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 a positive integer n representing the number of nodes in an undirected graph. The nodes are labeled from 1 to n.
You are also given a 2D integer array edges, where edges[i] = [ai, bi] indicates that there is a bidirectional edge between nodes ai and bi. Notice that the given graph may be disconnected.
Divide the nodes of the graph into m groups (1-indexed) such that:
[ai, bi], if ai belongs to the group with index x, and bi belongs to the group with index y, then |y - x| = 1.Return the maximum number of groups (i.e., maximum m) into which you can divide the nodes. Return -1 if it is impossible to group the nodes with the given conditions.
Example 1:
Input: n = 6, edges = [[1,2],[1,4],[1,5],[2,6],[2,3],[4,6]] Output: 4 Explanation: As shown in the image we: - Add node 5 to the first group. - Add node 1 to the second group. - Add nodes 2 and 4 to the third group. - Add nodes 3 and 6 to the fourth group. We can see that every edge is satisfied. It can be shown that that if we create a fifth group and move any node from the third or fourth group to it, at least on of the edges will not be satisfied.
Example 2:
Input: n = 3, edges = [[1,2],[2,3],[3,1]] Output: -1 Explanation: If we add node 1 to the first group, node 2 to the second group, and node 3 to the third group to satisfy the first two edges, we can see that the third edge will not be satisfied. It can be shown that no grouping is possible.
Constraints:
1 <= n <= 5001 <= edges.length <= 104edges[i].length == 21 <= ai, bi <= nai != biProblem summary: You are given a positive integer n representing the number of nodes in an undirected graph. The nodes are labeled from 1 to n. You are also given a 2D integer array edges, where edges[i] = [ai, bi] indicates that there is a bidirectional edge between nodes ai and bi. Notice that the given graph may be disconnected. Divide the nodes of the graph into m groups (1-indexed) such that: Each node in the graph belongs to exactly one group. For every pair of nodes in the graph that are connected by an edge [ai, bi], if ai belongs to the group with index x, and bi belongs to the group with index y, then |y - x| = 1. Return the maximum number of groups (i.e., maximum m) into which you can divide the nodes. Return -1 if it is impossible to group the nodes with the given conditions.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Union-Find
6 [[1,2],[1,4],[1,5],[2,6],[2,3],[4,6]]
3 [[1,2],[2,3],[3,1]]
binary-tree-level-order-traversal)is-graph-bipartite)shortest-cycle-in-a-graph)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2493: Divide Nodes Into the Maximum Number of Groups
class Solution {
public int magnificentSets(int n, int[][] edges) {
List<Integer>[] g = new List[n];
Arrays.setAll(g, k -> new ArrayList<>());
for (var e : edges) {
int a = e[0] - 1, b = e[1] - 1;
g[a].add(b);
g[b].add(a);
}
int[] d = new int[n];
int[] dist = new int[n];
for (int i = 0; i < n; ++i) {
Deque<Integer> q = new ArrayDeque<>();
q.offer(i);
Arrays.fill(dist, 0);
dist[i] = 1;
int mx = 1;
int root = i;
while (!q.isEmpty()) {
int a = q.poll();
root = Math.min(root, a);
for (int b : g[a]) {
if (dist[b] == 0) {
dist[b] = dist[a] + 1;
mx = Math.max(mx, dist[b]);
q.offer(b);
} else if (Math.abs(dist[b] - dist[a]) != 1) {
return -1;
}
}
}
d[root] = Math.max(d[root], mx);
}
return Arrays.stream(d).sum();
}
}
// Accepted solution for LeetCode #2493: Divide Nodes Into the Maximum Number of Groups
