Hey guys this course has no arrays concept

This it to keep track where I am in the video:
01:44:44

To know where I remained:
48:30

checkpoints for myself:
4. 2:33:29
3. 1:45:00
2. 1:14:00
1. 45:00

1:25:29

3:23:49

Is heaps/stacks included?

awesome teaching sir!

2:37:00

1:59:00

3:25:00

Keeping track of my progress:
1:04:33, 1:44:15

I like the problem solving template approach to coding! But I do wish the instructor spent less time explaining how to use the platform

Thank you so much !

Now ı will first do this gentelman’s project on flask and python then ı wil devour this course as many times it takes! I am really thankful to you guys. When ı find a job ı will be an avid supporter!!

Bro spent more time explaining his website than actually explaining the material

I don't think 7:36:00 is correct:
Memoization-exclusive time complexity doesn't seem to be 2^m+n, given m and n are the nums of elements in two sequences.

first, finding the longest common subsequence by quick sort would be meaningless if one of the lists/sequences is exhausted, regardless of the remainders within the other. That means, and as stated by the code itself, the code will stop getting rid of character/ element when there's no more comparison to take place, so making "all leaves end at 0, 0" unachievable. (They either stuck in a loop like (1, 0) would generate (1, 0) and (0, 0) or could not be proceeded further.)

So, based on the above, it looks like the tree is not a symmetrical binary and hence 2^x could no be induced from the statement "each element would cause two possible choices to be made". But when we tried that — 2^(m+n) — with small numbers, it does approximate what was stated and seemed to be true. Let's say [a, b] & [c, d]. but as the quantity of input gets exponentially larger, I'm no longer convinced that is still correct.

When we're given a fixed yet allocable nums of elements (you can see this as you gambling chips), the max num we could potentially get from them from a multiplication is through the minumum difference between the two numbers. (like how 2×2 > 1×3) So, the worst case where iteration would be at maximum, under the circumstance that total num of elements of both sequences are fixed, would be when seq1 and seq2 have the same number. And now what would be the other extreme case? I'm sure you can tell it's when there is only one element in seq1 and the remainder in seq2. The time complexity of that would be:
(2 x N of the longest sequence)+1
and the leaves would just be N+1.

Now let's get back to the "worst of worst" case. I've tried to construct the binary tree up to (4, 4) root and noticed it does resemble a perfect, symmetrical binary tree but with surplus leaves (the number of leaves after you filled out all the gaps left by a zero-sth pairs to make it a perfect binary). And my conclusion was that the time complexity should be of 2 to the power of the height of tree, instead of simply adding m and n. The increment or trend of the surplus leaves is a multiplication of 4 for every one element added to both sequences: I had 4 leaves at (2, 2), 8 at (3, 3), and 12 at (4, 4), so on and forth. And we know that exponent has a greater growing power than multiplication, so eventually the disparity between the number given by m+n and the surplus leaves would be exponentially greater.

From the sum of geometrical series, if we substitute a with 1 and r with 2, we know that the sum of all the nunber of nodes (including leaves) of a h-layer tree would be approximately 2^h. And for a same-number pair root, the height of the tree (excluding root and surplus leaves) is just 2n-2. On a side note, we can now deduce that if we toggle the number on the scale, let say from a fixed total of 10, the number of nodes would lie somewhere between 19 to ~256. Hence, it would be more intuitive (for me, personally) to say that the time complexity of the quick sort that without memoization would be ~ 2^h, for which h is the number of layers excluding root (or counting from 0 if you wanna do so). But, ultimately, it's just the matter of 2^n+2, so I assumed it doesn't affect much. In the end, worst case of memoization (which simply is m*n) would beat 2^(2n-2).

Is this a best cource for 2nd year student?

will this video taught me python also as i am a beginner ? please give reply….

Hi Akash, Just want to apprecite your work and the efforts you have put in this course, I am doing programming since long and currently working as an engineer, but this course changed the way of my programminng, have saw so many such course but this one is truly exceptional.

It would have been really nice to have chapter markers on this video, but this is gold.

FreeCodeCamp saving lives as always. Helped me with my college project with your extensive UI UX tutorials. And saving my exams with this