Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
You are given an undirected graph (the "original graph") with n nodes labeled from 0 to n - 1. You decide to subdivide each edge in the graph into a chain of nodes, with the number of new nodes varying between each edge.
The graph is given as a 2D array of edges where edges[i] = [ui, vi, cnti] indicates that there is an edge between nodes ui and vi in the original graph, and cnti is the total number of new nodes that you will subdivide the edge into. Note that cnti == 0 means you will not subdivide the edge.
To subdivide the edge [ui, vi], replace it with (cnti + 1) new edges and cnti new nodes. The new nodes are x1, x2, ..., xcnti, and the new edges are [ui, x1], [x1, x2], [x2, x3], ..., [xcnti-1, xcnti], [xcnti, vi].
In this new graph, you want to know how many nodes are reachable from the node 0, where a node is reachable if the distance is maxMoves or less.
Given the original graph and maxMoves, return the number of nodes that are reachable from node 0 in the new graph.
Example 1:
Input: edges = [[0,1,10],[0,2,1],[1,2,2]], maxMoves = 6, n = 3 Output: 13 Explanation: The edge subdivisions are shown in the image above. The nodes that are reachable are highlighted in yellow.
Example 2:
Input: edges = [[0,1,4],[1,2,6],[0,2,8],[1,3,1]], maxMoves = 10, n = 4 Output: 23
Example 3:
Input: edges = [[1,2,4],[1,4,5],[1,3,1],[2,3,4],[3,4,5]], maxMoves = 17, n = 5 Output: 1 Explanation: Node 0 is disconnected from the rest of the graph, so only node 0 is reachable.
Constraints:
0 <= edges.length <= min(n * (n - 1) / 2, 104)edges[i].length == 30 <= ui < vi < n0 <= cnti <= 1040 <= maxMoves <= 1091 <= n <= 3000Problem summary: You are given an undirected graph (the "original graph") with n nodes labeled from 0 to n - 1. You decide to subdivide each edge in the graph into a chain of nodes, with the number of new nodes varying between each edge. The graph is given as a 2D array of edges where edges[i] = [ui, vi, cnti] indicates that there is an edge between nodes ui and vi in the original graph, and cnti is the total number of new nodes that you will subdivide the edge into. Note that cnti == 0 means you will not subdivide the edge. To subdivide the edge [ui, vi], replace it with (cnti + 1) new edges and cnti new nodes. The new nodes are x1, x2, ..., xcnti, and the new edges are [ui, x1], [x1, x2], [x2, x3], ..., [xcnti-1, xcnti], [xcnti, vi]. In this new graph, you want to know how many nodes are reachable from the node 0, where a node is reachable if the distance is maxMoves or less. Given the original graph
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
[[0,1,10],[0,2,1],[1,2,2]] 6 3
[[0,1,4],[1,2,6],[0,2,8],[1,3,1]] 10 4
[[1,2,4],[1,4,5],[1,3,1],[2,3,4],[3,4,5]] 17 5
find-all-people-with-secret)paths-in-maze-that-lead-to-same-room)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #882: Reachable Nodes In Subdivided Graph
class Solution {
public int reachableNodes(int[][] edges, int maxMoves, int n) {
List<int[]>[] g = new List[n];
Arrays.setAll(g, e -> new ArrayList<>());
for (var e : edges) {
int u = e[0], v = e[1], cnt = e[2] + 1;
g[u].add(new int[] {v, cnt});
g[v].add(new int[] {u, cnt});
}
int[] dist = new int[n];
Arrays.fill(dist, 1 << 30);
PriorityQueue<int[]> q = new PriorityQueue<>((a, b) -> a[0] - b[0]);
q.offer(new int[] {0, 0});
dist[0] = 0;
while (!q.isEmpty()) {
var p = q.poll();
int d = p[0], u = p[1];
for (var nxt : g[u]) {
int v = nxt[0], cnt = nxt[1];
if (d + cnt < dist[v]) {
dist[v] = d + cnt;
q.offer(new int[] {dist[v], v});
}
}
}
int ans = 0;
for (int d : dist) {
if (d <= maxMoves) {
++ans;
}
}
for (var e : edges) {