func magnificentSets(n int, edges [][]int) (ans int) {
g := make([][]int, n)
for _, e := range edges {
a, b := e[0]-1, e[1]-1
g[a] = append(g[a], b)
g[b] = append(g[b], a)
}
d := make([]int, n)
for i := range d {
q := []int{i}
dist := make([]int, n)
dist[i] = 1
mx := 1
root := i
for len(q) > 0 {
a := q[0]
q = q[1:]
root = min(root, a)
for _, b := range g[a] {
if dist[b] == 0 {
dist[b] = dist[a] + 1
mx = max(mx, dist[b])
q = append(q, b)
} else if abs(dist[b]-dist[a]) != 1 {
return -1
}
}
}
d[root] = max(d[root], mx)
}
for _, x := range d {
ans += x
}
return
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
# Accepted solution for LeetCode #2493: Divide Nodes Into the Maximum Number of Groups
class Solution:
def magnificentSets(self, n: int, edges: List[List[int]]) -> int:
g = [[] for _ in range(n)]
for a, b in edges:
g[a - 1].append(b - 1)
g[b - 1].append(a - 1)
d = defaultdict(int)
for i in range(n):
q = deque([i])
dist = [0] * n
dist[i] = mx = 1
root = i
while q:
a = q.popleft()
root = min(root, a)
for b in g[a]:
if dist[b] == 0:
dist[b] = dist[a] + 1
mx = max(mx, dist[b])
q.append(b)
elif abs(dist[b] - dist[a]) != 1:
return -1
d[root] = max(d[root], mx)
return sum(d.values())
// Accepted solution for LeetCode #2493: Divide Nodes Into the Maximum Number of Groups
// 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 #2493: Divide Nodes Into the Maximum Number of Groups
// class Solution {
// public int magnificentSets(int n, int[][] edges) {
// List<Integer>[] g = new List[n];
// Arrays.setAll(g, k -> new ArrayList<>());
// for (var e : edges) {
// int a = e[0] - 1, b = e[1] - 1;
// g[a].add(b);
// g[b].add(a);
// }
// int[] d = new int[n];
// int[] dist = new int[n];
// for (int i = 0; i < n; ++i) {
// Deque<Integer> q = new ArrayDeque<>();
// q.offer(i);
// Arrays.fill(dist, 0);
// dist[i] = 1;
// int mx = 1;
// int root = i;
// while (!q.isEmpty()) {
// int a = q.poll();
// root = Math.min(root, a);
// for (int b : g[a]) {
// if (dist[b] == 0) {
// dist[b] = dist[a] + 1;
// mx = Math.max(mx, dist[b]);
// q.offer(b);
// } else if (Math.abs(dist[b] - dist[a]) != 1) {
// return -1;
// }
// }
// }
// d[root] = Math.max(d[root], mx);
// }
// return Arrays.stream(d).sum();
// }
// }
// Accepted solution for LeetCode #2493: Divide Nodes Into the Maximum Number of Groups
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2493: Divide Nodes Into the Maximum Number of Groups
// class Solution {
// public int magnificentSets(int n, int[][] edges) {
// List<Integer>[] g = new List[n];
// Arrays.setAll(g, k -> new ArrayList<>());
// for (var e : edges) {
// int a = e[0] - 1, b = e[1] - 1;
// g[a].add(b);
// g[b].add(a);
// }
// int[] d = new int[n];
// int[] dist = new int[n];
// for (int i = 0; i < n; ++i) {
// Deque<Integer> q = new ArrayDeque<>();
// q.offer(i);
// Arrays.fill(dist, 0);
// dist[i] = 1;
// int mx = 1;
// int root = i;
// while (!q.isEmpty()) {
// int a = q.poll();
// root = Math.min(root, a);
// for (int b : g[a]) {
// if (dist[b] == 0) {
// dist[b] = dist[a] + 1;
// mx = Math.max(mx, dist[b]);
// q.offer(b);
// } else if (Math.abs(dist[b] - dist[a]) != 1) {
// return -1;
// }
// }
// }
// d[root] = Math.max(d[root], mx);
// }
// return Arrays.stream(d).sum();
// }
// }
Use this to step through a reusable interview workflow for this problem.
Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.
With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.
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.