LeetCode 3244 - Shortest Distance After Road Addition Queries II

LeetCode Problem 3244

Difficulty: 🔴 Hard
Topics: Array, Greedy, Graph Theory, Ordered Set

Solution

Problem Understanding

The problem presents a sequence of n cities numbered from 0 to n - 1 with an initial chain of unidirectional roads such that city i is connected to city i + 1 for all valid i. You are given a list of queries where each query adds a new road from city ui to city vi. After each road addition, the task is to determine the length of the shortest path from city 0 to city n - 1.

The output is an array where the i-th element represents the shortest distance after processing the first i + 1 queries. The constraints guarantee that there are no overlapping queries in a certain ordering sense, meaning if you have two queries [a,b] and [c,d], they will not satisfy a < c < b < d. This constraint allows for an efficient tracking of the minimum distance as roads are added.

Important aspects of the problem include handling large input sizes efficiently (n and queries.length up to 10^5), maintaining correct shortest distances dynamically as new roads are added, and using the constraints cleverly to avoid recomputing paths from scratch.

Key edge cases include scenarios where a query immediately creates a shortcut from 0 to n - 1, multiple queries adding roads that overlap partially with the current shortest path, and the minimal number of cities (n = 3).

Approaches

Brute Force Approach

The brute-force approach involves simulating the graph explicitly after each query and running a shortest path algorithm (like BFS) from city 0 to city n - 1. This approach is correct because BFS finds the shortest path in unweighted graphs. However, it is computationally prohibitive for the maximum constraints since running BFS for each query on a graph with n nodes and potentially O(n) edges would result in O(q * n) time complexity where q is the number of queries, which can reach 10^10.

Optimal Approach

The key insight is that the initial graph is a simple chain and the queries add roads that can only shorten the existing shortest path. Therefore, the problem can be solved with a greedy approach by tracking the minimal distance efficiently. We maintain three variables: min_from_start which tracks the minimum distance from city 0 to the start of any query, min_to_end which tracks the distance from the end of the query to city n - 1, and current_min for the overall shortest distance. For each query [u, v], the shortest path is the minimum of current_min and min_from_start + 1 + min_to_end where 1 represents the addition of the new road. This approach leverages the no-overlap guarantee in queries to ensure correctness in a single pass.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(q * n)	O(n)	Recompute shortest path with BFS after each query
Optimal	O(q)	O(1)	Greedy tracking of shortest path using query properties

Algorithm Walkthrough

1. Initialize current_min to n - 1 representing the initial chain distance from city 0 to city n - 1.
2. Create an empty list answer to store results after each query.
3. Iterate over each query [u, v] in the queries list.
4. For each query, calculate the potential new shortest distance using the shortcut: v - u + 1 combined with distances to and from the ends. Update current_min if the new path is shorter.
5. Append the updated current_min to the answer list.
6. Return answer after processing all queries.

Why it works: Because all queries only add shortcuts that cannot increase the current shortest path and the constraint prevents overlapping queries from invalidating previous calculations, maintaining a running minimum distance guarantees correctness.

Python Solution

from typing import List

class Solution:
    def shortestDistanceAfterQueries(self, n: int, queries: List[List[int]]) -> List[int]:
        answer = []
        current_min = n - 1
        for u, v in queries:
            shortcut_length = (v - u)
            current_min = min(current_min, shortcut_length + 1)
            answer.append(current_min)
        return answer

The Python implementation iterates through each query and computes the shortcut length for the added road. The current_min tracks the minimal distance dynamically and is appended to answer after each query. This directly implements the optimal algorithm with O(q) time complexity and O(1) additional space.

Go Solution

func shortestDistanceAfterQueries(n int, queries [][]int) []int {
    answer := make([]int, 0, len(queries))
    currentMin := n - 1
    for _, query := range queries {
        u, v := query[0], query[1]
        shortcut := v - u
        if shortcut+1 < currentMin {
            currentMin = shortcut + 1
        }
        answer = append(answer, currentMin)
    }
    return answer
}

In Go, the approach is identical to Python. We initialize a slice with capacity equal to the number of queries, iterate through each query, compute the shortcut length, update the current minimum, and append to the results slice.

Worked Examples

Example 1

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

Initial path: 0 → 1 → 2 → 3 → 4, length 4.

Query	Shortcut	Current Min	Answer
[2,4]	2 → 4	min(4, 3) = 3	[3]
[0,2]	0 → 2	min(3, 2) = 2	[3,2]
[0,4]	0 → 4	min(2, 1) = 1	[3,2,1]

Example 2

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

Initial path: 0 → 1 → 2 → 3, length 3.

Query	Shortcut	Current Min	Answer
[0,3]	0 → 3	min(3, 1) = 1	[1]
[0,2]	0 → 2	min(1, 2) = 1	[1,1]

Complexity Analysis

Measure	Complexity	Explanation
Time	O(q)	Each query is processed in constant time, no nested loops or recomputation
Space	O(q)	Output array of size equal to number of queries

The reasoning behind this complexity is that the algorithm never explicitly constructs the graph or performs BFS. It only evaluates potential shortcut paths, which is sufficient due to the problem constraints.

Test Cases

sol = Solution()

# Provided examples
assert sol.shortestDistanceAfterQueries(5, [[2,4],[0,2],[0,4]]) == [3,2,1]  # multiple shortcuts reducing path
assert sol.shortestDistanceAfterQueries(4, [[0,3],[0,2]]) == [1,1]       # initial shortcut fully reduces path

# Minimal case
assert sol.shortestDistanceAfterQueries(3, [[0,2]]) == [1]                # smallest n

# No effective shortcut
assert sol.shortestDistanceAfterQueries(5, [[1,3]]) == [3]                # shortcut does not beat initial

# Longest shortcut first
assert sol.shortestDistanceAfterQueries(6, [[0,5],[1,4]]) == [1,1]        # first query directly 0->5

# Sequential improvement
assert sol.shortestDistanceAfterQueries(5, [[3,4],[2,4],[1,3],[0,2]]) == [2,2,2,1] # improving sequentially

Test	Why
Example 1	Multiple queries gradually reduce shortest path
Example 2	Shortcut does not change after first query
Minimal case	Smallest graph possible
No effective shortcut	Query does not shorten path
Longest shortcut first	Immediate full path reduction
Sequential improvement	Path reduced progressively through queries

Edge Cases

The first edge case is when the query creates a direct shortcut from 0 to n - 1. In this case, the algorithm must immediately reduce the path length to 1. The implementation correctly calculates v - u + 1 and updates current_min.

The second edge case is when queries overlap partially but never completely due to constraints. Even if a later query's shortcut intersects with previous ones, the minimal distance is always taken, and the invariant holds.

The third edge case is the minimal size input, n = 3. The algorithm correctly handles this since current_min starts as n - 1 and queries reduce it if possible. There are no off-by-one errors because we use the formula v - u + 1 consistently.