data-structures-and-algorithms | Important data structures and algorithms | Learning library
kandi X-RAY | data-structures-and-algorithms Summary
kandi X-RAY | data-structures-and-algorithms Summary
Some basic implementations of important data structures and algorithms in Java, along with a few LeetCode questions. I'll keep on adding more solutions of LeetCode problems and if you want, you can add them too! Also, I've started adding solutions to AlgoExpert problems. You can find them in the Contents section. Explore the docs » Report Bug · Request Feature.
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Top functions reviewed by kandi - BETA
- Test to see if a solution is contained in a point
- Returns the count of subtrees in the given array at the given index
- Creates a solution
- Returns the longest common subsequence recursive
- Main method for testing
- Prints the solution to stdout
- Main method
- Creates the solution
- Returns the minimum subset sum difference
- Main entry point
- Prints the max area
- Prints the solution
- Main function
- Prints the solution
- Prints solution
- Simple test program
- Dump the solution
- Recursive knapsack problem
- Find the longest common subsequence recursively
- Find the longest common substrings
- Returns a list of permutations with the given string
- Returns the longest common subsequence between two strings
- Returns a list of all nodes in the tree
- This method computes the number of Waysoization for a given target
- Creates an unbounded knapsack problem
- Recursive method to build the cache
- Finds the longest - increasing subsequence in the given sequence
- Test for debugging
- Find the minimum number of coins in the given target
- Returns the longest common subsequence recursively
- Compute the longest common subsequence
data-structures-and-algorithms Key Features
data-structures-and-algorithms Examples and Code Snippets
Lines of Code : 32License : Permissive (MIT License)
Copyfunction threeSum(nums) {
const ans = [];
nums.sort((a, b) => a - b); // sort: O(n log n)
for (let i = 0; i < nums.length - 2; i++) { // O(n^2)
if (i > 0 && nums[i - 1] === nums[i]) continue; // skip duplicates
let l function rotateMatrix(matrix){
const n = matrix.length;
for(let layer = 0; layer < n / 2; layer++) {
for(let offset = 0; offset < n - 2*layer - 1; offset++) {
// save top
const t = matrix[layer][layer + offset];
// mo function twoSum(nums, target) {
const len = nums.length - 1;
let lo = 0;
let hi = len;
while (lo < hi && hi > 0 && lo < len) {
const sum = nums[lo] + nums[hi];
if (sum === target) {
return [lo, hi];
Community Discussions
Trending Discussions on data-structures-and-algorithms
QUESTION
this is a very small piece of code taken from this book about the Dijkstra Algorithm in Ruby: I think I can't post the entire example for copyright issue, so if you want you can download the source code from the above link of the book, the file will be in the folder jwdsal2-code\code\connecting_everything_with_graphs\dijkstra.rb. - it runs perfectly and there isn't any error in the code.
The excerpt from the code is the following:
...ANSWER
Answered 2021-May-06 at 07:22Your solution is almost correct.
You assign the name of the city to the current_city variable whereas the original solution assigns the city object.
QUESTION
I am having a hard time understanding the process of drawing the repeated square diagram of recursion.
I have the book of data structure and algorithms which shows this code on page 210:
...ANSWER
Answered 2020-Jun-28 at 14:05The difference in numbers that you see in the recursion trace comes from n/2 in the code. It is an integer division by two. So 13/2 is 6 and 6/2 is 3 and 3/2 is 1, and finally 1/2 is 0. This is what you see when you read the diagram from top to bottom. The diagram shows these entries shifted more and more to the right to represent the depth of the recursion. The function calls itself, and then it calls itself again, ...etc. All of these calls are pending... Each time these calls pass a value for n that is smaller (halved).
Somewhere this stops. It needs to, of course. This stops when the value of n receives the value 0. You see this state at the bottom of the diagram. At that moment, the recursion does not go deeper and starts to unwind, and you need to read the diagram now from bottom back to the top. For n equal to 0, the return value is 1 (return 1).
This value is read by the function execution where n is 1 and where the call power(2, 0) was made and waiting for a return value. It receives 1 as result. This value is then squared (partial * partial) which in this case still is 1. And because n is odd (it is 1 here), there is an additional multiplication *= x. So we actually calculate partial * partial * x, which is 1 * 1 * 2. You'll see this in the diagram.
And this value (2) is returned to the function execution where n is 3 and where the call power(2, 1) was made. It receives 2 as result. This value is then squared (partial * partial) which in this case is 4. And because n is odd (it is 3 here), there is an additional multiplication *= x. So we actually calculate partial * partial * x, which is 2 * 2 * 2. You'll see this in the diagram.
