Thank you Keith

close to 100k lets gooo

NEAT!

Thank you, Keith! Best lessons , as usual!

Did he get a haircut?

Thank you so much

I love my own solution to the second question on Destination City. I simply returned the cityB that doesn't occur anywhere else in the whole list.

class Solution:
def destCity(self, paths: List[List[str]]) -> str:
for path in paths:
lister = [x for x in paths if x != path]
if not (any(path[1] in listeri for listeri in lister)):
return path[1]

Thanks keith. I've been looking for content like this.

Thank you Keith for great Videos. A couple of suggestions: If you can dissect the problems, that would be great. E.g. String command = string method in Python. In order to solve this problem, you need to know class, string methods, replace() and functions. Here are the links for class,…,functions.

Dude , people are gonna think you are u/deepfuckingvalue .
The guy who started the GME🚀🚀🚀

Loved this 🔥 please do more of such leetcode questions. Let's hit that 100k and 1k😁

For the binary search question, couldnt you just do something like

try:
return (nums.index(target))
except:
return "-1"

For smallerNumbersThanCurrent() could you get rid of the if statement and just overwrite the dictionary values, because the last overwrite will be the correct one to save. Or does writing to a dictionary take more time than the if statement to check?

loved how you have labeled each segment of the video with the corresponding problem title , keep up the great work !!

Great job Keith! Thanks a lot.
In case someone finds the maxCoin solution too confusing with all those i and indexes, here is simpler one I came up with:
def maxCoins(piles):

sorted_piles = sorted(piles)

# sorted = [1, 2, 3, 4, 5, 6, 7, 8, 9]

my_score = 0

while len(sorted_piles) > 0:

del sorted_piles[-1] #deleting the biggest

my_score += sorted_piles.pop(-1)

del sorted_piles[0] #deleting the smallest

return my_score

I have an interview as an intern at amazon on the 11th I will comment if I made it or not wish me luck!!! I haven't taken any alogirthm class before so I know this is going to be hard

The very first problem of Goal Parser Interpretation, I made a mess.
1. First I didn't read the question properly. My mind started thinking of different edge cases, wherein different words might be given as test inputs. (e.g. F()()t(ball), etc)
2. I started looping over the string, I'm much habituated to loop over things which is the worst approach ever.
3. I started trying to figure out to how to replace the single brackets without creating blank spaces.
4. I never read the constraints properly.

Thanks Keith for this video, it help me analyze my problems and I can start working on fixing those whilst attempting these interview questions.

a2
class Solution:

def destCity(self, paths: List[List[str]]) -> str:

cat = []

dog = []

for i in range(0,len(paths)):

cat.append(paths[i][0])

dog.append(paths[i][1])

for i in range(0,len(dog)):

if dog[i] not in cat:

return(dog[i])

For 1561 maximum number of coins:

piles.sort()
c = -2
result = 0
for i in range(len(piles)//3):
result += piles[ c ]
c -= 2
return result

Or in for loop
piles.pop()
result += piles.pop()

confused on how in the example [-1,3,4,6,8,9,12,15,18] if the target is 10, it would break out of the while loop instead of being stuck inside the while loop

interesting and tricky little prob:

def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
result = []
for i in range(len(nums)):
targets = [1 for (idx,x) in enumerate(nums) if idx != i and x < nums[i]]
result.append(sum(targets))
return result

Excellent video! Greatly helped me. Thank you very much!

If the language is Python and you are allowed to 'use' Python, for question 704. Binary Search my initial thought was 'return nums.index(target)'. Why reinvent the wheel unless it was specified that the solution should be faster than the built-in function. Great video and this did definitely get the brain thinking about possible solutions and ways to improve them.

Que vídeo incrível. Surreal. Muito didático até para mim, brasileiro.

Great video!
Please note the middle can overflow. Thus, we prefer to use mid=start+(end-start) / 2

Thank you very much for the video, it helped a lot !!(:

For Second Solution, I simply converted the nested list to the dictionary and iterated the dictionary to find the city which is not a key in that dictionary :

cities = dict(cities)

#{'London': 'New York', 'New York': 'Lima', 'Lima': 'Sao Paulo'}

for city in cities.values() :

if city not in cities.keys() :

return city

Thank you for doing this.

Ok, this one works in my condition

class Solution:
def destCity(self, paths):#: List[List[str]]) -> str:
going_count = {}
for i in paths:
[*cities] = i
for j in range(len(cities)-1):
going_count[cities[j]] = going_count.get(cities[j], 0) + 1
going_count[cities[-1]] = 0

print(going_count)

dest = []

for city in going_count:
if going_count[city] == 0:
dest.append(city)

return dest

paths = [['a', 'b', 'c', 'd'], ['a', 'c', 'e', 'f'], ['b', 'c', 'f'], ['d', 'j']]#, ['d', 'c']] # should not use dict

solution = Solution()
solution.destCity(paths)

For 33:13 – LeetCode 704 – Binary Search,this is the most optimal solution.
class Solution(object):
def search(self, nums, target):
return -1 if target not in nums else nums.index(target)

If we are appending the root to the output array in each iteration. How did we end up having subarrays in the output ? This confused me any help is appreciated

18:00 Mistakes are a part of the process.

Thanks Kieth for so many insights.
However, for LongestPalindrome, I think the solution won't work for input 'aaabbb', i.e., when more than one letter have odd count

This is very helpful. more of these please!!

destination city : (easier way) lazy way 😛

paths = [["London","New York"],["New York","Lima"],["Lima","Sao Paulo"]]

dic = dict(paths)

origin = list(dic.keys())

destination = list(dic.values())

ans = set(destination).difference(set(origin))