BFS Breadth First Search on Graph in DSA Go - Time & Space Complexity

Time Complexity: BFS Breadth First Search on Graph

O(n + e)

Understanding Time Complexity

We want to understand how the time needed to explore a graph grows as the graph gets bigger using BFS.

Specifically, how does BFS's work change when the graph has more nodes and edges?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func BFS(graph map[int][]int, start int) []int {
    visited := make(map[int]bool)
    queue := []int{start}
    visited[start] = true
    order := []int{}

    for len(queue) > 0 {
        node := queue[0]
        queue = queue[1:]
        order = append(order, node)

        for _, neighbor := range graph[node] {
            if !visited[neighbor] {
                visited[neighbor] = true
                queue = append(queue, neighbor)
            }
        }
    }
    return order
}
    

This code visits all nodes reachable from the start node in a 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 is visited once; each edge is checked once.

How Execution Grows With Input

As the number of nodes (n) and edges (e) grow, BFS visits each node once and checks all edges once.

Input Size (n, e)	Approx. Operations
10 nodes, 15 edges	About 10 node visits + 15 edge checks = 25 operations
100 nodes, 200 edges	About 100 + 200 = 300 operations
1000 nodes, 5000 edges	About 1000 + 5000 = 6000 operations

Pattern observation: Operations grow roughly in proportion to nodes plus edges.

Final Time Complexity

Time Complexity: O(n + e)

This means BFS takes time proportional to the number of nodes plus the number of edges in the graph.

Common Mistake

[X] Wrong: "BFS always 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 do not multiply to n squared.

Interview Connect

Understanding BFS time complexity helps you explain how graph traversal scales and why BFS is efficient for many problems.

Self-Check

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

Practice

1. What is the main purpose of BFS (Breadth First Search) on a graph?

easy

A. To sort nodes in descending order

B. To find the shortest path using depth-first traversal

C. To explore all nodes level by level starting from a source node

D. To find cycles in a graph using recursion

Solution

1. Step 1: Understand BFS traversal

   BFS explores nodes by visiting all neighbors of a node before moving to the next level.

2. Step 2: Compare BFS with other traversals

   Unlike DFS, BFS uses a queue to explore nodes level by level, which helps in shortest path problems.

3. Final Answer:

   To explore all nodes level by level starting from a source node -> Option C

4. Quick Check:

   BFS = Level-wise exploration [OK]

Hint: BFS uses a queue to explore nodes level by level [OK]

Common Mistakes:

• Confusing BFS with DFS
• Thinking BFS uses recursion
• Assuming BFS sorts nodes

2. Which of the following Go code snippets correctly initializes a queue for BFS traversal?

easy

A. queue := []int{} // empty slice as queue

B. queue := make(map[int]bool) // map as queue

C. queue := 0 // integer as queue

D. queue := make(chan int) // channel as queue

Solution

1. Step 1: Identify data structure for BFS queue

   BFS requires a FIFO structure; a slice can be used as a queue in Go.

2. Step 2: Check each option

   queue := []int{} // empty slice as queue uses a slice which can enqueue and dequeue elements. queue := make(map[int]bool) // map as queue is a map, not suitable as queue. queue := 0 // integer as queue is an int, not a queue. queue := make(chan int) // channel as queue is a channel, which can be used but requires goroutines and is complex.

3. Final Answer:

   queue := []int{} // empty slice as queue -> Option A

4. Quick Check:

   Slice as queue = queue := []int{} // empty slice as queue [OK]

Hint: Use slice for queue in BFS, not map or int [OK]

Common Mistakes:

• Using map instead of slice for queue
• Using int variable as queue
• Confusing channels with simple queue

3. What is the output of the following Go BFS traversal code on the graph below?

Graph edges:
0 -> 1, 2
1 -> 2
2 -> 0, 3
3 -> 3

package main
import "fmt"
func bfs(graph map[int][]int, start int) []int {
    visited := make(map[int]bool)
    queue := []int{start}
    order := []int{}
    visited[start] = true
    for len(queue) > 0 {
        node := queue[0]
        queue = queue[1:]
        order = append(order, node)
        for _, neighbor := range graph[node] {
            if !visited[neighbor] {
                visited[neighbor] = true
                queue = append(queue, neighbor)
            }
        }
    }
    return order
}
func main() {
    graph := map[int][]int{
        0: {1, 2},
        1: {2},
        2: {0, 3},
        3: {3},
    }
    fmt.Println(bfs(graph, 2))
}

medium

A. [2, 0, 1, 3]

B. [2, 0, 3, 1]

C. [2, 3, 0, 1]

D. [2, 1, 0, 3]

Solution

1. Step 1: Start BFS from node 2

   Initialize queue=[2], visited={2}. Dequeue 2, enqueue neighbors 0 and 3. order=[2]

2. Step 2: Process nodes in queue

   Queue=[0,3]. Dequeue 0, enqueue neighbor 1 (2 already visited). order=[2,0]. Queue=[3,1]. Dequeue 3, neighbor 3 already visited. order=[2,0,3]. Queue=[1]. Dequeue 1, neighbor 2 visited. order=[2,0,3,1]. Queue empty.

3. Final Answer:

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

4. Quick Check:

   BFS order from 2 = [2,0,3,1] [OK]

Hint: Track queue and visited sets stepwise [OK]

Common Mistakes:

• Not marking visited before enqueue
• Wrong order of neighbors enqueued
• Confusing BFS with DFS order

4. Identify the bug in the following BFS Go code snippet and select the correct fix.

func bfs(graph map[int][]int, start int) []int {
    visited := make(map[int]bool)
    queue := []int{start}
    order := []int{}
    for len(queue) > 0 {
        node := queue[0]
        queue = queue[1:]
        order = append(order, node)
        for _, neighbor := range graph[node] {
            if !visited[neighbor] {
                queue = append(queue, neighbor)
                visited[neighbor] = true
            }
        }
    }
    return order
}

medium

A. Mark start node as visited before loop

B. Initialize order slice with capacity

C. Use a stack instead of queue

D. Remove visited map entirely

Solution

1. Step 1: Check visited marking for start node

   The code never marks the start node as visited before starting, so it may revisit it.

2. Step 2: Fix by marking start visited before loop

   Adding visited[start] = true before the loop prevents revisiting the start node and infinite loops.

3. Final Answer:

   Mark start node as visited before loop -> Option A

4. Quick Check:

   Start visited before loop = Mark start node as visited before loop [OK]

Hint: Always mark start visited before BFS loop [OK]

Common Mistakes:

• Marking visited after enqueue
• Using stack instead of queue
• Ignoring visited map

5. Given a directed graph represented as adjacency list, how can BFS be adapted to find the shortest path length from a start node to a target node? Choose the best approach.

hard

A. Use BFS but enqueue nodes multiple times to find shortest path

B. Use DFS and backtrack to find all paths, then pick shortest

C. Use BFS but ignore visited nodes to explore all paths

D. Use BFS with a distance map to track levels from start, stop when target found

Solution

1. Step 1: Understand BFS for shortest path

   BFS explores nodes level by level, so the first time target is reached is the shortest path.

2. Step 2: Use a distance map to track levels

   Maintain a map to record distance from start for each node. When target is dequeued, return its distance.

3. Final Answer:

   Use BFS with a distance map to track levels from start, stop when target found -> Option D

4. Quick Check:

   BFS + distance map = shortest path [OK]

Hint: Track distance in BFS, stop at target [OK]

Common Mistakes:

• Using DFS for shortest path in unweighted graph
• Ignoring visited nodes causing infinite loops
• Enqueuing nodes multiple times unnecessarily