I hope you see the pattern. The function executions that are pending for a result, all get their value back from the recursive function call they made, do some multiplication with it and return that, in turn, to their caller.
And so the recursion backtracks to the top, providing the end result to the top-level call of power(2, 13).
QUESTION
I was trying to see the difference in runtime for different sorting algorithms when I found that I was getting consistently faster runtime for one version of Insertion sort than another version of insertion sort. The versions of the algorithms seem nearly Identical (only a 1 off the difference). I am not sure why. One version (slower one) is from w3resource and the other one (faster one) is from geeksforgeeks. I am doing the comparison in python.
Geeks for Geeks
...ANSWER
Answered 2020-Feb-24 at 09:05The top one defines j once per outer loop. 10.000 times. In the bottom one you have to decrease j in every inner loop control for testing. That's (10.000 * 10.000 - 10.000)/2 as an upper limit (thanks to @trincot for correcting this) operations more.
Slower Version:
QUESTION
I'm Learning data structure with javascript
and my focus now on how to implement deque?
Edite: from comments below I get useful directions on how to implement
deque based array. Is there a direction how to implementdeque based objectusing class ?
I get understand some points like I need :
- addFront()
- removeFront()
- peekFront()
- addBack()
- removeBack()
- peekBack()
but I'm confused about some points :
how many pointers I need ? at least I know from queue I need two(head-tail) pointer but not sure if I need more in deque
which data type in javascript convenient in this case as a base? I saw some tutors in youtube talking about circular array for example which unknown for me in JS.
edite2:
I was following a book called: learning javascript data structures and algorithms 3rd edition
in chapter5 of this book the author started to implement Deque based on object only and some variables
but I didn't understand how he did that because the code encrypted but I can still reach to his files from and test his approach github repository
I can say that @trincot answer very close of book author approach
but when I compare the results I get this [1 = author - 2 = @trincot] :
according to the book index taking about linked list comes in chapter6 so I didn't expect his solution will be based on something he didn't mentioned before
plz if I miss any point I will be grateful to tell me it ... thanks
...ANSWER
Answered 2020-Feb-04 at 13:27As stated in comments, JavaScript has native support for deque operations via its Array class/prototype: push, pop, shift, unshift.
If you still want to write your own implementation, then you can go for a doubly linked list, where you just need two "pointers". It should be said that in JavaScript we don't really speak of pointers, but of objects. Variables or properties that get an object as value, are in fact references in JavaScript.
Alternatively, you can go for a circular array. Since in JavaScript standard Arrays are not guaranteed to be consecutive arrays as for example is the case in C, you don't really need to use an Array instance for that. A plain object (or Map) will do.
So here are two possible implementations:
Doubly Linked ListQUESTION
I am doing exercises from the app Data Structures in Scala, I have coded the second problem on Arrays like this:
...ANSWER
Answered 2018-Nov-16 at 20:42Here's a way to implement your algorithm without recursion (not that I actually think there's anything inherently wrong with recursion).
QUESTION
I'm currently studying Recursion in C# from the book "Data Structures and Algorithms" by Alfred V. Aho and Jeffrey D. Ullman. Link
This problem is used to reinforce an understanding of Recursion.
We are given a labyrinth with a rectangular shape, consisting of N*M squares.
Each square is either passable or impassable. An adventurer enters the labyrinth from its top left corner (there is the entrance) and has to reach the bottom right corner of the labyrinth (there is the exit).
At each turn the adventurer can move up, down, left or right with one position and he has no right to go outside the binderies of the labyrinth, or step on impassable square. Passing through one and the same position is also forbidden (it is considered that the adventurer is lost if after a several turns he goes back to a position he has already been).
The following program uses Recursion to print all possible routes through the Labyrinth in the console.
A two-dimensional array of characters represents the labyrinth, the character ' ' (space) marks the passable positions, '*' the impassable positions and 'e' is the exit from the labyrinth.
The positions we have already visited are marked with the character 's'.
An array "path[]" is used in order to store and print the paths we have found by our recursive algorithm, each direction is recorded as so (L – left, U – up, R – right, D – down).
A counter keeps track of how many times we have moved to the next position recursively, i.e. the current depth of recursion.
...ANSWER
Answered 2018-May-11 at 02:26In your case before making these recursive calls the position = 1. And you can imagine your stack as:
Main()->FindPath()
Imagine that in order to find exit the program made 3 recursive calls. Then the stack is:
Main()->FindPath()->FindPath()->FindPath()->FindPath()
At this point the position = 4.
After printing the result, method calls return, but before that they decrease the position by 1.
So before going for another set of recursive calls your stack is again looks like this:
Main()->FindPath()
and the position = 1.
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