int u = e[0], v = e[1], cnt = e[2];
int a = Math.min(cnt, Math.max(0, maxMoves - dist[u]));
int b = Math.min(cnt, Math.max(0, maxMoves - dist[v]));
ans += Math.min(cnt, a + b);
}
return ans;
}
}
// Accepted solution for LeetCode #882: Reachable Nodes In Subdivided Graph
func reachableNodes(edges [][]int, maxMoves int, n int) (ans int) {
g := make([][]pair, n)
for _, e := range edges {
u, v, cnt := e[0], e[1], e[2]+1
g[u] = append(g[u], pair{cnt, v})
g[v] = append(g[v], pair{cnt, u})
}
dist := make([]int, n)
for i := range dist {
dist[i] = 1 << 30
}
dist[0] = 0
q := hp{{0, 0}}
for len(q) > 0 {
p := heap.Pop(&q).(pair)
d, u := p.v, p.i
for _, nxt := range g[u] {
v, cnt := nxt.i, nxt.v
if t := d + cnt; t < dist[v] {
dist[v] = t
heap.Push(&q, pair{t, v})
}
}
}
for _, d := range dist {
if d <= maxMoves {
ans++
}
}
for _, e := range edges {
u, v, cnt := e[0], e[1], e[2]
a := min(cnt, max(0, maxMoves-dist[u]))
b := min(cnt, max(0, maxMoves-dist[v]))
ans += min(cnt, a+b)
}
return
}
type pair struct{ v, i int }
type hp []pair
func (h hp) Len() int { return len(h) }
func (h hp) Less(i, j int) bool { return h[i].v < h[j].v }
func (h hp) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(v any) { *h = append(*h, v.(pair)) }
func (h *hp) Pop() any { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }
# Accepted solution for LeetCode #882: Reachable Nodes In Subdivided Graph
class Solution:
def reachableNodes(self, edges: List[List[int]], maxMoves: int, n: int) -> int:
g = defaultdict(list)
for u, v, cnt in edges:
g[u].append((v, cnt + 1))
g[v].append((u, cnt + 1))
q = [(0, 0)]
dist = [0] + [inf] * n
while q:
d, u = heappop(q)
for v, cnt in g[u]:
if (t := d + cnt) < dist[v]:
dist[v] = t
q.append((t, v))
ans = sum(d <= maxMoves for d in dist)
for u, v, cnt in edges:
a = min(cnt, max(0, maxMoves - dist[u]))
b = min(cnt, max(0, maxMoves - dist[v]))
ans += min(cnt, a + b)
return ans
// Accepted solution for LeetCode #882: Reachable Nodes In Subdivided Graph
/**
* [0882] Reachable Nodes In Subdivided Graph
*
* You are given an undirected graph (the "original graph") with n nodes labeled from 0 to n - 1. You decide to subdivide each edge in the graph into a chain of nodes, with the number of new nodes varying between each edge.
* The graph is given as a 2D array of edges where edges[i] = [ui, vi, cnti] indicates that there is an edge between nodes ui and vi in the original graph, and cnti is the total number of new nodes that you will subdivide the edge into. Note that cnti == 0 means you will not subdivide the edge.
* To subdivide the edge [ui, vi], replace it with (cnti + 1) new edges and cnti new nodes. The new nodes are x1, x2, ..., xcnti, and the new edges are [ui, x1], [x1, x2], [x2, x3], ..., [xcnti-1, xcnti], [xcnti, vi].
* In this new graph, you want to know how many nodes are reachable from the node 0, where a node is reachable if the distance is maxMoves or less.
* Given the original graph and maxMoves, return the number of nodes that are reachable from node 0 in the new graph.
*
* Example 1:
* <img alt="" src="https://s3-lc-upload.s3.amazonaws.com/uploads/2018/08/01/origfinal.png" style="width: 600px; height: 247px;" />
* Input: edges = [[0,1,10],[0,2,1],[1,2,2]], maxMoves = 6, n = 3
* Output: 13
* Explanation: The edge subdivisions are shown in the image above.
* The nodes that are reachable are highlighted in yellow.
*
* Example 2:
*
* Input: edges = [[0,1,4],[1,2,6],[0,2,8],[1,3,1]], maxMoves = 10, n = 4
* Output: 23
*
* Example 3:
*
* Input: edges = [[1,2,4],[1,4,5],[1,3,1],[2,3,4],[3,4,5]], maxMoves = 17, n = 5
* Output: 1
* Explanation: Node 0 is disconnected from the rest of the graph, so only node 0 is reachable.
