Find Eventual Safe States

🎯

Find Eventual Safe States

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Imagine a network of tasks where some tasks depend on others. You want to find which tasks are guaranteed to finish without getting stuck in a cycle.

💡 This problem involves understanding graph cycles and safe nodes. Beginners often struggle because detecting cycles and reasoning about node safety requires careful graph traversal and sometimes reversing edges to apply topological sorting.

📋

Problem Statement

Given a directed graph with n nodes labeled from 0 to n-1, where graph[i] is a list of nodes j such that there is a directed edge from node i to node j, return an array containing all the nodes that are eventually safe. A node is eventually safe if and only if all possible paths starting from that node lead to a terminal node (a node with no outgoing edges) or nodes that are themselves eventually safe.

1 ≤ n ≤ 10^50 ≤ graph[i].length ≤ n0 ≤ graph[i][j] < nThe graph may contain cycles

💡

Example

Input"[[1,2],[2,3],[5],[0],[5],[],[]]"

Output[2,4,5,6]

Nodes 2,4,5,6 are safe because starting from them, all paths lead to terminal nodes or safe nodes. For example, node 2 leads to node 5 which is terminal.

• Graph with no edges → all nodes are safe
• Graph where all nodes form a single cycle → no safe nodes
• Graph with multiple disconnected components → safe nodes only in acyclic components
• Graph with self-loop → node with self-loop is unsafe

⚠️

Common Mistakes

Not detecting cycles correctly in DFS

Incorrectly marking unsafe nodes as safe, leading to wrong output

✅ Use a recursion stack or onPath array to detect cycles during DFS

Not memoizing DFS results

Excessive repeated computations causing timeouts

✅ Cache safety results for each node to avoid redundant DFS calls

Confusing in-degree and out-degree when building reversed graph

Incorrect queue initialization and wrong safe nodes identified

✅ Remember in reversed graph, in-degree equals original out-degree

Not sorting final safe nodes list

Output order differs from expected, causing test failures

✅ Sort the safe nodes before returning

Using recursion without handling large input sizes

Stack overflow or runtime error on large graphs

✅ Use iterative approach like Kahn's algorithm for large inputs

🧠

Brute Force (DFS Cycle Detection)

💡 This approach exists to build intuition about cycles and node safety by directly exploring all paths. It is slow but helps understand why cycles make nodes unsafe.

Intuition

We try to detect if starting from a node, we can reach a cycle. If yes, the node is unsafe; otherwise, it is safe.

Algorithm

1. Initialize arrays to track visited nodes and recursion stack.
2. For each node, run DFS to detect cycles.
3. If a cycle is detected starting from a node, mark it unsafe.
4. Collect all nodes that are not part of any cycle as safe nodes.

💡 The recursion stack helps detect cycles by tracking nodes currently being explored, which is key to identifying unsafe nodes.

</>

Code

from typing import List

def eventualSafeNodes(graph: List[List[int]]) -> List[int]:
    n = len(graph)
    visited = [False] * n
    onPath = [False] * n
    safe = [False] * n

    def dfs(node: int) -> bool:
        if visited[node]:
            return safe[node]
        visited[node] = True
        onPath[node] = True
        for nei in graph[node]:
            if onPath[nei] or not dfs(nei):
                return False
        onPath[node] = False
        safe[node] = True
        return True

    return [i for i in range(n) if dfs(i)]

# Driver code
if __name__ == '__main__':
    graph = [[1,2],[2,3],[5],[0],[5],[],[]]
    print(eventualSafeNodes(graph))

Line Notes

visited = [False] * nTrack if node has been visited to avoid repeated work

onPath = [False] * nTrack nodes currently in recursion stack to detect cycles

if visited[node]:If already visited, return previously computed safety

for nei in graph[node]:Explore all neighbors to check for cycles

import java.util.*;

public class Solution {
    public List<Integer> eventualSafeNodes(int[][] graph) {
        int n = graph.length;
        boolean[] visited = new boolean[n];
        boolean[] onPath = new boolean[n];
        boolean[] safe = new boolean[n];
        List<Integer> res = new ArrayList<>();

