1]`

for all valid`i`

This question is about using binary search to find a peak element in an array.

I know we can think of the array as alternating ascending and descending sequences. Here is my solution

...ANSWER

Answered 2021-Oct-11 at 06:20When the following condition is true:

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## Install leetcode

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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