0609MediumDepth-First SearchBreadth-First SearchGraph TheoryTopological SortKosaraju's AlgorithmTarjan's SCC Algorithm

States From Which the Batch Always Halts

Tracked in this browser only

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 batch machine has n states numbered 0 through n - 1. graph[i] lists, in strictly increasing order, the states the machine may move to in one step from state i. When graph[i] is empty the machine has nowhere to go and stops there; call such a state a stopping state. A state may list itself.

The machine is free to pick any listed move at each step, so a run from a state is any sequence of moves that follows the lists. A state is settled when every run that starts there reaches a stopping state after finitely many steps, no matter which moves are picked along the way. Note that a stopping state is settled.

Return the list of all settled states, in increasing numerical order.

Examples

Example 1

Input: graph = [[1], [2], [1], []]
Output: [3]

State 3 lists no moves, so it is a stopping state and is settled. States 1 and 2 move into each other, so a run between them never stops. State 0 moves to state 1, so its run never stops either.

Example 2

Input: graph = [[0], []]
Output: [1]

State 0 lists itself, so the run from it can go on forever. State 1 lists no moves and is settled.

Example 3

Input: graph = [[1, 2], [], [2]]
Output: [1]

State 1 is a stopping state. State 2 lists only itself, so a run from it never stops. State 0 may move to state 2, and the definition requires every run from a state to stop, so state 0 is not settled.

Constraints

n == graph.length
1 <= n <= 10^4
0 <= graph[i].length <= n
0 <= graph[i][j] <= n - 1
graph[i] is sorted in strictly increasing order.
A state may list itself.
The total number of moves listed across all states is between 1 and 4 * 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 eventual_safe_nodes(graph: list[list[int]]) -> list[int]:

Java

public List<Integer> eventualSafeNodes(int[][] graph)

The intended approach

Time O(V + E): every state and every listed move is handled once.Space O(V + E) for the reversed lists, the counters and the queue.

A state is settled only if every single move it lists leads to a settled state, so work backwards from the stopping states. Build the reversed lists, which say for each state which states can move into it, and give every state a counter holding how many of its own moves are still unaccounted for. Stopping states start at zero and go into a work queue. Taking a state off the queue marks it settled, and each state that can move into it has its counter cut by one; when a counter reaches zero all of that state's moves lead to settled states, so it joins the queue. Anything still holding a positive counter when the queue empties can keep moving forever. Reading the settled flags in index order gives the answer already sorted.

Pythonentry point: eventual_safe_nodes

def eventual_safe_nodes(graph):
    # O(V + E) time and space. Peel the machine from its halting states inward.
    # A state settles only once every one of its outgoing moves has been shown to
    # settle, so build the reversed lists, count each state's outgoing moves, and
    # decrement that count whenever one of its targets settles. A state trapped in
    # or behind a loop never reaches a count of zero. No recursion is used.
    n = len(graph)
    remaining = [len(graph[i]) for i in range(n)]
    feeders = [[] for _ in range(n)]
    for i in range(n):
        for j in graph[i]:
            feeders[j].append(i)
    queue = []
    for i in range(n):
        if remaining[i] == 0:
            queue.append(i)
    settled = [False] * n
    head = 0
    while head < len(queue):
        node = queue[head]
        head += 1
        settled[node] = True
        for before in feeders[node]:
            remaining[before] -= 1
            if remaining[before] == 0:
                queue.append(before)
    answer = []
    for i in range(n):
        if settled[i]:
            answer.append(i)
    return answer

The obvious version

Start by calling only the stopping states safe. Then sweep the whole machine over and over, marking a state safe as soon as all of the moves it lists point at states already called safe, and repeat until a full sweep changes 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_eventual_safe_nodes(graph):
    n = len(graph)
    safe = [len(graph[i]) == 0 for i in range(n)]
    moved = True
    while moved:
        moved = False
        for i in range(n):
            if safe[i]:
                continue
            allgood = True
            for j in graph[i]:
                if not safe[j]:
                    allgood = False
                    break
            if allgood:
                safe[i] = True
                moved = True
    out = []
    for i in range(n):
        if safe[i]:
            out.append(i)
    return out

How this problem was checked

Reference solution executed against 9 test cases — 9 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in