        for (int i = 0; i < n; i++) {
            if (dfs(i, graph, visited, onPath, safe)) {
                res.add(i);
            }
        }
        return res;
    }

    private boolean dfs(int node, int[][] graph, boolean[] visited, boolean[] onPath, boolean[] safe) {
        if (visited[node]) return safe[node];
        visited[node] = true;
        onPath[node] = true;
        for (int nei : graph[node]) {
            if (onPath[nei] || !dfs(nei, graph, visited, onPath, safe)) {
                return false;
            }
        }
        onPath[node] = false;
        safe[node] = true;
        return true;
    }

    // Main method for testing
    public static void main(String[] args) {
        Solution sol = new Solution();
        int[][] graph = {{1,2},{2,3},{5},{0},{5},{}, {}};
        System.out.println(sol.eventualSafeNodes(graph));
    }
}

Line Notes

boolean[] visited = new boolean[n];Track visited nodes to avoid recomputation

boolean[] onPath = new boolean[n];Detect cycles by tracking recursion stack

if (visited[node]) return safe[node];Return cached safety result if visited

for (int nei : graph[node])Check all neighbors for cycles

#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    vector<int> eventualSafeNodes(vector<vector<int>>& graph) {
        int n = graph.size();
        vector<bool> visited(n, false), onPath(n, false), safe(n, false);
        vector<int> res;
        for (int i = 0; i < n; i++) {
            if (dfs(i, graph, visited, onPath, safe)) {
                res.push_back(i);
            }
        }
        return res;
    }

private:
    bool dfs(int node, vector<vector<int>>& graph, vector<bool>& visited, vector<bool>& onPath, vector<bool>& safe) {
        if (visited[node]) return safe[node];
        visited[node] = true;
        onPath[node] = true;
        for (int nei : graph[node]) {
            if (onPath[nei] || !dfs(nei, graph, visited, onPath, safe)) {
                return false;
            }
        }
        onPath[node] = false;
        safe[node] = true;
        return true;
    }
};

int main() {
    Solution sol;
    vector<vector<int>> graph = {{1,2},{2,3},{5},{0},{5},{}, {}};
    vector<int> safeNodes = sol.eventualSafeNodes(graph);
    for (int node : safeNodes) cout << node << ' ';
    cout << endl;
    return 0;
}

Line Notes

vector<bool> visited(n, false)Initialize visited array to track explored nodes

vector<bool> onPath(n, false)Track recursion stack to detect cycles

if (visited[node]) return safe[node];Use memoized result if node already processed

for (int nei : graph[node])Explore all neighbors to detect cycles

var eventualSafeNodes = function(graph) {
    const n = graph.length;
    const visited = new Array(n).fill(false);
    const onPath = new Array(n).fill(false);
    const safe = new Array(n).fill(false);

    function dfs(node) {
        if (visited[node]) return safe[node];
        visited[node] = true;
        onPath[node] = true;
        for (const nei of graph[node]) {
            if (onPath[nei] || !dfs(nei)) return false;
        }
        onPath[node] = false;
        safe[node] = true;
        return true;
    }

    const res = [];
    for (let i = 0; i < n; i++) {
        if (dfs(i)) res.push(i);
    }
    return res;
};

// Test
console.log(eventualSafeNodes([[1,2],[2,3],[5],[0],[5],[],[]]));

Line Notes

const visited = new Array(n).fill(false);Track nodes already visited to avoid repeated DFS

const onPath = new Array(n).fill(false);Detect cycles by tracking recursion stack

if (visited[node]) return safe[node];Return cached safety result if node processed

for (const nei of graph[node])Check all neighbors for cycles

⏱

Complexity

TimeO(V + E)

SpaceO(V)

Each node and edge is visited once during DFS. The recursion stack and visited arrays use O(V) space.

💡 For n=1000 nodes and m=2000 edges, this means roughly 3000 operations, which is efficient enough for medium-sized graphs.

