Dijkstra
Detailed guide and Python implementation for the 'Dijkstra' problem.
1. Concept Overview
The 'Dijkstra' problem is a key challenge in the Graphs section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Dijkstra.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Dijkstra carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Write a function dijkstra(graph, start) that calculates the shortest path from a starting node start to all other nodes in a weighted graph. The graph is represented as an adjacency list of dictionaries, where graph[u][v] is the weight of the directed edge from u to v. Return a dictionary of shortest distances.
- •1 <= V <= 1000
- •0 <= E <= 5000
- •Weights are non-negative integers.
Examples
graph = {0: {1: 4, 2: 1}, 1: {3: 1}, 2: {1: 2, 3: 5}, 3: {}}, start = 0{0: 0, 1: 3, 2: 1, 3: 4}Shortest distance from 0 to 1 is 3 (via 0->2->1).
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
import heapq
def dijkstra_opt(graph, start):
return dijkstra_brute(graph, start)Brute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
import heapq
def dijkstra_brute(graph, start):
distances = {node: float('infinity') for node in graph}
distances[start] = 0
pq = [(0, start)]
while pq:
curr_dist, curr_node = heapq.heappop(pq)
if curr_dist > distances[curr_node]: continue
for neighbor, weight in graph[curr_node].items():
dist = curr_dist + weight
if dist < distances[neighbor]:
distances[neighbor] = dist
heapq.heappush(pq, (dist, neighbor))
return distancesAlgorithm Pattern Checklist
When dealing with Graphs data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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