0014HardArrayDivide and ConquerHeap (Priority Queue)Merge Sort

Merge Shard Streams

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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 monitoring fleet writes its signed temperature readings into shards. You receive streams, a list of those shards. Each shard is already laid out from lowest reading to highest, and a shard may carry nothing at all.

Fold the shards into a single array that runs from lowest reading to highest. Every reading is kept, so a value stored in three shards shows up three times in the result.

The shard count can be far larger than the readings inside any one shard, so pull from all shard fronts together rather than flattening first.

Examples

Example 1

Input: streams = [[-8, -1, 4], [-3, 0], [2, 2, 9]]
Output: [-8, -3, -1, 0, 2, 2, 4, 9]

Eight readings from three shards line up as one run, and the value 2 stays twice because two shards recorded it.

Example 2

Input: streams = [[3, 4, 5], [1, 2]]
Output: [1, 2, 3, 4, 5]

The second shard supplies both of the two lowest readings before the first shard contributes anything.

Example 3

Input: streams = [[7, 7], [7]]
Output: [7, 7, 7]

Repeats spread across shards are all retained, so the result holds three copies.

Constraints

k == streams.length
0 <= k <= 10^4
0 <= streams[i].length <= 500
-10^4 <= streams[i][j] <= 10^4
Each streams[i] is already in non-decreasing order
The total number of readings across all shards is at most 10^4

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 merge_streams(streams: list[list[int]]) -> list[int]:

Java

public int[] mergeStreams(int[][] streams)

The intended approach

Time O(N log k)Space O(k)

Only the front reading of each shard can be the next smallest overall, so keep those k fronts in a min-heap keyed by value. Pop the winner, append it, and push the reading that follows it inside the same shard. Each of the N readings enters and leaves a heap bounded by k, which beats sorting the flattened data.

Pythonentry point: merge_streams

def merge_streams(streams):
    import heapq
    frontier = []
    for idx, shard in enumerate(streams):
        if shard:
            frontier.append((shard[0], idx, 0))
    heapq.heapify(frontier)
    merged = []
    while frontier:
        value, idx, pos = heapq.heappop(frontier)
        merged.append(value)
        if pos + 1 < len(streams[idx]):
            heapq.heappush(frontier, (streams[idx][pos + 1], idx, pos + 1))
    return merged

The obvious version

Hold a cursor per shard and, for each output slot, scan every cursor linearly to find the smallest available reading. O(N * k).

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_merge_streams(streams):
    cursors = [0] * len(streams)
    total = 0
    for shard in streams:
        total += len(shard)
    merged = []
    for _ in range(total):
        best = -1
        for i, shard in enumerate(streams):
            if cursors[i] >= len(shard):
                continue
            if best < 0 or shard[cursors[i]] < streams[best][cursors[best]]:
                best = i
        merged.append(streams[best][cursors[best]])
        cursors[best] += 1
    return merged

How this problem was checked

Reference solution executed against 11 test cases — 11 passed
3 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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