Interview Verdict: Accepted but can be slow on very large graphs

This approach is intuitive and correct but may cause stack overflow or timeout on very large inputs due to deep recursion.

🧠

Better Approach (DFS with Memoization)

💡 This approach improves the brute force by caching results to avoid repeated DFS calls, reducing redundant work.

Intuition

Once we know a node is safe or unsafe, we store that result to avoid recomputing it when encountered again.

Algorithm

1. Initialize a memo array to store safety status of nodes.
2. Run DFS from each node, returning cached results if available.
3. Mark nodes unsafe if a cycle is detected during DFS.
4. Collect all nodes marked safe.

💡 Memoization prevents exponential blowup by remembering previous results, making the algorithm efficient.

</>

Code

from typing import List

def eventualSafeNodes(graph: List[List[int]]) -> List[int]:
    n = len(graph)
    memo = [None] * n  # None=unvisited, True=safe, False=unsafe

    def dfs(node: int) -> bool:
        if memo[node] is not None:
            return memo[node]
        memo[node] = False  # mark as unsafe during exploration
        for nei in graph[node]:
            if not dfs(nei):
                return False
        memo[node] = True
        return True

    return [i for i in range(n) if dfs(i)]

# Driver code
if __name__ == '__main__':
    graph = [[1,2],[2,3],[5],[0],[5],[],[]]
    print(eventualSafeNodes(graph))

Line Notes

memo = [None] * nInitialize memo to track node safety status

if memo[node] is not None:Return cached result if node already processed

memo[node] = FalseMark node unsafe tentatively to detect cycles

for nei in graph[node]:Explore neighbors to check if any path leads to unsafe node

import java.util.*;

public class Solution {
    public List<Integer> eventualSafeNodes(int[][] graph) {
        int n = graph.length;
        Boolean[] memo = new Boolean[n];
        List<Integer> res = new ArrayList<>();

        for (int i = 0; i < n; i++) {
            if (dfs(i, graph, memo)) {
                res.add(i);
            }
        }
        return res;
    }

    private boolean dfs(int node, int[][] graph, Boolean[] memo) {
        if (memo[node] != null) return memo[node];
        memo[node] = false; // mark unsafe during exploration
        for (int nei : graph[node]) {
            if (!dfs(nei, graph, memo)) return false;
        }
        memo[node] = true;
        return true;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        int[][] graph = {{1,2},{2,3},{5},{0},{5},{}, {}};
        System.out.println(sol.eventualSafeNodes(graph));
    }
}

Line Notes

Boolean[] memo = new Boolean[n];Use Boolean object to allow null for unvisited nodes

if (memo[node] != null) return memo[node];Return cached safety status if available

memo[node] = false;Mark node unsafe during DFS to detect cycles

for (int nei : graph[node])Check neighbors recursively for unsafe paths

#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    vector<int> eventualSafeNodes(vector<vector<int>>& graph) {
        int n = graph.size();
        vector<int> memo(n, -1); // -1=unvisited, 0=unsafe, 1=safe
        vector<int> res;
        for (int i = 0; i < n; i++) {
            if (dfs(i, graph, memo)) res.push_back(i);
        }
        return res;
    }

private:
    bool dfs(int node, vector<vector<int>>& graph, vector<int>& memo) {
        if (memo[node] != -1) return memo[node] == 1;
        memo[node] = 0; // mark unsafe during exploration
        for (int nei : graph[node]) {
            if (!dfs(nei, graph, memo)) return false;
        }
        memo[node] = 1;
        return true;
    }
};

int main() {
    Solution sol;
    vector<vector<int>> graph = {{1,2},{2,3},{5},{0},{5},{}, {}};
    vector<int> safeNodes = sol.eventualSafeNodes(graph);
    for (int node : safeNodes) cout << node << ' ';
    cout << endl;
    return 0;
}

