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

Time Complexity: Shortest Path in Unweighted Graph Using BFS

O(n + m)

Understanding Time Complexity

We want to understand 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.

from collections import deque

def shortest_path_bfs(graph, start, end):
    queue = deque([start])
    visited = {start}
    distance = {start: 0}

    while queue:
        node = queue.popleft()
        if node == end:
            return distance[node]
        for neighbor in graph[node]:
            if neighbor not in visited:
                visited.add(neighbor)
                distance[neighbor] = distance[node] + 1
                queue.append(neighbor)
    return -1

This code finds the shortest path length between two 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.
• How many times: Each node is visited once, and each edge is checked once.

How Execution Grows With Input

As the graph grows, the time to visit nodes and edges grows roughly with the number of nodes plus edges.

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 work grows linearly with the number of nodes and edges combined.

Final Time Complexity

Time Complexity: O(n + m)

This means the time grows linearly with the total 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, avoiding quadratic time.

Interview Connect

Understanding BFS time complexity helps you explain how graph searches scale and why BFS is efficient for shortest paths in unweighted graphs.

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 shortest path from a start node to all other nodes

B. The longest path in the graph

C. The node with the highest degree

D. The cycle 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 distance, it finds the shortest path to each node.

3. Final Answer:

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

4. Quick Check:

   BFS = Shortest path in unweighted graph [OK]

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

Common Mistakes:

• Confusing BFS with DFS for shortest path
• Thinking BFS finds longest path
• Assuming BFS works only for weighted graphs

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

easy

A. list with append() and pop(0)

B. collections.deque for efficient pops from front

C. set to store unique nodes

D. dictionary to map nodes to distances

Solution

1. Step 1: Identify queue operations in BFS

   BFS requires adding to the back and removing from the front efficiently.

2. Step 2: Choose data structure with O(1) pops from front

   collections.deque supports fast append and popleft operations, unlike list pop(0) which is slow.

3. Final Answer:

   collections.deque for efficient pops from front -> Option B

4. Quick Check:

   Use deque for BFS queue [OK]

Hint: Use deque for fast queue operations in BFS [OK]

Common Mistakes:

• Using list pop(0) causing slow performance
• Using set which is unordered
• Using dict which is not a queue

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

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

medium

A. [0, 2, 2, 3]

B. [1, 0, 1, 2]

C. [0, 1, 1, 2]

D. [0, 1, 2, 3]

Solution

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

   Distance to 0 is 0, others are infinity initially.

2. Step 2: Visit neighbors level-wise

   Neighbors of 0 are 1 and 2, distance 1 each. Then neighbor of 1 is 3, distance 2.

3. Final Answer:

   [0, 1, 1, 2] -> Option C

4. Quick Check:

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

Hint: Distance increments by 1 per BFS level [OK]

Common Mistakes:

• Mixing node indices with distances
• Assigning wrong distances to neighbors
• Forgetting to update distances only once

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

def bfs(graph, start):
    from collections import deque
    queue = deque([start])
    distances = {start: 0}
    while queue:
        node = queue.pop()
        for neighbor in graph[node]:
            if neighbor not in distances:
                distances[neighbor] = distances[node] + 1
                queue.append(neighbor)
    return distances

medium

A. Using queue.pop() instead of queue.popleft() causes wrong order

B. Distances dictionary should be a list

C. Neighbors should be checked before popping from queue

D. Start node distance should be -1

Solution

1. Step 1: Check queue operations in BFS

   BFS requires removing from front (FIFO), but pop() removes from end (LIFO).

2. Step 2: Identify correct method to remove front element

   queue.popleft() removes from front, preserving BFS order.

3. Final Answer:

   Using queue.pop() instead of queue.popleft() causes wrong order -> Option A

4. Quick Check:

   Use popleft() for BFS queue [OK]

Hint: Use popleft() to dequeue in BFS, not pop() [OK]

Common Mistakes:

• Using pop() causing DFS-like behavior
• Confusing list and dict for distances
• Incorrect initialization of distances

5. Given a grid where 0 represents open cell and 1 represents blocked cell, how can BFS be used to find the shortest path from top-left to bottom-right?

grid = [
  [0, 0, 1],
  [1, 0, 1],
  [1, 0, 0]
]

hard

A. Use BFS only on rows, ignoring columns

B. Run DFS to explore all paths and pick shortest

C. Use BFS but ignore blocked cells as neighbors

D. Treat each cell as a node, edges connect adjacent open cells; run BFS from start to find shortest path

Solution

1. Step 1: Model grid as graph nodes and edges

   Each open cell is a node; edges connect up/down/left/right neighbors if open.

2. Step 2: Run BFS from top-left to bottom-right

   BFS finds shortest path in unweighted graph; blocked cells are excluded from edges.

3. Final Answer:

   Treat each cell as a node, edges connect adjacent open cells; run BFS from start to find shortest path -> Option D

4. Quick Check:

   Grid BFS = shortest path via open neighbors [OK]

Hint: Model grid as graph, BFS finds shortest path avoiding blocks [OK]

Common Mistakes:

• Using DFS which is not shortest path
• Ignoring blocked cells incorrectly
• Not connecting neighbors properly