For bestSum can we do BFS instead of DFS as it will find shorter sequence faster, also we can sort the array in reverse order if using DFS

This is a great course i recommend to watch

On 1:45:57, there should be n**m + m instead of n**m * m because the array is null if I dont have an answer and that is at max O(1) time, but If in any recursive call I have an answer, it reutrns and returns over not going to the next recursive call, so it should be m**2+ n**m.

On 1:49:25, the time complexity should be O(n*m + m**2) instead of O (n*m**2)

I just noticed. 33:27 is the same were of how morse code is done…start with E at the top, a short dit goes left a long dah goes right…rinse and repeat to get to letters you need….huh interesting

thanks for this explanation bro ✨
in the problem "howSum" for real I don't like return an array even if the language accept it by reference (in js the array is a special object as you know) and python is the same when you send any collection to a function it will accept by reference so is there another way without returning an array ?
I found this way using the stack so in each call if you didn't found an answer yet then add the number in the stack but when I finish the branch ? if you got no result then pop from the stack
the stack should be in the global scope but in python I used a nested function to make the function pure (without side effect)
another thing about my solution that's I solved the problem without watch you how you solved so I started from zero then add the number from the array to the sum and then recursive start
so in my code if you asked for 0 it will return true (specially in the canSum problem) so I checked if the target is zero and the array contains 0 I will return array of one element (zero) (this will go with O(n) as time complexity) without further talking.
Here is my memo solution for canSum in python:

def howSumDP(target, numbers):

memo = dict()

# Save the indices for numbers you got them, this will take a space O(m)

stack = []

# Indicating whether we found an answer or not

weFoundAnswer = False

def add(sumy):

nonlocal target

nonlocal numbers

nonlocal weFoundAnswer

nonlocal stack

# Base Case One: if we already has an answer

if weFoundAnswer:

return

# Base Case Two: if the sumy exceed the target

if sumy > target:

return

# Base Case Three: if the summy is equal to the target

if target == sumy:

weFoundAnswer = True # Save that we got an answer

return

if sumy in memo:

return

for i in range(len(numbers)):

memo[sumy] = 1

if not weFoundAnswer:

stack.append(numbers[i])

add(sumy + numbers[i])

if not weFoundAnswer:

stack.pop(-1)

return

# Maybe the user enter 0 and numbers doesn't contain zero then ?

if target == 0:

if 0 in numbers:

for num in numbers:

if num == 0:

return [0]

return None

# Normall way

add(0)

if len(stack) == 0:

return None

return stack

how in earth O(2^(n/2)) same as O(2^n) , just take the value of n = 20, its difference of 10^3 and 10^7

I watched it before like till 2:12:00 I am like super confident 'Yeah you got this DP'. Now after a while I hit this question Coin Change…. And really I have no clue how to do it.😂🤭

One correction: 2^n and 2^(n/2) are not the same complexity.

To be of the same complexity in this case, means that there exist positive constants A,B s.t. A*2^n <= 2^(n/2) <= B*2^n

Fantastic video @freeCodeCamp. I identified a small mistake at 46:58 in the tree diagram: the rightmost leaf node should be (2,0) and not (0,0) or am I missing something?

I can only say .. Thank you. If you want to learn DP, i don't know if there is a better video than this

1:52:00

He made it looo so simple ❤

Thank you so much ❤❤

For canConstruct, what about the indexOf call for time complexity? It seems like it has to search through the target linearly for the word, adding another m. Also, I think it'd be fast to use startsWith because it's constant time.

This video was awesome! I learned a ton from it. I really appreciate the time and effort you put into making this video. I can't thank you enough.

I love the "Over 1 million views" sticker on the thumbnail when the video now has x4 that amount.
There's some good stuff in this vid.

console.log(canSum(13,[2,5])); // true with memo how?