Sorts 8 Quick Sort | Thông tin về chủ đề quicksort |

Superb explanation 🙏🙏

This is the best video on quick sort! <3

How can we exactly count the number of comparisons? I establish a counter for 10000 number am I obtain 9999. If I consider the recursion -1 each time the numbers goes to 12K. Depending of the start or end or media election differs too.

Best explanation of quicksort I have found so far

Thank you professor!! This is really clear and understandable, I now understand how partition work and how it works with pivot point.

Thank you!

am too lazy in writing a comment but this tutorial made me do that! THANKSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS

# quick sort inplace
def quick_sort_inplace(arr):
if len(arr) <= 1:
return arr
else:
pivot = arr[-1] #consider last element as pivot
i,j = 0,0
while i < len(arr)-1:
# print(arr)
# print(arr[i], arr[j])
if arr[i] > pivot:
i = i+1
else:
arr[i],arr[j] = arr[j], arr[i]
j = j+1
i = i+1
# print(arr)
# print(arr[i], arr[j])
arr[j],arr[-1] = arr[-1], arr[j]
return quick_sort_inplace(arr[:j])+[arr[j]]+quick_sort_inplace(arr[j+1:])

l = [10,7,12,8,3,2,6]
l1 = [11,1,2,3,4,5,6,10,7,6,5,4,3,2,1,0,8,9,10]
print(quick_sort_inplace(l1))

Using the last element as a pivot is necessarily not a good option if there is even a slight chance that the list could be (nearly) sorted. A good option for pivot selection is the median-of-three method in which you choose the median of the first, middle and last element of the list as your pivot. This leads to efficient behaviour for all inputs.

4:08 if the numbers are randomly sorted then, any number in the list has the same probability of being in the middle of the sorted list… so you could just pick the number of the very right no? Edited: 8:38 lol he answered my question by the end of the video 🙂

'crystal' clear.

This should be at the top of the search list.