Understand
30-second explanation
Build the interview-ready mental model before comparing implementations.
You want the fewest edges between any two nodes that share the target color. Because the graph is unweighted, BFS gives shortest distances. The clean answer is a multi-source BFS: seed the queue with every node of the target color, tag each node with the source it came from, and the first time two different sources' waves touch, that meeting distance is the global minimum. O(V + E).
Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview contextHide interview context
What's being tested
Whether you recognize an unweighted shortest-path question as BFS, and whether you can avoid the naive 'BFS from every clone' by collapsing it into one multi-source pass with source tagging.
Where candidates slip
Running BFS from each colored node separately (correct but O(C·(V+E))), or in the multi-source version recording a collision between two waves that trace back to the same source — which is not a real pair.
How to stand out
Seed all clones at once, carry a source label per node, and explain why the first inter-source collision is provably the global minimum because BFS expands in non-decreasing distance order.
Choose
Compare the available approaches
Use constraints and trade-offs to select the route you can defend in an interview.
Interactive decision treeInteractiveChoose the right approachFirst decision previewShortest path between two same-color nodesEdges weighted?YesNoExplore the full treeCollapse the tree
How to read it:Nodes 1 and 4 are color 1. BFS waves from each meet at node 2; total hops 2.
Learn & trace
Build each solution step by step
Move from the core idea to a visual trace, implementation, proof, and edge cases.
Approach 1 of 2
BFS from Each Clone (Baseline)
1Hint 1Explore hintHide hint
Forget cleverness: from one clone, what is the cheapest way to find the nearest other clone in an unweighted graph?
2Hint 2Explore hintHide hint
A BFS from a clone reaches nodes in increasing hop order. The first OTHER clone it pops is the nearest clone to that start.
3Hint 3Explore hintHide hint
Do that BFS from every clone and take the smallest distance found. C clones each cost one O(V + E) BFS.
A · Core idea
The most direct reading: for each clone, find its nearest fellow clone, then take the best across all clones.
Why BFS and not DFS or Dijkstra. The graph is unweighted, so the shortest path is the one with the fewest edges. BFS explores in concentric layers — all 1-hop nodes, then all 2-hop nodes, and so on — so the first time it reaches a target it has reached it by the shortest route. DFS does not have that layer property; Dijkstra would work but is overkill (and slower) when every edge weighs the same.
Why the first other clone popped is the nearest. Within a single BFS, nodes leave the queue in non-decreasing distance order. So the moment you pop a node that is a different clone of the same color, its recorded distance is the minimum distance from this start to any other clone. You can stop that BFS immediately.
Why we must keep BFS iterative. BFS is queue-driven, so it is already iterative — no recursion depth to blow even at 10⁵ nodes. (Contrast a recursive DFS, which would risk a stack overflow on a long chain.)
Why this is the baseline, not the answer. Repeating a full BFS from every clone repeats a lot of traversal. If half the graph shares the target color, you run ~V/2 BFS passes — quadratic. It is correct and easy to defend, but the multi-source version does the same job in one pass.
B · Visual dry run
Interactive dry run
Select a frame to connect the decision with its visual state.
n = 4, edges = [[1,2],[1,3],[2,4]], ids = [1,2,3,1], val = 1Interactive diagram
One BFS from clone 1
BFS from 1Layers expand from node 1; node 4 (a clone) appears at distance 2.
2C · Code with teaching comments
Reference implementation previewpublic int findShortest(int n, int[][] edges, int[] ids, int val) { List<List<Integer>> adj = new ArrayList<>(); for (int i = 0; i <= n; i++) adj.add(new ArrayList<>()); for (int[] e : edges) { adj.get(e[0]).add(e[1]);Explore the full codeHide the full code
D · Why it works and complexityO(C · (V + E)) timeO(V + E) spaceOpen the invariant and proofExplore the proofHide the proof
A single BFS already answers 'nearest clone to this node'. The only inefficiency is repeating it; the optimal version removes the repetition by starting all the searches together.
ComplexityO(C · (V + E)) time, O(V + E) space — Each BFS visits every vertex and edge once → O(V + E). Running it from C clones gives O(C · (V + E)). When C is small this is near linear, but C can be as large as V, making it O(V·(V+E)) in the worst case.
E · Edge cases and pitfalls3Not stopping the BFS at the first other clone — wastes work, though still correct.Explore edge casesHide edge cases
Pitfalls
- Not stopping the BFS at the first other clone — wastes work, though still correct.
