วีดีโอ

great explanation

I never would have thought of this solution. Thank you!

you could check if count > k within while loop if so then return true. This will save some cpu cycles. Also, use for s in sweetness to reduce the dereference

谢谢

Very well explained. Thank you so much for this!

what extension do you use for drawing?

a mistake at 3:35, should be 9, 9, and 1 for first example

great video!

🔥🔥🔥🫡

the coding: class Solution: def findInMountainArray(self, target: int, mountainArr: 'MountainArray') -> int: def findPeak(): l, r = 1, mountainArr.length() - 2 res = 0 while l <= r: m = (l + r) // 2 if mountainArr.get(m) > mountainArr.get(m + 1): res = m r = m - 1 else: l = m + 1 return res pIndex = findPeak() def binary_search(l, r): while l <= r: m = (l + r) // 2 if mountainArr.get(m) > target: r = m - 1 elif mountainArr.get(m) < target: l = m + 1 else: return m return -1 def binary_search2(l, r): while l <= r: m = (l + r) // 2 if mountainArr.get(m) > target: l = m + 1 elif mountainArr.get(m) < target: r = m - 1 else: return m return -1 idx = -1 idx = binary_search(0, pIndex) if idx != -1: return idx return binary_search2(pIndex, mountainArr.length() - 1)

Exactly this was asked to me in microsoft . May i know How you got to know ?

some tech issues, here is the whole coding: class Solution: def minimumLevels(self, possible: List[int]) -> int: nums = [] for p in possible: if p == 1: nums.append(p) else: nums.append(-1) s = sum(nums) res = 0 for i, n in enumerate(nums): res += n if res > s - res and i < len(nums) - 1: return i + 1 return -1

哥好牛

沖！

for anyone who check the video may need the last code, or you can code by yourself. class Solution: def findLadders(self, beginWord: str, endWord: str, wordList: List[str]) -> List[List[str]]: wordSet = set(wordList) ans = [] if endWord not in wordSet: return ans l, s1, s2 = len(beginWord), {beginWord}, {endWord} flag, d = False, defaultdict(list) while s1: s = set() for w in s1: words = [w[:i] + c + w[i + 1:] for c in ascii_lowercase for i in range(l)] for word in words: if word in s2: flag = True if word in wordSet: s.add(word) d[word].append(w) for word in s: wordSet.remove(word) s1 = s if flag: break def backtrack(cur, res): if cur == beginWord: ans.append(res[::-1]) return for word in d[cur]: backtrack(word, res + [word]) return ans return backtrack(endWord, [endWord])

Great explanation thanks a lot!

Great explaination!!

thank you so much

clear explanation on the core idea !!

very good explanation = thank you so much

This is the best explanation and solution to the problem I could find on TH-cam! Thanks!

Great video! The most detailed explaination I saw! Please keep posting! It will be even better if you could provide your code, thanks!

keep posting BoSS

instand of going from 2 directions, do it as cherry pickup 1, start from 1 direction

nice!

very well sir. Thanks

Thanks

👍👍

it may have been easier to understand by keeping the/one example

Great explanation. Thank you so much. Keep this great work up !! 🥰

Keep up the good work!

good explanation = thank you so much + please continue❤❤

Thanks a lot this was helpful👍👍

给我讲懂了，太强了

What is your intution when you come across this problem?

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very easy and intuitive solution

Thanks!

Nice reading comprehension

Thanks for the solution, this would be very hard without sortedlist.

Hey buddy I have no idea about this but Im still watching..!

This was helpful thanks!

why do you draw that grid's border as a semi-circle? that should be a straight line god damn it!

this is a 4 color problem

class Solution: def findTheDistanceValue(self, arr1: List[int], arr2: List[int], d: int) -> int: myset = set(arr2) count = 0 def process(i,d): ans = [i] while d>=0: ans.extend([i-d,i+d]) d-=1 return ans for i in arr1: acceptable_values=process(i,d) ll=list(filter(lambda x:x in myset,acceptable_values)) if len(ll)>=1: continue else: count+=1 return count

Elegant

we can use return list(A.intersection(P)) to make it mush easier to write

If someone is good at codeforces problems can they easily become good at leetcode problems

Thanks for the video, helped me out. keep up the good work! 🌸