Line Notes

vector<int> memo(n, -1);Use integer states to represent unvisited, unsafe, and safe

if (memo[node] != -1) return memo[node] == 1;Return cached result if node processed

memo[node] = 0;Mark node unsafe during DFS to detect cycles

for (int nei : graph[node])Recursively check neighbors for unsafe paths

var eventualSafeNodes = function(graph) {
    const n = graph.length;
    const memo = new Array(n).fill(null); // null=unvisited, true=safe, false=unsafe

    function dfs(node) {
        if (memo[node] !== null) return memo[node];
        memo[node] = false; // mark unsafe during exploration
        for (const nei of graph[node]) {
            if (!dfs(nei)) return false;
        }
        memo[node] = true;
        return true;
    }

    const res = [];
    for (let i = 0; i < n; i++) {
        if (dfs(i)) res.push(i);
    }
    return res;
};

// Test
console.log(eventualSafeNodes([[1,2],[2,3],[5],[0],[5],[],[]]));

Line Notes

const memo = new Array(n).fill(null);Initialize memo array to cache node safety

if (memo[node] !== null) return memo[node];Return cached safety if node already processed

memo[node] = false;Mark node unsafe during DFS to detect cycles

for (const nei of graph[node])Check all neighbors recursively for unsafe paths

⏱

Complexity

TimeO(V + E)

SpaceO(V)

Memoization ensures each node is processed once, so total time is linear in nodes plus edges.

💡 For n=1000 and m=2000, this approach efficiently handles all nodes and edges once, making it scalable.

Interview Verdict: Accepted and efficient for large inputs

Memoization avoids redundant DFS calls, making this approach practical and interview-ready.

🧠

Optimal Approach (Reverse Graph + Kahn's Algorithm)

💡 This approach uses topological sorting on the reversed graph to identify safe nodes without recursion, making it scalable and iterative.

Intuition

By reversing edges, terminal nodes become sources. Using Kahn's algorithm, we iteratively remove safe nodes and identify all nodes that eventually lead to terminal nodes.

Algorithm

1. Build the reverse graph and compute in-degree for each node in the reversed graph.
2. Initialize a queue with nodes having zero in-degree (terminal nodes in original graph).
3. While queue not empty, pop node and add it to safe list.
4. For each predecessor in reversed graph, decrement in-degree; if it becomes zero, add to queue.
5. Sort and return the list of safe nodes.

💡 This approach transforms the problem into a topological sort on the reversed graph, efficiently identifying safe nodes without DFS.

</>

Code

from typing import List
from collections import deque

def eventualSafeNodes(graph: List[List[int]]) -> List[int]:
    n = len(graph)
    reverse_graph = [[] for _ in range(n)]
    in_degree = [0] * n

    for u in range(n):
        for v in graph[u]:
            reverse_graph[v].append(u)
        in_degree[u] = len(graph[u])

    queue = deque([i for i in range(n) if in_degree[i] == 0])
    safe = []

    while queue:
        node = queue.popleft()
        safe.append(node)
        for prev in reverse_graph[node]:
            in_degree[prev] -= 1
            if in_degree[prev] == 0:
                queue.append(prev)

    return sorted(safe)

# Driver code
if __name__ == '__main__':
    graph = [[1,2],[2,3],[5],[0],[5],[],[]]
    print(eventualSafeNodes(graph))

Line Notes

reverse_graph = [[] for _ in range(n)]Create adjacency list for reversed graph

in_degree[u] = len(graph[u])In-degree in reversed graph equals out-degree in original graph

queue = deque([i for i in range(n) if in_degree[i] == 0])Start with terminal nodes having zero out-degree

for prev in reverse_graph[node]Process predecessors to reduce their in-degree

import java.util.*;

public class Solution {
    public List<Integer> eventualSafeNodes(int[][] graph) {
        int n = graph.length;
        List<List<Integer>> reverseGraph = new ArrayList<>();
        int[] inDegree = new int[n];
        for (int i = 0; i < n; i++) reverseGraph.add(new ArrayList<>());

        for (int u = 0; u < n; u++) {
            inDegree[u] = graph[u].length;
            for (int v : graph[u]) {
                reverseGraph.get(v).add(u);
            }
        }

