Sort the following array a using quicksort,
[6, 11, 4, 9, 8, 2, 5, 8, 13, 7]
The pivot should be chosen as the arithmetic mean of the first
Your pivot is 6. Your pivot is NOT the 6th element Now you can apply the following algorith.
function quicksort(array)
var list less, greater
if length(array) ≤ 1
return array // an array of zero or one elements is already sorted
select and remove a pivot value pivot from array
for each x in array
if x ≤ pivot then append x to less
else append x to greater
return concatenate(quicksort(less), pivot, quicksort(greater))
By the way the question is worded, the pivot should just be 6 and not necessarily the 6th item in the array.
This is most definitely the case because if there were only 3 items in the array, for example, and the arithmetic mean came out to be greater than 3, you would have no pivot to choose because there is no item with that index.
Note: Be careful with the way you index elements in your array. You said the 6th element is '2', when it may be '5' if your programming language starts indices at 0.
For quicksort, the pivot can be whatever element you want. Check out Wikipedia.
The problem was easily solved by choosing either a random index for the pivot, choosing the middle index of the partition or (especially for longer partitions) choosing the median of the first, middle and last element of the partition for the pivot
Three choices thus :
And in you case using the mean of first and last element value would give you :
6 + 7 = 13
13 / 2 = 6.5
6.5 rounded down = 6
For the given array [6, 11, 4, 9, 8, 2, 5, 8, 13, 7]
Calculate something like this:
6
Select last element of list which is 7
Select mid element which is mid = (length%2==0) ? (length/2)-1 : (length-1)/2
, which is 8
[6,7,8]
, now mid element in your array is mid.The position of the pivot from that calculation is not important, quicksort sorts the elements based on whether they are more or less than the pivot. Then the pivot is placed in between the two sets (more and less).