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Best Case: If the list is sorted already then A[i] <= key at line 4. Thus, rest of the lines in the inner loop will not execute. Then,
T (n) = c1n + c2 (n -1) + c3(n -1) + c4 (n -1) = O (n), which indicates that the time complexity is linear.
Worst Case: This case arises while the list is sorted in reverse order. Thus, for execution of line 1, the Boolean condition at line 4 will be true.
So, step line 4 is executed
T (n) = c1n + c2(n -1) + c3(n -1) + c4 (n(n+1)/2 - 1) + c5(n(n -1)/2) + c6(n(n-1)/2) + c7 (n -1)
= O (n2).
Average case: In mostly cases, the list will be into some random order. That is, it neither sorted in descending or ascending order and the time complexity will lie somewhere among the best & the worst case.
T (n) best < T(n) Avg. < T(n) worst
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In this sorting algorithm, multiple swapping occurs in one pass. Smaller elements move or 'bubble' up to the top of the list, so the name given to the algorithm. In this method,
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