Shortest Path in Unweighted Graph Using BFS in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind using BFS to find the shortest path in an unweighted graph?

BFS explores nodes level by level, so the first time it reaches a node, it has found the shortest path to that node because all edges have equal weight.

beginner

In BFS for shortest path, what data structure is commonly used to keep track of nodes to visit next?

A queue is used to keep track of nodes to visit in the order they were discovered, ensuring level-by-level exploration.

intermediate

How do you keep track of the actual shortest path from the start node to a target node in BFS?

By storing the parent (or predecessor) of each visited node, you can backtrack from the target node to the start node to reconstruct the path.

intermediate

Why can't BFS be used directly to find shortest paths in weighted graphs?

Because BFS assumes all edges have equal weight; in weighted graphs, edges have different costs, so BFS might not find the true shortest path.

intermediate

What is the time complexity of BFS when used to find the shortest path in an unweighted graph?

The time complexity is O(V + E), where V is the number of vertices and E is the number of edges, because each node and edge is processed once.

What data structure does BFS use to explore nodes in an unweighted graph?

A Queue

B Stack

C Priority Queue

D Hash Map

BFS uses a queue to explore nodes level by level.

When BFS first visits a node, what does it guarantee about the path to that node?

A It is the heaviest path

B It is the longest path

C It is the shortest path

D It is a random path

BFS finds the shortest path in an unweighted graph because it explores neighbors level by level.

How do you reconstruct the shortest path after BFS completes?

A By using a stack to store visited nodes

B By following parent pointers from target to start

C By sorting nodes by distance

D By running DFS again

Parent pointers let you backtrack from the target node to the start node to get the path.

Which of these graphs is BFS NOT suitable for finding shortest paths?

A Weighted graph with different edge costs

B Unweighted graph

C Graph with equal edge weights

D Undirected graph

BFS does not handle weighted graphs with different edge costs correctly for shortest paths.

What is the time complexity of BFS in terms of vertices (V) and edges (E)?

A O(V^2)

B O(log V)

C O(E^2)

D O(V + E)

BFS visits each vertex and edge once, so the complexity is O(V + E).

Explain how BFS finds the shortest path in an unweighted graph and how to reconstruct that path.

Think about how BFS visits nodes and how parents help build the path.

Why is BFS not suitable for shortest path in weighted graphs and what algorithm should be used instead?

Consider edge weights and how BFS treats them.

Practice

1. What does BFS (Breadth-First Search) help find in an unweighted graph?

easy

A. The node with the highest degree

B. The longest path in the graph

C. The shortest path from a start node to all other nodes

D. The minimum spanning tree

Solution

1. Step 1: Understand BFS traversal

   BFS explores nodes level by level starting from the source node, visiting all neighbors before moving deeper.

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

   Because BFS visits nodes in order of increasing distance, the first time it reaches a node is via the shortest path.

3. Final Answer:

   The shortest path from a start node to all other nodes -> Option C

4. Quick Check:

   BFS = Shortest path in unweighted graph [OK]

Hint: BFS visits nodes by distance layers to find shortest paths [OK]

Common Mistakes:

• Confusing BFS with DFS for shortest path
• Thinking BFS finds longest paths
• Mixing BFS with minimum spanning tree algorithms

2. Which Go data structure is best to implement the queue needed for BFS?

easy

A. map

B. channel

C. struct

D. slice

Solution

1. Step 1: Identify queue requirements in BFS

   BFS requires a FIFO (first-in, first-out) structure to process nodes in order.

2. Step 2: Use slice as queue in Go

   Slices can be used to append at the end and remove from the front, effectively acting as a queue.

