Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
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].
Consider the following process on the graph:
x and keep visiting other nodes through edges until you reach a node that you have already visited before on this same process.Return an array answer where answer[i] is the number of different nodes that you will visit if you perform the process starting from node i.
Example 1:
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.
Example 2:
Input: edges = [1,2,3,4,0] Output: [5,5,5,5,5] Explanation: Starting from any node we can visit every node in the graph in the process.
Constraints:
n == edges.length2 <= n <= 1050 <= edges[i] <= n - 1edges[i] != iProblem summary: 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]. Consider the following process on the graph: You start from a node x and keep visiting other nodes through edges until you reach a node that you have already visited before on this same process. Return an array answer where answer[i] is the number of different nodes that you will visit if you perform the process starting from node i.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming
[1,2,0,0]
[1,2,3,4,0]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2876: Count Visited Nodes in a Directed Graph
class Solution {
public int[] countVisitedNodes(List<Integer> edges) {
int n = edges.size();
int[] ans = new int[n];
int[] vis = new int[n];
for (int i = 0; i < n; ++i) {
if (ans[i] == 0) {
int cnt = 0, j = i;
while (vis[j] == 0) {
vis[j] = ++cnt;
j = edges.get(j);
}
int cycle = 0, total = cnt + ans[j];
if (ans[j] == 0) {
cycle = cnt - vis[j] + 1;
}
j = i;
while (ans[j] == 0) {
ans[j] = Math.max(total--, cycle);
j = edges.get(j);
}
}
}
return ans;
}
}
// Accepted solution for LeetCode #2876: Count Visited Nodes in a Directed Graph
func countVisitedNodes(edges []int) []int {
n := len(edges)
ans := make([]int, n)
vis := make([]int, n)
for i := range ans {
if ans[i] == 0 {
cnt, j := 0, i
for vis[j] == 0 {
cnt++
vis[j] = cnt
j = edges[j]
}
cycle, total := 0, cnt+ans[j]
if ans[j] == 0 {
cycle = cnt - vis[j] + 1
}
j = i
for ans[j] == 0 {
ans[j] = max(total, cycle)
total--
j = edges[j]
}
}
}
return ans
}
# Accepted solution for LeetCode #2876: Count Visited Nodes in a Directed Graph
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
// Accepted solution for LeetCode #2876: Count Visited Nodes in a Directed Graph
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #2876: Count Visited Nodes in a Directed Graph
// class Solution {
// public int[] countVisitedNodes(List<Integer> edges) {
// int n = edges.size();
// int[] ans = new int[n];
// int[] vis = new int[n];
// for (int i = 0; i < n; ++i) {
// if (ans[i] == 0) {
// int cnt = 0, j = i;
// while (vis[j] == 0) {
// vis[j] = ++cnt;
// j = edges.get(j);
// }
// int cycle = 0, total = cnt + ans[j];
// if (ans[j] == 0) {
// cycle = cnt - vis[j] + 1;
// }
// j = i;
// while (ans[j] == 0) {
// ans[j] = Math.max(total--, cycle);
// j = edges.get(j);
// }
// }
// }
// return ans;
// }
// }
// Accepted solution for LeetCode #2876: Count Visited Nodes in a Directed Graph
function countVisitedNodes(edges: number[]): number[] {
const n = edges.length;
const ans: number[] = Array(n).fill(0);
const vis: number[] = Array(n).fill(0);
for (let i = 0; i < n; ++i) {
if (ans[i] === 0) {
let [cnt, j] = [0, i];
while (vis[j] === 0) {
vis[j] = ++cnt;
j = edges[j];
}
let [cycle, total] = [0, cnt + ans[j]];
if (ans[j] === 0) {
cycle = cnt - vis[j] + 1;
}
j = i;
while (ans[j] === 0) {
ans[j] = Math.max(total--, cycle);
j = edges[j];
}
}
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.