document1.txt

/hi
    # Last element will be the pivot and the first element the pointer
    pivot, ptr = nums[r], l
    for i in range(l, r):
        if nums[i] <= pivot:
            # Swapping values smaller than the pivot to the front
            nums[i], nums[ptr] = nums[ptr], nums[i]
            ptr += 1
    # Finally swapping the last element with the pointer indexed number
    nums[ptr], nums[r] = nums[r], nums[ptr]
    return ptr
 
 
 
def quicksort(l, r, nums):
    if len(nums) == 1:  # Terminating Condition for recursion. VERY IMPORTANT!
        return nums
    if l < r:
        pi = partition(l, r, nums)
        quicksort(l, pi-1, nums)  # Recursively sorting the left values
        quicksort(pi+1, r, nums)  # Recursively sorting the right values
    return nums
 
 
 
example = [2, 5, 6, 1, 4, 6, 2, 4, 7, 8]
result = [1, 2, 2, 4, 4, 5, 6, 6, 7, 8]
# As you can see, it works for duplicates too
print(quicksort(0, len(example)-1, example))