No. Partition fixes one rank; only one remaining side can contain the desired rank.

Translate kth largest into sorted index n-k. Partition around a pivot. The pivot’s final index tells whether to stop or discard one whole side.

What the function promises: Return the kth largest value, with k starting at one. Mutates the list. Raise ValueError for invalid k.

def kth_largest(values, k):
    if not 1 <= k <= len(values):
        raise ValueError("Invalid k")
    target, left, right = len(values) - k, 0, len(values) - 1
    while True:
        pivot, boundary = values[right], 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]
        if boundary == target:
            return values[boundary]
        if boundary < target:
            left = boundary + 1
        else:
            right = boundary - 1

assert kth_largest([3,2,1,5,6,4], 2) == 5

Work and storage: Average O(n), worst-case O(n²), O(1) auxiliary storage.

When to recognize it: Partition fixes one rank; only one remaining side can contain the desired rank.

Check before applying it: Deterministic poor pivots can make quadratic work, and mutation may be unacceptable.

Maintain a min-heap of the largest k values seen.