Why Shortest Path in Unweighted Graph Using BFS in DSA C++?

The Big Idea

What if you could find the shortest route without trying every possible path?

The Scenario

Imagine you are in a city with many streets and intersections. You want to find the shortest way to walk from your home to a friend's house without knowing the map well.

If you try to guess the path by walking randomly or checking every street one by one, it will take a lot of time and effort.

The Problem

Manually checking every possible path is slow and confusing. You might get lost or take a longer route without realizing it.

Writing code to check all paths without a clear method can cause mistakes and waste computer time.

The Solution

Using Breadth-First Search (BFS) helps you explore all nearby intersections first before moving farther. This way, you find the shortest path quickly and without missing any closer options.

BFS works like spreading out in waves, checking all neighbors step by step, ensuring the first time you reach your friend's house is the shortest route.

Before vs After

✗ Before

void findShortestPath(int start, int end) {
  // Try all paths manually, very complex
  // No clear order, may repeat nodes
}

✓ After

void findShortestPath(int start, int end) {
  std::queue<int> toVisit;
  toVisit.push(start);
  // BFS ensures shortest path in unweighted graph
}

What It Enables

You can quickly find the shortest route between two points in a network without guessing or checking every path.

Real Life Example

Finding the quickest way to send a message through a social network where each connection counts as one step.

Key Takeaways

Manual path search is slow and error-prone.

BFS explores neighbors level by level to find shortest paths.

This method guarantees the shortest path in unweighted graphs efficiently.

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