        Queue<Integer> queue = new LinkedList<>();
        for (int i = 0; i < n; i++) {
            if (inDegree[i] == 0) queue.offer(i);
        }

        List<Integer> safe = new ArrayList<>();
        while (!queue.isEmpty()) {
            int node = queue.poll();
            safe.add(node);
            for (int prev : reverseGraph.get(node)) {
                inDegree[prev]--;
                if (inDegree[prev] == 0) queue.offer(prev);
            }
        }

        Collections.sort(safe);
        return safe;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        int[][] graph = {{1,2},{2,3},{5},{0},{5},{}, {}};
        System.out.println(sol.eventualSafeNodes(graph));
    }
}

Line Notes

List<List<Integer>> reverseGraph = new ArrayList<>();Build reversed adjacency list

inDegree[u] = graph[u].length;In-degree in reversed graph equals out-degree in original

if (inDegree[i] == 0) queue.offer(i);Initialize queue with terminal nodes

for (int prev : reverseGraph.get(node))Update in-degree of predecessors

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;

class Solution {
public:
    vector<int> eventualSafeNodes(vector<vector<int>>& graph) {
        int n = graph.size();
        vector<vector<int>> reverseGraph(n);
        vector<int> inDegree(n, 0);

        for (int u = 0; u < n; u++) {
            inDegree[u] = graph[u].size();
            for (int v : graph[u]) {
                reverseGraph[v].push_back(u);
            }
        }

        queue<int> q;
        for (int i = 0; i < n; i++) {
            if (inDegree[i] == 0) q.push(i);
        }

        vector<int> safe;
        while (!q.empty()) {
            int node = q.front(); q.pop();
            safe.push_back(node);
            for (int prev : reverseGraph[node]) {
                inDegree[prev]--;
                if (inDegree[prev] == 0) q.push(prev);
            }
        }

        sort(safe.begin(), safe.end());
        return safe;
    }
};

int main() {
    Solution sol;
    vector<vector<int>> graph = {{1,2},{2,3},{5},{0},{5},{}, {}};
    vector<int> safeNodes = sol.eventualSafeNodes(graph);
    for (int node : safeNodes) cout << node << ' ';
    cout << endl;
    return 0;
}

Line Notes

vector<vector<int>> reverseGraph(n);Create reversed graph adjacency list

inDegree[u] = graph[u].size();In-degree in reversed graph equals out-degree in original

if (inDegree[i] == 0) q.push(i);Start with terminal nodes having zero out-degree

for (int prev : reverseGraph[node])Decrease in-degree of predecessors and enqueue if zero

var eventualSafeNodes = function(graph) {
    const n = graph.length;
    const reverseGraph = Array.from({length: n}, () => []);
    const inDegree = new Array(n).fill(0);

    for (let u = 0; u < n; u++) {
        inDegree[u] = graph[u].length;
        for (const v of graph[u]) {
            reverseGraph[v].push(u);
        }
    }

    const queue = [];
    for (let i = 0; i < n; i++) {
        if (inDegree[i] === 0) queue.push(i);
    }

    const safe = [];
    while (queue.length > 0) {
        const node = queue.shift();
        safe.push(node);
        for (const prev of reverseGraph[node]) {
            inDegree[prev]--;
            if (inDegree[prev] === 0) queue.push(prev);
        }
    }

    return safe.sort((a,b) => a - b);
};

// Test
console.log(eventualSafeNodes([[1,2],[2,3],[5],[0],[5],[],[]]));

Line Notes

const reverseGraph = Array.from({length: n}, () => []);Build reversed adjacency list

inDegree[u] = graph[u].length;In-degree in reversed graph equals out-degree in original

if (inDegree[i] === 0) queue.push(i);Initialize queue with terminal nodes

for (const prev of reverseGraph[node])Update in-degree of predecessors and enqueue if zero

⏱

Complexity

TimeO(V + E)

SpaceO(V + E)

Building the reversed graph and running Kahn's algorithm each take linear time and space in nodes and edges.

