Shortest Path in Unweighted Graph Using BFS in DSA C++ - Time & Space Complexity

Time Complexity: Shortest Path in Unweighted Graph Using BFS

O(n + m)

Understanding Time Complexity

We want to know how long it takes to find the shortest path in an unweighted graph using BFS.

Specifically, how the time grows as the graph gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void bfsShortestPath(int start, vector<vector<int>> &graph, vector<int> &dist) {
    int n = graph.size();
    queue<int> q;
    dist.assign(n, -1);
    dist[start] = 0;
    q.push(start);

    while (!q.empty()) {
        int node = q.front(); q.pop();
        for (int neighbor : graph[node]) {
            if (dist[neighbor] == -1) {
                dist[neighbor] = dist[node] + 1;
                q.push(neighbor);
            }
        }
    }
}
    

This code finds the shortest distance from a start node to all other nodes in an unweighted graph using BFS.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node and checking its neighbors in the graph.
• How many times: Each node is visited once, and each edge is checked once during the BFS traversal.

How Execution Grows With Input

As the graph grows, the BFS visits every node and edge once.

Input Size (n nodes, m edges)	Approx. Operations
10 nodes, 15 edges	About 25 operations (10 nodes + 15 edges)
100 nodes, 200 edges	About 300 operations (100 + 200)
1000 nodes, 5000 edges	About 6000 operations (1000 + 5000)

Pattern observation: The operations grow roughly in proportion to the number of nodes plus edges.

Final Time Complexity

Time Complexity: O(n + m)

This means the time grows linearly with the number of nodes and edges in the graph.

Common Mistake

[X] Wrong: "BFS takes O(n²) time because it uses nested loops."

[OK] Correct: BFS visits each node and edge only once, so it does not repeatedly check all pairs. The loops are not nested over all nodes but over neighbors, which sums to edges.

Interview Connect

Understanding BFS time complexity helps you explain why it is efficient for shortest paths in unweighted graphs, a common interview topic.

Self-Check

"What if the graph was represented using an adjacency matrix instead of adjacency lists? How would the time complexity change?"

Practice

1. What is the main purpose of using BFS (Breadth-First Search) in an unweighted graph?

easy

A. To sort the nodes in topological order

B. To find the maximum depth of the graph

C. To find the shortest path from a source node to all other nodes

D. To detect cycles in the graph

Solution

1. Step 1: Understand BFS traversal

   BFS explores nodes level by level starting from the source node.

2. Step 2: BFS finds shortest paths in unweighted graphs

   Because BFS visits nodes in order of increasing distance, it finds shortest paths from source to all nodes.

3. Final Answer:

   To find the shortest path from a source node to all other nodes -> Option C

4. Quick Check:

   BFS shortest path = D [OK]

Hint: BFS explores neighbors level-wise to find shortest paths [OK]

Common Mistakes:

• Confusing BFS with DFS for shortest path
• Thinking BFS finds maximum depth
• Assuming BFS sorts nodes topologically

2. Which of the following is the correct way to initialize a queue for BFS in C++?

easy

A. queue<int> q; q.push(0);

B. stack<int> q; q.push(0);

C. queue<int> q; q.insert(0);

D. queue<int> q; q.add(0);

Solution

1. Step 1: Identify correct queue declaration

   In C++, BFS uses std::queue with push() to add elements.

2. Step 2: Check method names

   push() is correct for queue; insert() and add() are invalid; stack is wrong data structure.

3. Final Answer:

   queue<int> q; q.push(0); -> Option A

4. Quick Check:

   queue push() method = C [OK]

Hint: Use std::queue and push() to add elements in BFS [OK]

Common Mistakes:

• Using stack instead of queue
• Using insert() or add() instead of push()
• Confusing queue methods with vector methods

3. Consider the following C++ code snippet for BFS shortest path in an unweighted graph:

int n = 4;
vector adj = {{1,2}, {0,2,3}, {0,1,3}, {1,2}};
vector<int> dist(n, -1);
queue<int> q;
dist[0] = 0;
q.push(0);
while(!q.empty()) {
  int u = q.front(); q.pop();
  for(int v : adj[u]) {
    if(dist[v] == -1) {
      dist[v] = dist[u] + 1;
      q.push(v);
    }
  }
}
for(int i=0; iWhat is the output?

medium

A. 0 1 1 2

B. 0 1 2 3

C. 0 2 1 1

D. 0 1 1 1

Solution

1. Step 1: Initialize distances and start BFS

   dist[0]=0, others -1; start from node 0.

2. Step 2: BFS visits neighbors and updates distances

   From 0: neighbors 1,2 get dist=1; from 1 and 2: neighbor 3 gets dist=2.

3. Final Answer:

   0 1 1 2 -> Option A

4. Quick Check:

   BFS distance array = 0 1 1 2 [OK]

Hint: BFS distance increments by 1 per level from source [OK]

Common Mistakes:

• Mixing up node indices and distances
• Not initializing distances to -1
• Forgetting to check if node was visited

4. The following BFS code snippet is intended to find shortest paths in an unweighted graph. Identify the bug:

int n = 3;
vector adj = {{1}, {0,2}, {1}};
vector<int> dist(n, 0);
queue<int> q;
dist[0] = 0;
q.push(0);
while(!q.empty()) {
  int u = q.front(); q.pop();
  for(int v : adj[u]) {
    if(dist[v] == 0) {
      dist[v] = dist[u] + 1;
      q.push(v);
    }
  }
}
for(int i=0; i

medium

A. dist[u] is not updated inside the loop

B. Queue is not used properly

C. Adjacency list is incorrect

D. dist vector initialized with 0 instead of -1

Solution

1. Step 1: Check distance initialization

   dist initialized with 0 means unvisited nodes are indistinguishable from source.

2. Step 2: Identify impact on BFS logic

   Condition dist[v] == 0 fails to detect unvisited nodes correctly, causing wrong or no updates.

3. Final Answer:

   dist vector initialized with 0 instead of -1 -> Option D

4. Quick Check:

   dist init must be -1 to mark unvisited [OK]

Hint: Initialize distances with -1 to mark unvisited nodes [OK]

Common Mistakes:

• Using 0 as unvisited marker
• Not checking visited condition properly
• Misusing queue operations

5. You have an unweighted graph with 5 nodes and edges: 0-1, 0-2, 1-3, 2-3, 3-4. Using BFS from node 0, what is the shortest path distance to node 4? Also, which nodes are on one shortest path from 0 to 4?

hard

A. Distance: 2; Path: 0 -> 1 -> 3 -> 4

B. Distance: 3; Path: 0 -> 2 -> 3 -> 4

C. Distance: 2; Path: 0 -> 3 -> 4

D. Distance: 4; Path: 0 -> 2 -> 3 -> 4

Solution

1. Step 1: Build graph and run BFS from node 0

   Edges: 0-1,0-2; 1-3; 2-3; 3-4. BFS levels: 0(dist=0), 1 and 2(dist=1), 3(dist=2), 4(dist=3).

2. Step 2: Identify shortest path to node 4

   Shortest distance to 4 is 3. Paths: 0->1->3->4 or 0->2->3->4. Distance: 3; Path: 0 -> 2 -> 3 -> 4 shows correct distance and one valid path.

3. Final Answer:

   Distance: 3; Path: 0 -> 2 -> 3 -> 4 -> Option B

4. Quick Check:

   Shortest path length = 3 via node 2 [OK]

Hint: Trace BFS levels to find shortest path and distance [OK]

Common Mistakes:

• Assuming direct edge from 0 to 3
• Confusing distance with number of edges
• Ignoring multiple shortest paths