Leetcode #2876: Count Visited Nodes in a Directed Graph
In this guide, we solve Leetcode #2876 Count Visited Nodes in a Directed Graph in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
There is a directed graph consisting of n nodes numbered from 0 to n - 1 and n directed edges. You are given a 0-indexed array edges where edges[i] indicates that there is an edge from node i to node edges[i].
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Graph, Memoization, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: edges = [1,2,0,0]
Output: [3,3,3,4]
Explanation: We perform the process starting from each node in the following way:
- Starting from node 0, we visit the nodes 0 -> 1 -> 2 -> 0. The number of different nodes we visit is 3.
- Starting from node 1, we visit the nodes 1 -> 2 -> 0 -> 1. The number of different nodes we visit is 3.
- Starting from node 2, we visit the nodes 2 -> 0 -> 1 -> 2. The number of different nodes we visit is 3.
- Starting from node 3, we visit the nodes 3 -> 0 -> 1 -> 2 -> 0. The number of different nodes we visit is 4.
Python Solution
class Solution:
def countVisitedNodes(self, edges: List[int]) -> List[int]:
n = len(edges)
ans = [0] * n
vis = [0] * n
for i in range(n):
if not ans[i]:
cnt, j = 0, i
while not vis[j]:
cnt += 1
vis[j] = cnt
j = edges[j]
cycle, total = 0, cnt + ans[j]
if not ans[j]:
cycle = cnt - vis[j] + 1
total = cnt
j = i
while not ans[j]:
ans[j] = max(total, cycle)
total -= 1
j = edges[j]
return ans
Complexity
The time complexity is , and the space complexity is , where is the length of the array edges. The space complexity is , where is the length of the array edges.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.