How do you solve Shortest Distance After Road Addition Queries II on LeetCode?

Shortest Distance After Road Addition Queries II (LeetCode #3244) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to efficiently update paths in a graph is crucial for optimizing solutions.

What is the brute force solution for Shortest Distance After Road Addition Queries II?

The brute force approach for Shortest Distance After Road Addition Queries II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach calculates the shortest path from city 0 to city n-1 after each query by performing a breadth-first search (BFS) or depth-first search (DFS). This method is straightforward but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Shortest Distance After Road Addition Queries II?

Shortest Distance After Road Addition Queries II uses the following concepts: Array, Greedy, Graph Theory, Ordered Set. The recommended patterns to study are: Dynamic Programming, Graph Theory.

What are the interview tips for Shortest Distance After Road Addition Queries II?

Always explain your thought process while coding; it helps interviewers understand your approach. Consider edge cases, such as when no queries are provided or when all queries lead to the same destination. Practice explaining the difference between brute force and optimal solutions clearly.

What are common mistakes when solving Shortest Distance After Road Addition Queries II?

Failing to properly initialize the distance array, leading to incorrect results. Not considering the implications of multiple updates on the same destination city.

#3244

Shortest Distance After Road Addition Queries II

Hard

ArrayGreedyGraph TheoryOrdered SetDynamic ProgrammingGraph Theory

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a dynamic programming technique to maintain the shortest distances from city 0 to all other cities. By updating only the necessary distances after each query, we achieve a more efficient solution.

⚙️

Algorithm

3 steps

1Step 1: Initialize an array to store the shortest distances from city 0 to all cities, starting with distances to direct neighbors.
2Step 2: For each query, update the distance for the destination city if the new road provides a shorter path.
3Step 3: After processing each query, store the distance to city n-1 in the results.

solution.py9 lines

1def shortest_distance_optimal(n, queries):
2    dist = [float('inf')] * n
3    dist[0] = 0
4    results = []
5    for u, v in queries:
6        if dist[u] + 1 < dist[v]:
7            dist[v] = dist[u] + 1
8        results.append(dist[n - 1])
9    return results

ℹ

Complexity note: The time complexity is linear because we only process each query once and update the distances in constant time. The space complexity is linear due to the distance array.

1Understanding how to efficiently update paths in a graph is crucial for optimizing solutions.
2Dynamic programming can significantly reduce the time complexity when dealing with cumulative updates.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.