- Re-using a single dist array across starts without resetting — stale distances corrupt later searches.
- Forgetting the u != start guard — the start node itself has the target color and would falsely report distance 0.
Approach 2 of 2 · Optimal
Multi-Source BFS with Source Tags (Optimal)
1Hint 1Explore hintHide hint
Instead of restarting BFS from each clone, what if all clones started spreading at the same instant?
2Hint 2Explore hintHide hint
Give each node a 'source' label = which clone's wave first reached it, plus its distance from that clone. Seed all clones at distance 0 with their own label.
3Hint 3Explore hintHide hint
When relaxing an edge u→v and v already belongs to a DIFFERENT source, the two waves have met: candidate = dist[u] + dist[v] + 1. The smallest such candidate is the answer.
A · Core idea
The repetition in the baseline comes from launching C independent searches. Multi-source BFS launches them together and lets the fronts collide.
Seeding and the source tag. Put every clone into the queue at distance 0, each labeled with itself as its source. As BFS proceeds, when a wave first reaches a node v, that node inherits the source of the wave that arrived. So source[v] answers 'which clone is closest to v', and dist[v] is that closest distance.
Detecting a meeting. Consider an edge u → v. If v is unseen, it joins u's wave. But if v was already claimed by a different source, then the wave from source[u] and the wave from source[v] are about to overlap. A path runs source[u] → … → u → v → … → source[v] with length dist[u] + 1 + dist[v]. Every such collision yields a candidate distance between two distinct clones; the minimum over all collisions is the answer.
Why the first collision found is essentially the smallest. BFS dequeues nodes in non-decreasing distance order, so collisions surface roughly shortest-first. To be safe we do not stop at the very first collision — we take the min over all of them, because two collisions can be discovered in the same layer. That keeps the logic airtight while staying O(V + E).
Why same-source collisions are ignored. If source[u] == source[v], both nodes trace back to the same clone — connecting them does not join two distinct clones, so it is not a valid pair. The source[u] != source[v] guard is what enforces 'two distinct clones'.
Why it stays iterative. Like all BFS, the search runs off a queue with no recursion, so a graph of 10⁵ nodes never threatens the call stack.
B · Visual dry run
Interactive dry run
Select a frame to connect the decision with its visual state.
n = 4, edges = [[1,2],[1,3],[2,4]], ids = [1,2,3,1], val = 1Interactive diagram
Two waves colliding at the bridge edge
seedWave from clone 1 (green) and wave from clone 4 (orange) meet on edge 2-4: dist 1 + dist 0 + 1 = 2.
2C · Code with teaching comments
D · Why it works and complexityO(V + E) timeO(V + E) spaceOpen the invariant and proofExplore the proofHide the proof
Tagging each node with the source that reached it first turns a set of independent BFS runs into one. The collision test source[u] != source[v] is the entire trick: it is how a single sweep discovers the closest pair across the whole set of clones.
ComplexityO(V + E) time, O(V + E) space — A single BFS touches each vertex and edge once → O(V + E). The source and distance arrays add O(V). Building the adjacency list is O(V + E). No factor of C — every clone is processed inside the one pass.
E · Edge cases and pitfalls7ids has one node of color val — Only one seed enters the queue; no second wave ever exists, so no collision → best stays infinity → -1.Explore edge casesHide edge cases
Edge cases handled
ids has one node of color valOnly one seed enters the queue; no second wave ever exists, so no collision → best stays infinity → -1.two clones directly adjacentTheir waves collide on the shared edge at dist 0 + 0 + 1 = 1 → answer 1.clones in different componentsWaves never reach each other; no inter-source collision → -1.
Pitfalls
- Stopping at the first collision without taking a min — two collisions can occur in the same layer; keep the minimum.
- Allowing same-source collisions to count — they connect a clone to itself, not two distinct clones.
- Initializing dist to a sentinel that clashes with a real 0 distance — use the source array (-1) as the 'unseen' marker.
- Using recursion instead of a queue — unnecessary and risks stack overflow; BFS is inherently iterative.
Practice
Try the solution yourself
Edit the optimal implementation, test another input, and observe the result.
Editable code previewpublic int findShortest(int n, int[][] edges, int[] ids, int val) { List<List<Integer>> adj = new ArrayList<>(); for (int i = 0; i <= n; i++) adj.add(new ArrayList<>()); for (int[] e : edges) { adj.get(e[0]).add(e[1]);Explore the playgroundClose the playground
Avoid mistakes
Recognize the interview traps
Understand why common attempts fail before reviewing the corrected implementation.
