Why Bubble sort complexity is O(n^2)?

Question

As I understand, the complexity of an algorithm is a maximum number of operations performed while sorting. So, the complexity of Bubble sort should be a sum of arithmmetic progr

Answer 1

Let's do a worst case analysis.

In the worst case, the if (a[i] > a[j]) test will always be true, so the next 3 lines of code will be executed in each loop step. The inner loop goes from j=i+1 to n-1, so it will execute Sum_{j=i+1}^{n-1}{k} elementary operations (where k is a constant number of operations that involve the creation of the temp variable, array indexing, and value copying). If you solve the summation, it gives a number of elementary operations that is equal to k(n-i-1). The external loop will repeat this k(n-i-1) elementary operations from i=0 to i=n-1 (ie. Sum_{i=0}^{n-1}{k(n-i-1)}). So, again, if you solve the summation you see that the final number of elementary operations is proportional to n^2. The algorithm is quadratic in the worst case.

As you are incrementing the variable operationsCount before running any code in the inner loop, we can say that k (the number of elementary operations executed inside the inner loop) in our previous analysis is 1. So, solving Sum_{i=0}^{n-1}{n-i-1} gives n^2/2 - n/2, and substituting n with 10 gives a final result of 45, just the same result that you got by running the code.

Answer 2

This is because big-O notation describes the nature of the algorithm. The major term in the expansion (n-1) * (n-2) / 2 is n^2. And so as n increases all other terms become insignificant.

You are welcome to describe it more precisely, but for all intents and purposes the algorithm exhibits behaviour that is of the order n^2. That means if you graph the time complexity against n, you will see a parabolic growth curve.