1.5.1 Time Complexity #1 

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When you pause after a statement, it feels like I am actually physically sitting in the class and you are giving me a few seconds to grasp the information and proceeding after I have said Yes Sir! This feels so real! THANK YOU for doing this :) God bless you.

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Honestly everybody commented so accurately my feelings are exactly the same as every body in the comment section. I wish I could have got a teacher like you sir, when I was still in my college which ended the last month. Completed my graduation but this 'Time complexity' was not something which I understood for all my graduation years. But then discovering your channel and this lecture series ... make me feel blessed . Thanks a lot

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So beautiful explained. I am preparing fir exams & interview watching your videos. A little mistake at 5:30 f(n)= (n^2 + n) /2

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When I get stuck on solution. Sir : Let us analyse!!!! Me: 😇😌😌

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*Key Takeaways:* - time complexity analysis is basically trying to find out *how the number of instructions/steps required changes as a function of the input size* - you do this by basically seeing how many times each instruction/step of the algorithm/code executes for each input - if there is an instruction inside a for loop, and the for loop goes from 0 to n (n being the size of your input), then the instruction gets executed n times...this is pretty straightforward - if there is a for loop nested within a for loop, and both loops go from 0 to n, then the instruction in the *nested* loop will be executed n x n times. This is because the inner loop is executed n times, and each time the inner loop is executed, the instruction in it is executed n times, so in total, that instruction is executed n by n times - if you have an outter for loop that goes from 0 to n, but the inner for loop goes from 0 to i of the outter loop, let's take a look at how many times the inner loop code gets executed for each iteration of the outer loop. During the first iteration of the outter loop, the code in the inner loop gets executed 1 time (from 0 to i, which is 1). During the second iteration of the outter loop, the code in the inner loop gets executed 2 times (from 0 to i, which is now 2). During the third iteration, the code in the inner loop gets executed 3 times (from 0 to i, which is now 3), and so on. You basically have a little triangle here, with height n, and base n. To find the area, you do (n x n / 2) I may be off by 1, but I wanted to keep the explanation simple. Thanks again for the amazing videos

"Let us Analyze" every time whenever sir says this, there is a curiosity in my mind. Awesome Explanation

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Wow I think this is the by far one of the best videos on time complexity in code. Thank you so much for helping!

Sir, I have seen the first person explaining this topic in so much detail with easy examples

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What a Teacher! What a Teaching! In order to help my son understand O(n) fully well, I was searching for some videos. I was not satisfied with any of them until I found this video [and the series of videos on O(n) to follow this]. Mighty helpful - all of the videos in this series. My son (CSE first year) very quickly understood O(n) to the hilt. Once again, I reiterate: "What a Teacher! What a Teaching!" Pranams ever.

Exam tomorrow, 1.5x speed

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7:37 for who did not understand the difference between k and n,assume n=10,then the loop runs for k times=4 times;p=1+2+3+4=10.That means k(value=4)

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5:53 -->(n^2 +n)/2

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Nsquare +n by 2 at 5:52 . However degree of polynomial remains same

Thankyou sir... I was struggling with these concepts for a long time. but now all clear... thankyou very much

Thank you so much sir for explaining it so clearly. This is so good for the beginners.....

Do you think that there's an error in the third example that starts around 2:50? In the time units calculation, in my opinion, instead of 1 + 2 +3 + ... + n, there should be 1 + 2 + 3 + ... + n-1. Because when i = n, the for loop will be terminated. What do you guys think?

7:45 for those who were lost like I was when he immediately goes into setting P = k(k+1)/2... Look up the "Gauss Addition Trick". Essentially this is a recognized addition sequence (1+2+3... etc) that is represented by that equation. He's assuming we know how he got there, or forgot to explain it.

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Here's my explanation for the order of n^(1/2). So, when i=1, the program control goes inside the loop and the assignment statement (p=p+i) runs. This is the first time this statement executed. Let's keep track of i and p values. At first, when i =1, p =1. Now, move on to the second iteration. This time i=2 and p= (1+2). Now for the 3rd iteration, i=3 and p=1+2+3. Let's move on to fourth iteration. Now i=4 and p=1+2+3+4. Let's observe a pattern here. In total, we have run the assignment FOUR times till now which is exactly equal to the value of i. So when i= 10 (suppose) , the assignment statement will run 10 times. So when the value of i reaches k, the assignment will run for k times. We have to be aware that we are looking for how many times the assignment statement will run. So, that's k times. We found it but it's not as a function of n. So when i=k, p=(k(k+1))/2. It's a series formula for summing natural numbers upto k. This loop will stop when p>n. So (k^2+K)/2 > n . We solve this inequality and get k = (2n-k)^0.5. So, that's O(n^0.5).

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it was wonderful lesson for me .although my english's bad I understand time complexity on loops with your clear explanation

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I got scared after he paused for few second. I thought he was gonna pick on me and I was thinking about the answer.

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Clear, simple explanations, well done. Many thanks, this video is helpful in comprehending programming algorithm notation.

5:20 correction: (n(n+1))/2 = (n^2 + n)/2

at 5:49, actually F(n) = (n^2 + n) / 2 [instead of (n^2 +1) / 2], the result is the same since we only take care of the n with biggest exponential

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for(int i=0; i

@ 5:27 it should be (n^2 + n)/2 and not (n^2+1)/2. Thanks for making these freely available on youtube

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5:25 why it's not n* (n(n+1)/2)? why first loop's n executions not considered?

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The best time complexity for his videos is watching in 1.5x speed😌

Sir in last program, you said i will run for k times, but actually value of p is deciding whether stmt has to be executed or not, when value of p is 6(which is greater than n) the stmt will not be executed. So if we see, i is executed for (1,2,3,4) i.e 4 times(n+1) and the statement p=p+i will execute for 3 times only. So the time complexity should be O(n).. is this correct or not?

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5:30 The value of f(n) should be (n^2 + n) / 2. and NOT (n^2 + 1)/2

sir shouldn't the function be n(n-1)/2 because 'i' cannot have value 'n'

In the first example The loop runs n times rather than ( n + 1) times. The first loop iteration has i = 0; the second has i = 1, and so on until the last loop iteration has i = (n - 1). When i = (n - 1), we reach the end of the loop. At this point, we go to the top and run i++. As a result, i = n. The loop condition is then checked. Because I

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I had noticed a problem in 5:54 since it is a nested loop then the inner loop should run f(n) = n((n^2 + n) /2) making O(n) = n^3

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5:54 f(n) = ( (n^2) + n ) / 2 , shouldn't n be mulitpled to 1 or am I missing something ?

5:30 ,sir, you have made incorrect multiplication of f(n). It should be f(n) = (n^2 + n )/2.

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Thanks!

@5:50 see something?

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Comprehensive and great lecture. Thanks a lot Sir.

R u watching in 2020

Really awesome explanation, thank you sir making it too clear...

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For a loop starting from 0 and following the condition of i>n with an incrementation of 1 is basically 'n'times and not 'n+1' times. It were 'n+1' only if condition would have been either i>=n or i>n+1.