💡 For large graphs with 10^5 nodes and edges, this approach efficiently processes all edges once without recursion.

Interview Verdict: Accepted and optimal for large inputs

This iterative approach avoids recursion limits and is the best choice for large-scale graphs in interviews.

📊

All Approaches - One-Glance Tradeoffs

💡 In interviews, code the optimal approach (reverse graph + Kahn's). Mention brute force and memoization to show understanding.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(V + E)	O(V)	Yes (deep recursion)	N/A	Mention only - never code due to inefficiency
2. DFS with Memoization	O(V + E)	O(V)	Yes (deep recursion)	N/A	Good to mention and code if recursion depth is manageable
3. Reverse Graph + Kahn's Algorithm	O(V + E)	O(V + E)	No (iterative)	N/A	Best approach to code in interviews for scalability

💼

Interview Strategy

💡 Use this guide to understand the problem deeply before interviews. Start with brute force to grasp cycles, then optimize with memoization, and finally master the topological sort approach for scalability.

How to Present

Step 1: Clarify the problem and confirm understanding of 'eventual safety'.Step 2: Describe the brute force DFS cycle detection approach.Step 3: Explain how memoization improves efficiency.Step 4: Present the optimal solution using reversed graph and Kahn's algorithm.Step 5: Code the optimal approach and test with examples.

Time Allocation

Clarify: 3min → Approach: 5min → Code: 10min → Test: 5min. Total ~23min

What the Interviewer Tests

The interviewer tests your understanding of graph cycles, ability to detect safe nodes, and knowledge of topological sorting techniques.

Common Follow-ups

• What if the graph is very large? → Use iterative topological sort to avoid recursion stack overflow.
• Can you find safe nodes without reversing the graph? → It's possible but less efficient and more complex.

💡 These follow-ups check your ability to optimize and adapt your solution to constraints and alternative methods.

🔍

Pattern Recognition

When to Use

1) Problem involves directed graph nodes and edges 2) Need to identify nodes that do not lead to cycles 3) Terminal nodes or safe nodes are mentioned 4) Problem can be transformed using topological sort or reverse graph

Signature Phrases

eventual safe statesnodes that lead to terminal nodesno cycles reachable from node

NOT This Pattern When

Problems that only require simple DFS or BFS without cycle detection or topological sorting

Similar Problems

Course Schedule - also uses cycle detection and topological sortCourse Schedule II - finds order of courses using topological sort

Practice

1. Consider the following code snippet implementing the optimal solution for the largest color value in a directed graph. Given colors = "abaca" and edges = [[0,1],[0,2],[2,3],[3,4]], what is the value of max_color_value after processing all nodes?

easy

A. 3

B. 4

C. 2

D. 5

Solution

1. Step 1: Trace dp for node 0 (color 'a')

   dp[0]['a'] = 1, max_color_value = 1

2. Step 2: Propagate to nodes 1 and 2

   Node 1 (color 'b'): dp[1]['a'] = 1, dp[1]['b'] = 1; Node 2 (color 'a'): dp[2]['a'] = 2, max_color_value = 2

3. Step 3: Propagate to node 3 (color 'c')

   dp[3]['a'] = 2, dp[3]['c'] = 1, max_color_value = 2

4. Step 4: Propagate to node 4 (color 'a')

   dp[4]['a'] = 3, max_color_value = 3

5. Final Answer:

   Option A -> Option A

6. Quick Check:

   Maximum color count along path is 3 for 'a' [OK]

Hint: Track dp updates per node carefully [OK]

Common Mistakes:

• Off-by-one in dp increments or max updates

2. You are given a directed acyclic graph where each node has a character label. You want to find the longest path such that no two adjacent nodes on the path share the same character. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Greedy DFS that always picks the next node with a different character without backtracking

B. Simple DFS with memoization ignoring character constraints

C. Bellman-Ford algorithm treating characters as weights

D. Topological sort followed by iterative dynamic programming to compute longest valid paths

Solution

1. Step 1: Understand problem constraints

   The problem requires finding longest paths in a DAG with an additional character adjacency constraint, which prevents naive longest path algorithms.

