Shortest Path in Unweighted Graph Using BFS in DSA Python - Build from Scratch

Shortest Path in Unweighted Graph Using BFS

📖 Scenario: You are helping a delivery robot find the shortest route between locations in a small city. The city map is shown as connections between places.Each connection is a two-way street, and all streets take the same time to cross.

🎯 Goal: Build a program that finds the shortest path from a starting location to a destination using Breadth-First Search (BFS) on an unweighted graph.The program will print the shortest path as a list of places visited in order.

📋 What You'll Learn

Create a graph as a dictionary with exact connections between places

Set the start and end locations as variables

Implement BFS to find the shortest path between start and end

Print the shortest path as a list of place names

💡 Why This Matters

🌍 Real World

Finding shortest routes in maps, social networks, or any system where connections have equal cost.

💼 Career

Understanding BFS and shortest path algorithms is essential for roles in software development, data analysis, and network engineering.

1

Create the city map graph

Create a dictionary called graph representing the city map with these exact connections:
'A': ['B', 'C'], 'B': ['A', 'D', 'E'], 'C': ['A', 'F'], 'D': ['B'], 'E': ['B', 'F'], 'F': ['C', 'E']

DSA Python

# Your code here

Hint

Use a dictionary where keys are place names and values are lists of connected places.

2

Set start and end locations

Create two variables: start set to 'A' and end set to 'F'

DSA Python

graph = {
    'A': ['B', 'C'],
    'B': ['A', 'D', 'E'],
    'C': ['A', 'F'],
    'D': ['B'],
    'E': ['B', 'F'],
    'F': ['C', 'E']
}
# Your code here

Hint

Just assign the strings 'A' and 'F' to variables named start and end.

3

Implement BFS to find shortest path

Write a function called bfs_shortest_path(graph, start, end) that uses a queue to find the shortest path from start to end. Use a list as a queue and keep track of visited nodes and paths. Return the path as a list of places.

DSA Python

graph = {
    'A': ['B', 'C'],
    'B': ['A', 'D', 'E'],
    'C': ['A', 'F'],
    'D': ['B'],
    'E': ['B', 'F'],
    'F': ['C', 'E']
}
start = 'A'
end = 'F'

# Your code here

Hint

Use a queue to store paths. Pop the first path, check last node. If end, return path. Else add neighbors to queue with new paths.

4

Print the shortest path

Call bfs_shortest_path(graph, start, end) and print the returned path.

DSA Python

graph = {
    'A': ['B', 'C'],
    'B': ['A', 'D', 'E'],
    'C': ['A', 'F'],
    'D': ['B'],
    'E': ['B', 'F'],
    'F': ['C', 'E']
}
start = 'A'
end = 'F'

def bfs_shortest_path(graph, start, end):
    queue = [[start]]
    visited = set()
    while queue:
        path = queue.pop(0)
        node = path[-1]
        if node == end:
            return path
        if node not in visited:
            visited.add(node)
            for neighbor in graph[node]:
                new_path = list(path)
                new_path.append(neighbor)
                queue.append(new_path)
    return []

# Your code here

Hint

Call the function with graph, start, and end. Print the result.

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