3. Final Answer:

   slice -> Option D

4. Quick Check:

   Queue for BFS = slice [OK]

Hint: Use slice with append and slicing for queue in Go [OK]

Common Mistakes:

• Using map which is unordered
• Using struct without queue logic
• Using channel without concurrency context

3. What is the output distance array after running BFS from node 0 on this graph?

graph := [][]int{
  {1, 2},
  {0, 3},
  {0, 3},
  {1, 2, 4},
  {3},
}

medium

A. [0, 1, 1, 2, 3]

B. [0, 1, 2, 3, 4]

C. [0, 1, 1, 1, 2]

D. [0, 2, 2, 3, 4]

Solution

1. Step 1: Initialize distances and start BFS from node 0

   Distance[0] = 0, others = -1 (unvisited). Queue starts with 0.

2. Step 2: Visit neighbors level by level

   From 0: neighbors 1 and 2 get distance 1. From 1 and 2: neighbor 3 gets distance 2. From 3: neighbor 4 gets distance 3.

3. Final Answer:

   [0, 1, 1, 2, 3] -> Option A

4. Quick Check:

   Distances = [0,1,1,2,3] [OK]

Hint: Track BFS layers to assign distances incrementally [OK]

Common Mistakes:

• Assigning wrong distances to neighbors
• Confusing node indices
• Not initializing distances properly

4. Identify the bug in this BFS code snippet for shortest path:

func BFS(graph [][]int, start int) []int {
  dist := make([]int, len(graph))
  for i := range dist {
    dist[i] = -1
  }
  dist[start] = 0
  queue := []int{start}
  for len(queue) > 0 {
    node := queue[0]
    queue = queue[1:]
    for _, neighbor := range graph[node] {
      if dist[neighbor] == 0 {
        dist[neighbor] = dist[node] + 1
        queue = append(queue, neighbor)
      }
    }
  }
  return dist
}

medium

A. dist[neighbor] == 0 should be dist[neighbor] == -1

B. queue initialization is wrong

C. dist[start] should be -1

D. for loop over neighbors is incorrect

Solution

1. Step 1: Check distance initialization and visit condition

   Distances start at -1 for unvisited nodes. The condition to visit neighbors should check for -1, not 0.

2. Step 2: Fix condition to correctly detect unvisited neighbors

   Change if dist[neighbor] == 0 to if dist[neighbor] == -1 to avoid skipping nodes.

3. Final Answer:

   dist[neighbor] == 0 should be dist[neighbor] == -1 -> Option A

4. Quick Check:

   Check unvisited with -1, not 0 [OK]

Hint: Use -1 to mark unvisited nodes in BFS distance array [OK]

Common Mistakes:

• Checking visited nodes incorrectly
• Initializing distances with zero
• Modifying queue incorrectly

5. Given a graph with nodes 0 to 5 and edges:

0-1, 0-2, 1-3, 2-3, 3-4, 4-5

What is the shortest path from node 0 to node 5 using BFS?

hard

A. [0, 2, 3, 4, 5]

B. [0, 1, 3, 4, 5]

C. [0, 1, 2, 3, 4, 5]

D. [0, 2, 4, 5]

Solution

1. Step 1: Explore neighbors from node 0

   From 0, neighbors are 1 and 2. Both at distance 1.

2. Step 2: Explore neighbors from 1 and 2

   From 1: neighbor 3; from 2: neighbor 3. Node 3 is at distance 2.

3. Step 3: Explore neighbors from 3, then 4, then 5

   From 3: neighbor 4 (distance 3). From 4: neighbor 5 (distance 4). Path is 0->1->3->4->5 or 0->2->3->4->5.

4. Step 4: Choose shortest path with earliest discovery

   BFS discovers 1 before 2, so path is [0,1,3,4,5].

5. Final Answer:

   [0, 1, 3, 4, 5] -> Option B

6. Quick Check:

   Shortest path BFS = [0,1,3,4,5] [OK]

Hint: BFS finds shortest path by earliest node discovery [OK]

Common Mistakes:

• Choosing longer paths ignoring BFS order
• Skipping nodes in path reconstruction
• Confusing path with distance array