2. Step 2: Identify suitable algorithm

   Topological sort linearizes the DAG, enabling dynamic programming to compute longest paths while enforcing character difference constraints efficiently.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Topological sort + DP handles DAG and character constraints correctly [OK]

Hint: Longest path in DAG with constraints -> topological sort + DP [OK]

Common Mistakes:

• Assuming greedy DFS always works without backtracking
• Ignoring character constraints in memoization
• Using shortest path algorithms like Bellman-Ford incorrectly

3. You are given a list of sequences, each representing partial orderings of elements. Your task is to determine if there exists exactly one sequence that can be constructed which is consistent with all these partial orderings and matches a given original sequence. Which algorithmic approach best guarantees an efficient and correct solution for this problem?

easy

A. Use brute force to generate all permutations of the original sequence and check which ones are supersequences of all given sequences.

B. Apply a greedy approach by always choosing the smallest available element not yet placed, hoping to build the unique sequence.

C. Use dynamic programming to count the number of ways to interleave the sequences and verify if only one matches the original.

D. Construct a directed graph from the sequences and perform a DFS postorder topological sort with cycle detection and uniqueness verification.

Solution

1. Step 1: Understand problem constraints

   The problem requires verifying if exactly one sequence matches all partial orderings, which is a classic unique topological order problem.

2. Step 2: Identify suitable algorithm

   Brute force is inefficient; greedy and DP approaches do not guarantee correctness or uniqueness checks. DFS postorder topological sort with cycle detection and uniqueness verification efficiently solves the problem.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   DFS topological sort with uniqueness check is standard for unique sequence reconstruction [OK]

Hint: Unique topo order requires graph + DFS with uniqueness check [OK]

Common Mistakes:

• Assuming greedy or DP can guarantee uniqueness
• Trying brute force despite factorial complexity

4. Examine the following code snippet for finding Minimum Height Trees. Which line contains a subtle bug that causes incorrect results on single-node trees?

medium

A. Missing explicit check for n == 1 at the start

B. Line initializing leaves with nodes of degree 1

C. Line decrementing remaining_nodes by leaves_count

D. Line updating degree[nei] after removing a leaf

Solution

1. Step 1: Identify edge case handling

   The code lacks a special case for n=1, so leaves initialization is empty and the loop never runs, returning an empty list.

2. Step 2: Understand impact of missing check

   Without returning [0] for single-node tree, the function returns incorrect empty result.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Single-node tree must be handled explicitly [OK]

Hint: Check for single-node tree before main logic [OK]

Common Mistakes:

• Forgetting n=1 case
• Incorrect leaves initialization
• Not updating degrees properly

5. Suppose the problem is modified so that some courses can be taken multiple times (repeatedly) and prerequisites must be satisfied each time before retaking. Which of the following statements about adapting the BFS topological sort approach is correct?

hard

A. We should switch to a dynamic programming approach because BFS topological sort cannot handle repeated courses.

B. The original BFS topological sort can be used without changes since it handles cycles by returning -1.

C. We can use BFS topological sort by resetting indegrees after each semester and reprocessing courses as needed.

D. The problem becomes invalid because repeated courses create cycles; BFS topological sort cannot be applied directly.

Solution

1. Step 1: Understand repeated courses and prerequisites

   Allowing repeated courses with prerequisites means cycles can form because a course depends on itself indirectly.

2. Step 2: Analyze BFS topological sort applicability

   BFS topological sort requires a DAG; cycles break this assumption, so it cannot be applied directly.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Repeated courses induce cycles -> BFS topological sort invalid [OK]

Hint: Repeated courses create cycles, breaking DAG assumption [OK]

Common Mistakes:

• Assuming BFS topological sort handles cycles by returning -1
• Thinking resetting indegrees can simulate repeats correctly
• Believing DP is always needed without analyzing graph structure