Відео

even though I am watching this video, still can't get it..

Hi bro.. do u have the solution to "Best time to buy and sell stock III"... I am in dire need of it!!... The other solutions in the internet are totally going above my head!!

U a God

Can you share the solution for "688. Knight Probability in Chessboard " using dp

This is wrinkling my brain.

After doing a few of these challenges, coding feels like tricks to manipulate stuff to get what you want. So different from other subjects

acid compliance😅

This code throws an error. You can't append to a int.

How come this O (n) complexity? there are two internal loops. For an example: for nums = [8,7,6 ,5, 4, 3, 2, 1] could be worst case time complexity scenario ie o(n^2)

So at 25 y.o, now I get how multiplication is done.This proves the whole education system is fucked.

Nice solution i used a hash map instead

yea everything you told was fine..except you gotta know that Java handles the user input in a more efficient way than the other 2

wouldnt it be easier to have a left array and a right array. Left contains the products of all elements left of that element except itself. Same goes for the right. leftArray = [1,1,2,6] rightArray = [24,12,4,1]. Then just multiply each corresponding element of leftArray to the rightArray to get outpur array = [24,12,8,6]

Thanks!

For the first problem, I guess the only optimization (beyond knowing the math eq) would be selecting the smaller of m and n, so that your memory complexity is minimized.

why my first thought is tying to write the dictionary mapper from letter to number.....

Please do more SQL 🙏

Can we improve the efficiency at 2:24:28 by mutating wordDict into the form where the key is len(word) and the value is a list of all words of that length (no extra space). Then we can check: for key in wordDict: if i+key in dp and dp[i + key] = True: for word in wordDict[key]: #check the word That way, you eliminate the vast majority of building dp.

2:08:59 why is the third value in the min necessary (just n). cur_max can never be less than 1. If cur_max is 1, its the same as n, if its larger, it's bigger than n.

var checkInclusion = function(s1, s2) { if (s2.length<s1.length) return false; let map1 = makeHashmap(s1, 0, s1.length); let map2 = makeHashmap(s2, 0, s1.length); let matches = 0; for (let [key, val] of map1.entries()) { if(map2.get(key) == val) matches++; } if(matches == map1.size) return true; for (let i = 1; i < s2.length-s1.length+1; i++) { let oldVal = s2[i-1]; let newVal = s2[i+s1.length-1]; if (map1.has(oldVal)) { if(map2.get(oldVal) === map1.get(oldVal)) matches--; map2.set(oldVal, map2.get(oldVal) - 1); if(map2.get(oldVal) === map1.get(oldVal)) matches++; } if (map1.has(newVal)) { if(map2.get(newVal) === map1.get(newVal)) matches--; map2.set(newVal, (map2.has(newVal) ? map2.get(newVal) : 0) + 1) if(map2.get(newVal) === map1.get(newVal)) matches++; } if(matches === map1.size) return true; } return false; }; var makeHashmap = function(string, start, end) { const hash = new Map(); for(let i = start; i < end; i++){ hash.set(string[i], (hash.has(string[i]) ? hash.get(string[i]) : 0) + 1) } return hash }

Very good video!!! Thanks!

thank you it is very helpful🤩🤩🤩

BFS , More intuitive solution class Solution(object): def canFinish(self, numCourses, prerequisites): """ :type numCourses: int :type prerequisites: List[List[int]] :rtype: bool """ graph = defaultdict(list) in_degree = [0] * numCourses for course, prereq in prerequisites: graph[prereq].append(course) in_degree[course] += 1 # Step 2: Initialize the queue with courses having in-degree of 0 queue = deque([i for i in range(numCourses) if in_degree[i] == 0]) # Step 3: Process the courses using BFS processed_courses = 0 while queue: current_course = queue.popleft() processed_courses += 1 # Reduce the in-degree of neighboring courses for neighbor in graph[current_course]: in_degree[neighbor] -= 1 # If in-degree becomes 0, add it to the queue if in_degree[neighbor] == 0: queue.append(neighbor) # Step 4: Check if all courses are processed return processed_courses == numCourses

does the goat respond

Thank you right man right your explanation right is very right clear right but I wish right you use right the word of "right" right less right.

I don't think you can say O(log26) for the maxHeap, since it's not just one appearance per letter, it is task.length < 10^4, so there could be 10^4 pops * idle queue

It’s actually possible to construct the adjacency list in linear time. First create a set of the word list, then for every char in every word try all the letters from the alphabet and check if it’s in the set. If it is then add it to the adjacency list for that word. Great video!

I finally understand it! Thank you!

