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The Lowest Working Well in the Cluster

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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 farm draws from count wells numbered 1 through count, and pipes lists the pairs of wells a pipe joins. Two wells belong to the same cluster when a chain of pipes runs from one to the other, so a well with no pipes is a cluster on its own. Every well is working to begin with.

Work through orders in turn. An order comes in one of two kinds:

[1, well] calls for water at that well. If the well is working the answer is that well itself. Otherwise the answer is the lowest-numbered working well of its cluster, and -1 when its cluster has none working.
[2, well] takes that well out of service. A well already out of service is left as it is, and a well out of service never comes back.

Return the answers to the [1, well] orders, in the order they were called.

Examples

Example 1

Input: count = 4, pipes = [[1, 2], [3, 4]], orders = [[1, 1], [2, 1], [1, 1], [1, 3], [2, 3], [1, 3], [2, 4], [1, 4]]
Output: [1, 2, 3, 4, -1]

Two clusters, wells 1 and 2 in one and wells 3 and 4 in the other. The first call finds well 1 working. Once it goes out of service the same call falls back on well 2, its only clustermate. Wells 3 and 4 go the same way, and the last call comes after both have gone, so nothing is left in that cluster.

Example 2

Input: count = 5, pipes = [[1, 2], [2, 3], [4, 5]], orders = [[2, 2], [1, 2], [2, 1], [1, 2], [2, 3], [1, 1], [1, 5]]
Output: [1, 3, -1, 5]

Wells 1, 2 and 3 share a cluster. With well 2 out of service the call at it falls back on well 1, and once well 1 has gone too the same call falls back on well 3. After well 3 goes the cluster is empty, while well 5 in the other cluster is untouched and answers itself.

Example 3

Input: count = 5, pipes = [], orders = [[2, 3], [1, 3], [2, 5], [1, 5], [1, 4]]
Output: [-1, -1, 4]

No pipes at all, so every well stands alone. A call at a well out of service has nowhere to fall back on, while well 4 is still working and answers itself.

Constraints

1 <= count <= 10^5
0 <= pipes.length <= 10^5
1 <= orders.length <= 2 * 10^5
pipes[i].length == 2
orders[i].length == 2
pipes[i] joins two different wells, each numbered from 1 to count
each orders[i] reads [1, well] or [2, well], with well numbered from 1 to count

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 process_queries(count: int, pipes: list[list[int]], orders: list[list[int]]) -> list[int]:

Java

public int[] processQueries(int count, int[][] pipes, int[][] orders)

The intended approach

Time O((count + p) alpha + q log count) over the run of orders.Space O(count + p) for the clusters and their heaps.

The pipes never change, so the clusters never change: settle them once with a disjoint-set structure, joining the two wells of every pipe, and every well then carries the name of its cluster. Give each cluster a min-heap of its wells. A call at a working well is answered by the well itself with no searching at all. A call at a well out of service goes to its cluster's heap, where the wells that have gone out of service are thrown off the top until either a working well shows or the heap empties, which is the `-1` case. Throwing them off for good is safe because a well out of service never comes back, so a well discarded from a heap could never have been an answer later either. That makes the discarding pay for itself across the whole run of orders rather than per call: every well leaves its heap at most once. Taking a well out of service is just a flag; the heap is left alone and tidied lazily on the next call that reaches it.

Pythonentry point: process_queries

def process_queries(count, pipes, orders):
    import heapq
    owner = list(range(count + 1))

    def root(well):
        while owner[well] != well:
            owner[well] = owner[owner[well]]
            well = owner[well]
        return well

    for a, b in pipes:
        ra = root(a)
        rb = root(b)
        if ra != rb:
            owner[ra] = rb
    pools = {}
    for well in range(1, count + 1):
        pools.setdefault(root(well), []).append(well)
    for name in pools:
        heapq.heapify(pools[name])
    working = [True] * (count + 1)
    answers = []
    for kind, well in orders:
        if kind == 2:
            working[well] = False
            continue
        if working[well]:
            answers.append(well)
            continue
        pool = pools[root(well)]
        while pool and not working[pool[0]]:
            heapq.heappop(pool)
        answers.append(pool[0] if pool else -1)
    return answers

The obvious version

Search the cluster afresh on every call. Hold the pipes as neighbour lists and a flag per well. A call at a working well answers itself. Otherwise walk the whole cluster from that well with a stack, keeping the lowest-numbered working well met along the way, and answer that or `-1` when the walk finds none. Costs a walk over the cluster for every call.

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_process_queries(count, pipes, orders):
    links = [[] for _ in range(count + 1)]
    for a, b in pipes:
        links[a].append(b)
        links[b].append(a)
    working = [True] * (count + 1)
    answers = []
    for kind, well in orders:
        if kind == 2:
            working[well] = False
            continue
        if working[well]:
            answers.append(well)
            continue
        seen = [False] * (count + 1)
        seen[well] = True
        stack = [well]
        best = -1
        while stack:
            at = stack.pop()
            if working[at] and (best < 0 or at < best):
                best = at
            for nb in links[at]:
                if not seen[nb]:
                    seen[nb] = True
                    stack.append(nb)
        answers.append(best)
    return answers

How this problem was checked

Reference solution executed against 12 test cases — 12 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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