Solving Coding Interview Questions in Python on LeetCode (easy & medium problems)

Hey everyone! If you enjoy this video, I just created a new channel where I'll be posting problems like this pretty frequently. Check it out here: https://www.youtube.com/channel/UCge9kgmp38FIjGK4xdJsThw/about (will start posting when it hits 1000 subs)

Also, we're getting so close to 100K!! So crazy. I really appreciate everyone's support and am super thankful to have a platform where the content can reach so many. Love ya'll!

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

I applied for an SRE role with a company but they also wanted someone with some Python experience, other than building speech to text chatbots and local Jarvis virtual assistants I've never worked with Python on this type of level. I did horribly on the assessment which lead me to this video. Don't know why but I cannot wrap my head around this binary, data parsing or algorithm stuff with Python, it's next level type stuff… Big ups to all that understand this, I think I'll just stick with bash scripting and IAC- Orchestration/Automation and API's using Terraform, Ansible, Packer 😒

18:00 Mistakes are a part of the process.

im not smart enough for this : ( are these really easy?

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

you're brilliant bro. The way you think really helps me understand how to solve these questions

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)

Very informative content. It’d be even more awesome if you could explain a little more in depth for beginners

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)

We love the way you did this. I prefer this approach over youtubers that clearly comes prepared and solve it with no hesitancy and in a bit of complex way

Thank you for doing this.

SIMPLE SOLUTION:

a=[8,1,2,2,3]

m=[]

b=0

for i in range(0,len(a)):#0 to 4

for j in range(0,len(a)):#0 to 4

if a[i]>a[j]:

b+=1

m.append(b)

b*=0

print(m)

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

You miss the use of a lot of simple, easy to read builtins

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

Hi, I tried the sharing between friends question but got no output
data = [2, 4, 1, 2, 7, 8]
Alice = 0
Roso = 0
Bob =0

def split_data(data, Alice, Roso, Bob):
share = len(data)//2
while len(data) != 0:
Alice += data[0]
Roso += data[1]
Bob += data[2]
return Alice
return Roso
return Bob

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

My solution was:
If the second index of any pair does not exist as the source of any other pairs, then that's the city.

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

src=[]

dst=[]

for x in paths:

src.append(x[0])

dst.append(x[1])

for d in dst:

if not (d in src):

return d

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

for the 1436, the key point is using hash table.
def destCity(self, paths: List[List[str]]) -> str:
citiesA = {path[0] for path in paths}
return next(path[1] for path in paths if path[1] not in citiesA)

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.

Destination City

out_city = {path[0] for path in paths}

for path in paths:
if path[1] in out_city.keys():
out_city.pop(path[1])

The only remaining key would be the answer

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

Thank you, learned a lot from this video, Please post more videos

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

Love the video, but I dont think the longest palindrome problem is fully correct because if you have 2 instances of odd characters, you would need to only add the count of the largest number of a specific odd character. EX: "abbccccddeee" would need to ignore the odd "a" character but instead would include the 3 odd "e" characters.

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

He got drake eye brows

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()

a3
class Solution:

def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:

cat = sorted(nums)

lol = []

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

lol.append(cat.index(nums[i]))

return(lol)

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])

Hey Keith or anyone care to explain what is going on in the function parameter(smallerNumberThanCurrent(self,nums …) I dont understand what is going on there.Thanks

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.

Good Video.

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

Hey Keith, well done video!
For the palindrome problem, your solution wouldn't work properly if you have 2 different characters with odd number of appearances, you would add both of them to the count while only one could be used as the palindrome "center"

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

Thank you, you are the best

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

Hi Keith,

This is nit picking but in for i in range(len(x)) loops are anti-patterns, for index,value in enumerate(iterable), is the pythonic way to access the index in a for loop

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?

Hey Keith more videos on leetcode please, this video is super awesome, love the way how you explain each and every step, please do more videos on leetcode thank you

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

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

Love your videos. Each time you upload a new one, its just like you read my mind and give me exactly what I need to become a data scientist. Please do a SQL tutorial or even better a "real world" example of how using SQL together with Python. Greetings from Sweden

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

Do you have any suggestions for the GME thing for me? Im getting started on this crap