Video

4:00 another way to look at it is to see that ALL the solutions that have a 1 will be in the first branch. That way you don't need to check again in the branch that starts with 2

Got asked a harder variation of this in a Google interview. No real hint was also provided by the interviewer. I suppose luck is also a factor. If you've seen the problem before or if you're able to fit whatever problem, you get into the pattern that you've seen before.

This is awesome, solution I tried without using left & right pointer approach `def maxProfit(self, prices: List[int]) -> int: min_price = float('inf') max_profit = 0 for price in prices: min_price = min(min_price, price) profit = price - min_price max_profit = max(max_profit, profit) return max_profit`

On the profile picture updating async, you could always solve that by referencing a latest.png and then only updating the object store to change latest.png to reference to the newest image and hence the changes would propagate much faster than updating 1000x video documents

“And of course everyone’s favourite: Dynamic programming”

You always explain such this it is one of the best explanation🤩

This is the most vague problem I have seen on NC and LC. I have a solution but since not much is said about the decoder other than it's a string, I'll assume this is valid. The problem with using any sort of delimiter is determined on the constraints. In neetcode it says any utf-8 character is valid. In leetcode it has simplified the problem for any ascii character. The harder problem of any utf-8 character in neetcode, you could write a test case to break it with any normal character like ':', it is all banking on obscurity or not commonly used. For this reason I don't think the video provides a valid solution. So based on the constraints, one solution is to convert the string into utf-16 and then I used '\xFFFF' as the delimiter, since it is outside of utf-8. Then in the decode split on '\xFFFF' then convert the strings back to utf-8.

beautiful drawing and great explanation!!!!!!

Simple solution in Java🙂⬇ class Solution { List<Integer> indexes = new ArrayList<>(); public String encode(List<String> strs) { StringBuilder stringsCombined = new StringBuilder(); for(String word : strs ) { stringsCombined.append(word); indexes.add(stringsCombined.length()); } return stringsCombined.toString(); } public List<String> decode(String str) { int previousIndex = 0; List<String> answer = new ArrayList<>(); for(int index : indexes) { String word = str.substring(previousIndex, index); answer.add(word); previousIndex = index; } return answer; } }

Thanks!

I just solved this question in neetcode but I'm not sure this is actually a correct solution in all cases. Maybe I'm missing something, but what happens if we have an array like [9, 2, 3, 10, 4, 7]? In that case the best solution would be to pick 9, 10, and 7, jumping over both 2 and 3. I'm not sure this solution would pick up those houses. This is what I came up with (in C#): public class Solution { public int Rob(int[] nums) { var accum = new int[] {0, 0, 0}; for (int i = nums.Length - 1; i >= 0; i--) { var curr = nums[i] + Math.Max(accum[1], accum[2]); accum[2] = accum[1]; accum[1] = accum[0]; accum[0] = curr; } return Math.Max(accum[0], accum[1]); } }

fk it word1=word2

It’s a reasonable story. I had a similar experience at Meta but there was nothing to point at that was wrong so it took me 3.5 years to quit. Sigh. Thanks for sharing.

Amen

This shit does not make sense bruh. I'm cooked.

i did the trie for the matrix instead of given words lol

@NeetCode What are you using for a tablet/whiteboarding software?

Why creation of adjacency list take O(n.m.m) when it is O(n.m). The operation adj[pattern].append(word) should be in O(1) time right?

using a stack is over kill.. just travel inorder, and set k = k - 1. When k == 0 it's the correct value

See I was solving this problem on the actual neetcode website which does not have the hint about each node in the stack having a minimum value. That hint was all I needed, I actually solved it on my own with that! Would be great if the site included any hints that are on Leetcode.

I love you man! ❤ Keep up the great work

I am sorry but Time complexity looks wrong because you have n selection in each step because you do not increment index. In subset problem we have 2^n because we increment index even if we dont choose current element

Perhaps he will fail on behavior interview... he's disrespectful to the interviewer, which is a big bad sign. Words come too quick from his mouth.

perfect solution and illustration, thanks so much

what if we come to know that 0 is child of 6 at the end

having 2 while loop in this case == O(n²) time?

it isn't intuitive, if you have to forcefully apply binary search, given you have the optimal time complexity known.

u only need to care if indexes before value 0 can get pass it or not

Thanks! Wonderful explanation

A possible way to solve the TLE issue is using a dict to store the count of flight ticket. It aviods the list insertion operation which is O(n). The main logic reminds the same. class Solution: def findItinerary(self, tickets: List[List[str]]) -> List[str]: adjList = {} self.ticketSize = len(tickets) path = [] for pair in tickets: start = pair[0] dest = pair[1] if start not in adjList: adjList[start] = {dest:1} else: if dest not in adjList[start]: adjList[start][dest] = 1 else: adjList[start][dest] += 1 for key in adjList: adjList[key] = dict(sorted(adjList[key].items(), key=lambda item: item[0])) def traverse(cur): if self.ticketSize == 0: return True elif cur not in adjList: return False for nextStop in adjList[cur].keys(): if adjList[cur][nextStop] > 0: adjList[cur][nextStop] -= 1 self.ticketSize -= 1 path.append(nextStop) if travese(nextStop): return True adjList[cur][nextStop] += 1 self.ticketSize += 1 path.pop(-1) return False path.append("JFK") travese("JFK") return path

aw < 3

what is the time complexity for this solution?

i dont understand the "if dp[amount] != amount + 1 else return -1". Why is this if statement valid?

dude i got an approach before but i wasn't sure of its implementation when i saw urs it's not totally same but the iteration part was missing in my approach now i got that thx dude appreciated it

How does this solution take into account the fact that |n - m| <= 1 ?

public int missingNumber(int[] nums) { int xor = nums.length; for(int i=0;i<nums.length;i++){ xor ^= i; xor ^= nums[i]; } return xor; }

Easy 🤡

Why is both "visit" and "output" required? One of them should be enough right?

everyone at first talking about market. Sad reality but

well, the hiring didn't pick up since then... :D But doing leetcode for fun is a much better reason than doing it just to get a job :P Especially since a mediocre problem base for doing the actual interviews... :D

Instead of optimizing it with cache, I've got an almost identical result (1300ms->50ms) by simply ensuring you don't process subsequent stars. So if there is a*a*a*, you'd only check it once. Now the cache seems obvious, but that addressed directly the most compute intensive cases :) Funny to see how it's same efficient. Using both would probably improve it even further.

You don't need to check the condition res==MAX//10 && digit>=MAX%10 cause this would mean that the input should be 7463847412 ( reverse of 2147483647). This is outside the int range as in the question they mentioned the input is an integer. Similarly you also can avoid the res==MIN//10 && digit<=MIN%10 condition .

so Mr. Beast was a software engineer so amazed by youtube architecture that he started youtubing

I think returning -1 and making 2 + left + right makes it more complicated or at least for me. The below is working perfectly fine as well. def dfs(root: Optional[TreeNode]) -> int: if not root: return 0 left = dfs(root.left) right = dfs(root.right) result[0] = max(result[0], left + right) return 1 + max(left, right)

Can we all agree coding interviews are a form of hazing?

what is the space complexity?