No. A partition invariant can put one pivot in its final rank without sorting the entire range.

During partition, everything before the boundary is at most the pivot; everything between boundary and scan is known to be greater. Each inspected value extends one region.

What the function promises: Mutate a list of comparable values into nondecreasing order. This teaching version uses the final item as pivot and can degrade on ordered input.

def quicksort(values):
    def sort(left, right):
        if left >= right:
            return
        pivot = values[right]
        boundary = left
        for scan in range(left, right):
            if values[scan] <= pivot:
                values[boundary], values[scan] = values[scan], values[boundary]
                boundary += 1
        values[boundary], values[right] = values[right], values[boundary]
        sort(left, boundary - 1)
        sort(boundary + 1, right)
    sort(0, len(values) - 1)
    return values

assert quicksort([4,1,3,2,1]) == [1,1,2,3,4]

Work and storage: Average O(n log n), worst-case O(n²); recursion can reach O(n).

When to recognize it: A partition invariant can put one pivot in its final rank without sorting the entire range.

Check before applying it: Repeatedly choosing an extreme pivot creates quadratic work and deep recursion.

Randomize the pivot or use an introspective production sort with a worst-case fallback.