#3244

Hard

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Shortest Distance After Road Addition Queries II

The problem involves calculating the shortest path from city 0 to city n-1 after each road addition, leveraging greedy choices.

array greedy graph ordered set

Problem Statement

You are given an integer n representing the number of cities numbered from 0 to n-1. Initially, there is a unidirectional road from city i to city i + 1 for all 0 <= i < n - 1.

For each query [ui, vi], a new unidirectional road from city ui to city vi is added. After each road addition, you need to calculate the shortest path from city 0 to city n-1 and output the result.

Examples

Example 1

Input: n = 5, queries = [[2,4],[0,2],[0,4]]

Output: [3,2,1]

After the addition of the road from 2 to 4, the length of the shortest path from 0 to 4 is 3.

After the addition of the road from 0 to 2, the length of the shortest path from 0 to 4 is 2.

After the addition of the road from 0 to 4, the length of the shortest path from 0 to 4 is 1.

Example 2

Input: n = 4, queries = [[0,3],[0,2]]

Output: [1,1]

After the addition of the road from 0 to 3, the length of the shortest path from 0 to 3 is 1.

After the addition of the road from 0 to 2, the length of the shortest path remains 1.

Constraints

3 <= n <= 105
1 <= queries.length <= 105
queries[i].length == 2
0 <= queries[i][0] < queries[i][1] < n
1 < queries[i][1] - queries[i][0]
There are no repeated roads among the queries.
There are no two queries such that i != j and queries[i][0] < queries[j][0] < queries[i][1] < queries[j][1].

Solution Approach

Graph Representation with Optimized Updates

Represent the cities and roads as a graph, initially having a direct road from each city i to i + 1. As each query is processed, update the graph efficiently and calculate the shortest path using a greedy approach to minimize traversal cost.

Greedy Path Optimization with Invariant Validation

After each road addition, check for the shortest valid path from city 0 to city n-1. Greedily validate the effect of the newly added road by checking how it reduces the distance. Invariant validation ensures that each addition is processed optimally.

Efficient Path Calculation with Priority Queue

Utilize a priority queue to track the shortest distance dynamically as roads are added. This helps efficiently compute the shortest path after each road addition without recomputing all paths from scratch.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution depends on how the graph is updated and how the shortest path is recalculated after each query. Efficient graph traversal using priority queues or Union-Find can reduce the time complexity significantly. Space complexity depends on the number of cities and roads, which may grow dynamically with the queries.

What Interviewers Usually Probe

Check if the candidate can handle graph traversal problems with dynamic updates.
Look for understanding of greedy algorithms and how they can be applied to graph-related problems.
Assess how efficiently the candidate manages graph updates to avoid excessive recomputation.

Common Pitfalls or Variants

Common pitfalls

Not updating the graph efficiently, leading to redundant recalculations of the shortest path.
Forgetting to optimize the process for larger inputs, leading to timeouts.
Misunderstanding the greedy choice strategy, which may result in suboptimal path calculations.

Follow-up variants

Add more roads at once rather than one at a time, requiring batch processing.
Allow roads to be bi-directional, adding complexity to the shortest path calculations.
Introduce more constraints such as varying road lengths, instead of assuming unit length roads.

FAQ

What is the core pattern of the Shortest Distance After Road Addition Queries II problem?

The core pattern is greedy choice optimization combined with invariant validation, where each road addition optimally adjusts the shortest path.

How do I efficiently update the graph in Shortest Distance After Road Addition Queries II?

You can update the graph efficiently using priority queues or dynamic Union-Find data structures, which help manage graph changes after each query.

What is the time complexity of solving this problem?

The time complexity depends on the graph update and shortest path calculation method. Efficient implementations using priority queues or Union-Find can reduce complexity significantly.

What makes this problem different from a standard shortest path problem?

This problem involves dynamically updating the graph after each query, requiring an approach that efficiently recalculates the shortest path without recomputing from scratch.

Can the solution be optimized further for larger inputs?

Yes, using priority queues or advanced graph algorithms can optimize the solution for larger inputs, ensuring the problem is solved within time limits.

terminal

Solution

Solution 1: Greedy + Recording Jump Positions

We define an array $\textit{nxt}$ of length $n - 1$, where $\textit{nxt}[i]$ represents the next city that can be reached from city $i$. Initially, $\textit{nxt}[i] = i + 1$.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

class Solution:
    def shortestDistanceAfterQueries(
        self, n: int, queries: List[List[int]]
    ) -> List[int]:
        nxt = list(range(1, n))
        ans = []
        cnt = n - 1
        for u, v in queries:
            if 0 < nxt[u] < v:
                i = nxt[u]
                while i < v:
                    cnt -= 1
                    nxt[i], i = 0, nxt[i]
                nxt[u] = v
            ans.append(cnt)
        return ans

Continue Practicing

#3534 path existence queries in a graph ii #3607 power grid maintenance #2659 make array empty