how to calculate Bubble sort Time Complexity

Question

I was trying to understand the Data Structure and different algorithm, then i got confused to measure the Bubble sort time complexity.

for (c = 0; c < ( n - 1 ); c++) {
  for (d = 0; d < n - c - 1; d++) {
    if (array[d] > array[d+1]) /* For descending order use < */
    {
      swap       = array[d];
      array[d]   = array[d+1];
      array[d+1] = swap;
    }
  }
}

Now every Big O tells Best case O(n), Avg case(n2) and Worst Case(n2).But when i see the code, found in first phase inner loop run n time then in second phase n - 1, and n - 2 and so on. That means in every iteration its value goes down. For example if i have a[] = {4, 2, 9, 5, 3, 6, 11} so the total number of comparison will be -

1st Phase - 7 time
2nd phase - 6 time
3rd Phase - 5 time
4th Phase - 4 time
5th Phase - 3 time
6th Phase - 2 time
7th Phase - 1 time

so when i calculate the time it looks like = (7 + 6 + 5 + 4 + 3 + 2 + 1) + 7 = 35, but the worst time complexity is n2 as per doc.

so can Someone tell me how to calculate the correct value.

Answer 1

Let's go through the cases for Big O for Bubble Sort

Case 1) O(n) (Best case) This time complexity can occur if the array is already sorted, and that means that no swap occurred and only 1 iteration of n elements

Case 2) O(n^2) (Worst case) The worst case is if the array is already sorted but in descending order. This means that in the first iteration it would have to look at n elements, then after that it would look n - 1 elements (since the biggest integer is at the end) and so on and so forth till 1 comparison occurs. Big-O = n + n - 1 + n - 2 ... + 1 = (n*(n + 1))/2 = O(n^2)

In your example, it may not examine these many elements in each phase as the array is not in descending order.

Answer 2

So you've noticed that the total number of comparisons done is (n - 1) + ... + 2 + 1. This sum is equal to n * (n - 1) / 2 (see Triangular Numbers) which is equal to 0.5 n^2 - 0.5 n which is clearly O(n^2).

Answer 3

it does comparison between two elements. so in 1st Phase - n-1 comparison. i.e, 6 2nd phase - n-2 comparison. i.e, 5 and so on till 1. and thus, sum = n(n-1)/2 i.e O(n^2).

if there is any error you can correct.....

Answer 4

O(n^2) = n(n-1)/2 is the right one.

As in the above example of 5 elements.

5(5-1)/2 == 10.

5(5+1)/2 != 10.