0978HardArrayTreeGraph TheoryTopological Sort

Shortest Round Trip to Sweep Every Parcel

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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 network of hubs is given as a tree: edges lists the links, and coins[i] is 1 when hub i holds a parcel and 0 otherwise.

A van starts at any hub it likes and must finish back where it started. A parcel is swept when the van visits some hub no more than two links away from the parcel's hub.

Every link travelled counts, and travelling the same link again counts again. Return the fewest links the round trip can travel while sweeping every parcel.

Examples

Example 1

Input: coins = [1, 0, 0, 1], edges = [[0, 1], [1, 2], [2, 3]]
Output: 0

The hubs run in a line of four with parcels at the two ends. Standing at either middle hub sweeps both ends, since each is within two links, so the van need not move at all.

Example 2

Input: coins = [1, 0, 0, 0, 0, 0, 0, 1], edges = [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7]]
Output: 6

The line is eight hubs long with parcels at the ends. The van must reach within two links of each end, which means covering the two middle links and walking them both ways.

Example 3

Input: coins = [0], edges = []
Output: 0

A single hub with no parcel means there is nothing to sweep.

Constraints

1 <= coins.length <= 3 * 10^4
0 <= coins[i] <= 1
edges.length == coins.length - 1
edges[i].length == 2
0 <= edges[i][0] <= coins.length - 1
0 <= edges[i][1] <= coins.length - 1
The two hubs on a link are different
The links form a tree

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 collect_the_coins(coins: list[int], edges: list[list[int]]) -> int:

Java

public int collectTheCoins(int[] coins, int[][] edges)

The intended approach

Time O(n) for n hubs: each hub is snipped at most once.Space O(n) for the link lists and the degrees.

A round trip that returns to its start walks every link it uses exactly twice, so the answer is twice the number of links in the smallest connected piece of the network the van needs to cover. The whole job is therefore working out which hubs are worth visiting. Trim the network in two stages. First, a hub at the end of a branch holding no parcel is useless, and so is whatever it leaves stranded, so keep snipping parcel-free ends until none remain. What is left is the smallest piece that reaches every parcel. Then, because the van sweeps from two links away, the outermost two rings of that piece need not be entered at all, so snip the ends twice more, all the ends of one ring together before moving to the next. Twice the links still standing is the answer, and nothing standing means the van never has to move.

Pythonentry point: collect_the_coins

def collect_the_coins(coins, edges):
    from collections import deque
    n = len(coins)
    if n == 1:
        return 0
    adj = [[] for _ in range(n)]
    degree = [0] * n
    for u, v in edges:
        adj[u].append(v)
        adj[v].append(u)
        degree[u] += 1
        degree[v] += 1
    gone = [False] * n
    # snip parcel-free ends until none are left
    queue = deque()
    for i in range(n):
        if degree[i] == 1 and coins[i] == 0:
            queue.append(i)
    while queue:
        node = queue.popleft()
        if gone[node]:
            continue
        gone[node] = True
        for nxt in adj[node]:
            if gone[nxt]:
                continue
            degree[nxt] -= 1
            if degree[nxt] == 1 and coins[nxt] == 0:
                queue.append(nxt)
    # then snip the outermost ring twice over
    for _ in range(2):
        ring = []
        for i in range(n):
            if not gone[i] and degree[i] <= 1:
                ring.append(i)
        for node in ring:
            gone[node] = True
            for nxt in adj[node]:
                if not gone[nxt]:
                    degree[nxt] -= 1
    left = 0
    for i in range(n):
        if not gone[i]:
            left += 1
    if left <= 1:
        return 0
    return 2 * (left - 1)

The obvious version

Try every set of hubs, keep those that form one connected piece, check that every parcel sits within two links of the piece, and take twice the piece's link count at its smallest. Exponential in the hub count and it snips nothing.

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_collect_the_coins(coins, edges):
    from collections import deque
    n = len(coins)
    adj = [[] for _ in range(n)]
    for u, v in edges:
        adj[u].append(v)
        adj[v].append(u)
    near = []
    for c in range(n):
        reach = set()
        dist = {c: 0}
        queue = deque([c])
        while queue:
            node = queue.popleft()
            reach.add(node)
            if dist[node] == 2:
                continue
            for nxt in adj[node]:
                if nxt not in dist:
                    dist[nxt] = dist[node] + 1
                    queue.append(nxt)
        near.append(reach)
    best = None
    for mask in range(1, 1 << n):
        picked = [i for i in range(n) if mask >> i & 1]
        # connected?
        seen = set([picked[0]])
        queue = deque([picked[0]])
        while queue:
            node = queue.popleft()
            for nxt in adj[node]:
                if (mask >> nxt & 1) and nxt not in seen:
                    seen.add(nxt)
                    queue.append(nxt)
        if len(seen) != len(picked):
            continue
        ok = True
        for c in range(n):
            if coins[c] == 0:
                continue
            if not (near[c] & set(picked)):
                ok = False
                break
        if not ok:
            continue
        cost = 2 * (len(picked) - 1)
        if best is None or cost < best:
            best = cost
    return best

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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