1Matching a node with itself
The start node has the target color, so a check that does not exclude it reports a path of length 0 — but a clone cannot pair with itself.
Explore the code comparisonHide the code comparison
Matching a node with itself
The start node has the target color, so a check that does not exclude it reports a path of length 0 — but a clone cannot pair with itself.
Don't write this
if ids[u - 1] == val:
best = min(best, dist[u]) # fires immediately on the start nodeWrite this instead
if u != start and ids[u - 1] == val:
best = min(best, dist[u]) # only a DIFFERENT clone counts2Same-source collision in multi-source BFS
If both endpoints of an edge trace to the same clone, joining them does not connect two distinct clones. Only an inter-source meeting is a valid pair.
Explore the code comparisonHide the code comparison
Same-source collision in multi-source BFS
If both endpoints of an edge trace to the same clone, joining them does not connect two distinct clones. Only an inter-source meeting is a valid pair.
Don't write this
if (source[v] != -1) {
best = Math.min(best, dist[u] + dist[v] + 1); // also counts same-source
}Write this instead
if (source[v] != -1 && source[v] != source[u]) {
best = Math.min(best, dist[u] + dist[v] + 1); // distinct clones only
}3Using Dijkstra on an unweighted graph
Every edge costs the same here, so BFS already yields shortest hop counts in O(V + E). A min-heap adds a log factor and complexity for no benefit.
Explore the code comparisonHide the code comparison
Using Dijkstra on an unweighted graph
Every edge costs the same here, so BFS already yields shortest hop counts in O(V + E). A min-heap adds a log factor and complexity for no benefit.
Don't write this
import heapq
heap = [(0, start)]
while heap:
d, u = heapq.heappop(heap) # log factor wasted on equal weightsWrite this instead
from collections import deque
queue = deque([start])
while queue:
u = queue.popleft() # plain BFS, O(V + E)Revise
Carry the pattern into the interview
Pattern to remember
Multi-source BFS, meet-in-the-middle on same-color nodes
Code pattern to memorize
seed queue with every node of target color, dist 0, tag source = self
BFS layer by layer:
for neighbor v of u:
if v unseen: source[v]=source[u]; dist[v]=dist[u]+1; enqueue
elif source[v] != source[u]: # waves from two clones met
answer = min(answer, dist[u] + dist[v] + 1)
return answer or -1Rules to remember
- 1Two clones must be distinct nodes — a single node of the target color can never pair with itself, so fewer than two clones means
-1. - 2The graph is unweighted, so BFS (not Dijkstra) gives shortest distances by counting edges in layers.
- 3Tag each BFS node with the source clone it spread from; an answer appears only when two different sources collide.
- 4BFS uses a queue and is naturally iterative — no recursion, so no stack overflow even at 10⁵ nodes.
- 5Return
-1when no two same-color nodes are connected — initialize the answer to infinity and check it at the end.
Use this pattern when
- Shortest number of hops in an **unweighted** graph
- Distance between any two members of a *set* of special nodes, not one fixed pair
- You can start the search from many seeds at once and let the waves meet
- Reachability plus hop-count, where all edges cost the same
Choose another pattern when
- Edges carry differing weights → Dijkstra, BFS layers no longer equal distance
- You need the path between one fixed pair only → a single-source BFS is enough
- You need every pairwise distance → BFS from each node (or Floyd-Warshall on small graphs)
- Negative weights → Bellman-Ford, never BFS
Interview takeaway
Multi-source BFS turns 'closest pair among a set of nodes' into a single linear sweep: launch waves from all candidates simultaneously and the first collision between two different waves is your answer. The source tag is what makes one BFS replace many.
Continue
Transfer the pattern to another problem
Use a direct variation or pattern sibling to prove the idea is reusable.
Transferable patternMulti-Source BFS for Closest-Pair-in-a-Set. Recognize it when you need the shortest unweighted distance between any two members of a set of special nodes. Seed all of them, tag the source, and detect inter-source collisions. The same multi-source seeding powers Rotting Oranges and 01 Matrix.
Practice queueDirect variations and pattern siblings
Direct follow-ups
- BFS Shortest ReachSingle-source version: distance from one start to every node in an unweighted graph.
- Rotting OrangesMulti-source BFS on a grid — all rotten cells seed the queue at once, just like all clones here.
- Word LadderShortest transformation chain — BFS on an implicit unweighted graph of words.
- Dijkstra's Shortest PathWhen edges carry differing weights, swap the BFS queue for a min-heap.