1.5.1 Time Complexity #1

Finding Time Complexity of Different kind of snippets

Courses on Udemy
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Java Programming

Data Structures using C and C++

C++ Programming

[achajackson]

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[SunilSamantaSinghar]

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

[ankitabasu]

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[shubhamsaini]

The way he stops and look is a teacher look. Sir you nailed it...

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[VirulentVendetta]

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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[Noushin]

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[Abdullah-mgzl]

*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