0296HardArrayHash TableSliding WindowHeap (Priority Queue)Treap

Rolling Frame Middle

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A monitor logs one integer per tick into readings. A frame of span consecutive readings starts at the left end and slides right one tick at a time, so the first frame covers positions 0 through span - 1, the next covers 1 through span, and so on until the frame reaches the right end.

The middle value of a frame is defined on the frame's readings put in sorted order. When span is odd it is the single reading in the centre. When span is even it is the mean of the two central readings, so it can land halfway between two integers and must be reported that way, for example 5.5.

Return the middle value of every frame, in the order the frames appear.

Examples

Example 1

Input: readings = [4,1,7,1,8], span = 3
Output: [4.0,1.0,7.0]

The three frames sort to [1,4,7], [1,1,7] and [1,7,8], and the centre entry of each is 4, 1 and 7.

Example 2

Input: readings = [6,2,9,4], span = 2
Output: [4.0,5.5,6.5]

Every frame holds two readings, so each answer is their mean: 4 for 6 and 2, 5.5 for 2 and 9, and 6.5 for 9 and 4.

Constraints

1 <= span <= readings.length <= 10^5
-2^31 <= readings[i] <= 2^31 - 1

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def rolling_medians(readings: list[int], span: int) -> list[float]:

Java

public double[] rollingMedians(int[] readings, int span)

The intended approach

Time O(n log n)Space O(n) beyond the returned list of n - span + 1 values

Map the distinct readings to ranks so counts can live in a fixed-size Fenwick tree. Advancing the frame adds one at the incoming rank and, once the frame is full, removes one at the outgoing rank. The centre entry is then read off by descending the tree for the r-th smallest live rank, taking one descent for an odd span and two for an even one.

Pythonentry point: rolling_medians

def rolling_medians(readings, span):
    # Compress the distinct readings to ranks, then keep a Fenwick tree of how many
    # copies of each rank sit inside the window.  Sliding costs one increment and
    # one decrement, and the middle entry is found by a binary descent over the tree
    # for the r-th smallest.
    # O(n log n) time, O(n) space for the ranks and the tree, plus the returned list
    # of n - span + 1 medians.
    order = sorted(set(readings))
    rank = {v: i + 1 for i, v in enumerate(order)}
    m = len(order)
    tree = [0] * (m + 1)
    top = m.bit_length()

    def bump(i, delta):
        while i <= m:
            tree[i] += delta
            i += i & -i

    def nth(r):
        pos = 0
        rem = r
        step = 1 << top
        while step:
            nxt = pos + step
            if nxt <= m and tree[nxt] < rem:
                pos = nxt
                rem -= tree[pos]
            step >>= 1
        return order[pos]

    out = []
    for i, value in enumerate(readings):
        bump(rank[value], 1)
        if i >= span:
            bump(rank[readings[i - span]], -1)
        if i >= span - 1:
            if span % 2:
                out.append(float(nth(span // 2 + 1)))
            else:
                lo = nth(span // 2)
                hi = nth(span // 2 + 1)
                out.append((lo + hi) / 2)
    return out

The obvious version

Slice out each frame, sort the copy, and read the centre entry or the mean of the two central entries directly.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_rolling_medians(readings, span):
    out = []
    for i in range(len(readings) - span + 1):
        window = sorted(readings[i:i + span])
        if span % 2:
            out.append(float(window[span // 2]))
        else:
            out.append((window[span // 2 - 1] + window[span // 2]) / 2)
    return out

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
2 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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