Shortest Path in Unweighted Graph Using BFS in DSA Go - 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.

func shortestPathBFS(graph map[int][]int, start, target int) int {
    visited := make(map[int]bool)
    queue := []int{start}
    distance := make(map[int]int)
    distance[start] = 0
    visited[start] = true

    for len(queue) > 0 {
        node := queue[0]
        queue = queue[1:]
        if node == target {
            return distance[node]
        }
        for _, neighbor := range graph[node] {
            if !visited[neighbor] {
                visited[neighbor] = true
                distance[neighbor] = distance[node] + 1
                queue = append(queue, neighbor)
            }
        }
    }
    return -1
}
    

This code finds the shortest path length from start to target 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.
• How many times: Each node and each edge is processed at most once.

How Execution Grows With Input

As the graph grows, the algorithm visits more nodes and edges.

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

Pattern observation: The operations grow roughly with the sum of nodes and edges.

Final Time Complexity

Time Complexity: O(n + m)

This means the time grows linearly with the number of nodes plus 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 repeat work for the same parts, making it linear in nodes and edges.

Interview Connect

Understanding BFS time complexity helps you explain efficient graph traversal clearly and confidently in interviews.

Self-Check

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

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