*
*
* Constraints:
*
* 0 <= edges.length <= min(n * (n - 1) / 2, 10^4)
* edges[i].length == 3
* 0 <= ui < vi < n
* There are no multiple edges in the graph.
* 0 <= cnti <= 10^4
* 0 <= maxMoves <= 10^9
* 1 <= n <= 3000
*
*/
pub struct Solution {}
// problem: https://leetcode.com/problems/reachable-nodes-in-subdivided-graph/
// discuss: https://leetcode.com/problems/reachable-nodes-in-subdivided-graph/discuss/?currentPage=1&orderBy=most_votes&query=
// submission codes start here
impl Solution {
// Credit: https://leetcode.com/problems/reachable-nodes-in-subdivided-graph/solutions/1459584/rust-solution/
pub fn reachable_nodes(edges: Vec<Vec<i32>>, max_moves: i32, n: i32) -> i32 {
let mut graph = vec![Vec::new(); n as usize];
for edge in &edges {
graph[edge[0] as usize].push((edge[1] as usize, edge[2]));
graph[edge[1] as usize].push((edge[0] as usize, edge[2]));
}
let mut remains = vec![None; n as usize];
let mut bh = std::collections::BinaryHeap::new();
bh.push((std::cmp::Reverse(0), 0));
let mut result = 0;
while let Some((std::cmp::Reverse(min), u)) = bh.pop() {
if min > max_moves {
break;
}
if remains[u].is_some() {
continue;
}
result += 1;
remains[u] = Some(max_moves - min);
for &(v, c) in graph[u].iter().filter(|(v, _)| remains[*v].is_none()) {
bh.push((std::cmp::Reverse(min + c + 1), v));
}
}
for edge in &edges {
result += edge[2].min(
remains[edge[0] as usize].unwrap_or(0) + remains[edge[1] as usize].unwrap_or(0),
);
}
result
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn test_0882_example_1() {
let edges = vec![vec![0, 1, 10], vec![0, 2, 1], vec![1, 2, 2]];
let max_moves = 6;
let n = 3;
let result = 13;
assert_eq!(Solution::reachable_nodes(edges, max_moves, n), result);
}
#[test]
fn test_0882_example_2() {
let edges = vec![vec![0, 1, 4], vec![1, 2, 6], vec![0, 2, 8], vec![1, 3, 1]];
let max_moves = 10;
let n = 4;
let result = 23;
assert_eq!(Solution::reachable_nodes(edges, max_moves, n), result);
}
#[test]
fn test_0882_example_3() {
let edges = vec![
vec![1, 2, 4],
vec![1, 4, 5],
vec![1, 3, 1],
vec![2, 3, 4],
vec![3, 4, 5],
];
let max_moves = 17;
let n = 5;
let result = 1;
assert_eq!(Solution::reachable_nodes(edges, max_moves, n), result);
}
}
// Accepted solution for LeetCode #882: Reachable Nodes In Subdivided Graph
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #882: Reachable Nodes In Subdivided Graph
// class Solution {
// public int reachableNodes(int[][] edges, int maxMoves, int n) {
// List<int[]>[] g = new List[n];
// Arrays.setAll(g, e -> new ArrayList<>());
// for (var e : edges) {
// int u = e[0], v = e[1], cnt = e[2] + 1;
// g[u].add(new int[] {v, cnt});
// g[v].add(new int[] {u, cnt});
// }
// int[] dist = new int[n];
// Arrays.fill(dist, 1 << 30);
// PriorityQueue<int[]> q = new PriorityQueue<>((a, b) -> a[0] - b[0]);
// q.offer(new int[] {0, 0});
// dist[0] = 0;
// while (!q.isEmpty()) {
// var p = q.poll();
// int d = p[0], u = p[1];
// for (var nxt : g[u]) {
// int v = nxt[0], cnt = nxt[1];
// if (d + cnt < dist[v]) {
// dist[v] = d + cnt;
// q.offer(new int[] {dist[v], v});
// }
// }
// }
// int ans = 0;
// for (int d : dist) {
// if (d <= maxMoves) {
// ++ans;
// }
// }
// for (var e : edges) {
// int u = e[0], v = e[1], cnt = e[2];
// int a = Math.min(cnt, Math.max(0, maxMoves - dist[u]));
// int b = Math.min(cnt, Math.max(0, maxMoves - dist[v]));
// ans += Math.min(cnt, a + b);
// }
// return ans;
// }
// }
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.