bubble sort algorithm find time complexity

i am trying to find the time complexity of the bubble sort

n=length[A]
for j <- n-1 to 1
for i <- 0 to j-1
if A[i]>a[i+1]
temp=A[i]
A[i]=A[i+1]
A[i+1]=temp

return A

please any one can help thanks

Solution

In line 1 we are assigning length of array to n so constant time
In line 2 we have a for loop that decrements j by 1 every iteration until j=1 and in total will iterate n-2 times.
Inside the first for loop we have a second for loop that increments i by 1 every iteration until i=j-1 and will iterate j-1 times. On each iteration of the inner for loop we have lines 4,5,6,7 which are all just assignments and array access which cost, in total, constant time.
We can think about the two for loops in the following way: For every iteration of the outer for loop, the inner for loop will iterate j-1 times.
Therefore on the first iteration of the outer for loop, we have j = n-1. That means the inner for loop will iterate (n-1)-1 = (n-2) times. Then on the second iteration of the outer for loop we have j= n-2 so the inner for loop will iterate (n-2)-1 = (n-3) times and so on. And we do this until j = 1.
We will then have the equation: (n-2) + (n-3) + ... + 2 + 1 which is the total number of times the inner loop will iterate after the entire algorithm executes. We know that 1 + 2 + ... + n-1 + n = n(n-1)/2 so our expression can be simplified to this: n(n-1)/2 -(n-1) -n = n(n-1)/2 -2n + 1 = O(n^2).
Since our inner for loop will iterate O(n^2) times, and on each iteration do constant work, then that means our runtime will be O(cn^2) where c is the amount of constant work done by lines 4,5,6,7. Combine O(cn^2) with line 1 which is O(1) we have O(cn^2) + O(1) which is just O(n^2).
Therefore runtime of BubbleSort is O(n^2).

If you are still confused then maybe this will help: https://www.youtube.com/watch?v=Jdtq5uKz-w4