Hi Neet, loved the solution- it was very concise and simple. One thing I would add to the explanation is that you made it seem like it was a breadth first solution instead of a depth first solution inside your diagram segment. The for-loop executes after each backtracking return, not when they are called. In other words, the next iteration of the for loop is only executed after the first completed string.

after seeing the thumbnail i thought its a fireship video

I kept waiting for him to actually describe the work and the process of how he got disengaged at Amazon. Seems like he was more eager to talk about his drug use. Becoming a great software engineer takes a lot of dedication, focus, and stamina. Forget the drugs and learn to really dig writing code.

Farmers daily life is far mor excited and spontaneous than CS these days, and the gains enough to feed a big family alone.With modern tech farming is not tough like it was a the hell. I know this.Farming is more usefull too and more ethical than CS, and more freedom adn social interactions, i know that!

I thought i might not work in some scenario, but seem i am wrong.

Java Code without causing a Time limit Exceeded in Leetcode class Solution { public List<String> findItinerary(List<List<String>> tickets) { LinkedList<String> iternary = new LinkedList<>(); Map<String,PriorityQueue<String>> graph = new HashMap<>(); Stack<String> stack = new Stack<>(); stack.push("JFK"); for(List<String> ticket : tickets){ graph.computeIfAbsent(ticket.get(0), k->new PriorityQueue<String>()); graph.get(ticket.get(0)).add(ticket.get(1)); } while(!stack.isEmpty()){ String destination = stack.peek(); if(!graph.getOrDefault(destination, new PriorityQueue<String>()).isEmpty()){ stack.push(graph.get(destination).poll()); } else{ iternary.addFirst(stack.pop()); } } return iternary; } }

yoy desreve 10M followers . Your explanation is simply superb

wow. simply fkn wow with the pos diag neg diag.. such a nice pattern to this.. holy moly. neet! 1337!!

brilliant!!!!

literally this is insaneeee

I still don't get why does (self.isSameTree(p.left, q.left)) works. Could someone explain a bit?

Could someone explain a little bit what does 0 and 1 in "balanced = left[0] and right [0] and abs(left[1] - right[1] <= 1)" mean?

Guys you can solve this problem using hash set very easily

Similar events happened to me at the level of detail that it is uncanny (and at around the same time 2018-2019ish.) I actually started my CS journey after the darker paranoia stuff. Funny how we both ended up on LeetCode at the end of the day. Cheers to you friend for sharing your story. Time for me to lock in and do some LeetCode!

Dogg, I have been waiting for someone to explain the learning process using analogy to ML models--it's something I think about a lot, it seems *so* many people just have no idea how to teach themselves anything, how to think about learning itself, and a lot could really benefit from recognizing it is a fairly predictable process that need not generate any anxiety at all. People talk about banging their head against the wall for hours trying to figure a single problem out, and I just feel bad for them. I never, ever do that, and yet always (eventually) succeed at learning anything I want to--because I understand that it simply isn't necessary, any more than a neural network learning how to, say, complete a trackmania level *needs* to reach the finish line every single iteration--the whole point is to fail repeatedly with incremental improvement. Waste no time beating yourself up over something you don't know, just note "okay there's still a gap there" and move on. Come back to it later armed with a little more surrounding knowledge. Explore, approach the topic from multiple angles, pursue points of personal curiosity that provide precious motivation--as long as you continue processing relevant information, *your understanding of a topic WILL eventually gel into a functional model without ever needing to look complete at any point earlier along the way!*. That could be worded much more clearly but I don't feel like it right now. However, by making this one attempt to verbalize this concept now, I am certain I have made incremental progress toward being capable of explaining it very articulately in the future. No stress!

import java.util.*; class Solution { public int maxSubArray(int[] nums) { int sum=nums[0]; int max=Integer.MIN_VALUE; max=Math.max(max,sum); if(sum<0) sum=0; for(int i=1;i<nums.length;i++){ sum+=nums[i]; max=Math.max(max,sum); if(sum<0) sum=0; } return max; } }

Once the logic explanation was done...code looks easy peasy Great work.

Liked immediately after seeing Naruto reference

def numDecodings(self, s: str) -> int: if s[0] == '0': return 0 n = len(s) dp = [0] * (n + 1) dp[0], dp[1] = 1, 1 for i in range(2, n + 1): one = int(s[i - 1]) two = int(s[i - 2:i]) if 1 <= one <= 9: dp[i] += dp[i - 1] if 10 <= two <= 26: dp[i] += dp[i - 2] return dp[n] this is more like understandable dp tbh

Great explanation, with ur explanation feeling like it was so easy. Thank u so much

Why can't I just reverse the list using slicing? I'm still pretty new to coding, but it seems hella easier to just type out head[::-1] and call it a day. I'm guessing it has something to do with the time complexity?

A recursive solution is more intuitive but of course, is not O(1) in terms of extra space.