Since the worst case scenario is that the array is in reverse order, and that the first element in sorted array is the last element in the starting array, the most exchanges that will be necessary is equal to the length of the array.
Example:
Let us take the array of numbers "5 1 4 2 8".
First Pass:
( 5 1 4 2 8 ) ( 1 5 4 2 8 ), Here, algorithm compares the first two elements, and swaps them.
( 1 5 4 2 8 ) ( 1 4 5 2 8 ), Swap since 5 > 4
( 1 4 5 2 8 ) ( 1 4 2 5 8 ), Swap since 5 > 2
( 1 4 2 5 8 ) ( 1 4 2 5 8 ), Now, since these elements are already in order (8 > 5), algorithm does not swap them.
Second Pass:
( 1 4 2 5 8 ) ( 1 4 2 5 8 )
( 1 4 2 5 8 ) ( 1 2 4 5 8 ), Swap since 4 > 2
( 1 2 4 5 8 ) ( 1 2 4 5 8 )
( 1 2 4 5 8 ) ( 1 2 4 5 8 )
Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted.
Third Pass:
( 1 2 4 5 8 ) ( 1 2 4 5 8 )
( 1 2 4 5 8 ) ( 1 2 4 5 8 )
( 1 2 4 5 8 ) ( 1 2 4 5 8 )
( 1 2 4 5 8 ) ( 1 2 4 5 8 )
Finally, the array is sorted, and the algorithm can terminate.
( 5 1 4 2 8 ) ( 1 5 4 2 8 ), Here, algorithm compares the first two elements, and swaps them.
( 1 5 4 2 8 ) ( 1 4 5 2 8 ), Swap since 5 > 4
( 1 4 5 2 8 ) ( 1 4 2 5 8 ), Swap since 5 > 2
( 1 4 2 5 8 ) ( 1 4 2 5 8 ), Now, since these elements are already in order (8 > 5), algorithm does not swap them.
Second Pass:
( 1 4 2 5 8 ) ( 1 4 2 5 8 )
( 1 4 2 5 8 ) ( 1 2 4 5 8 ), Swap since 4 > 2
( 1 2 4 5 8 ) ( 1 2 4 5 8 )
( 1 2 4 5 8 ) ( 1 2 4 5 8 )
Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted.
Third Pass:
( 1 2 4 5 8 ) ( 1 2 4 5 8 )
( 1 2 4 5 8 ) ( 1 2 4 5 8 )
( 1 2 4 5 8 ) ( 1 2 4 5 8 )
( 1 2 4 5 8 ) ( 1 2 4 5 8 )
Finally, the array is sorted, and the algorithm can terminate.
dei...quick sort ku oru nalla example kudu da...
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