3243. Shortest Distance After Road Addition Queries I

Description

You are given an integer n and a 2D integer array queries.

There are n 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.

queries[i] = [u_i, v_i] represents the addition of a new unidirectional road from city u_i to city v_i. After each query, you need to find the length of the shortest path from city 0 to city n - 1.

Return an array answer where for each i in the range [0, queries.length - 1], answer[i] is the length of the shortest path from city 0 to city n - 1 after processing the first i + 1 queries.

Example 1:

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

Output: [3,2,1]

Explanation:

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]

Explanation:

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 <= 500
1 <= queries.length <= 500
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.

Solutions

Solution 1: BFS

We first build a directed graph $\textit{g}$, where $\textit{g}[i]$ represents the list of cities that can be reached from city $i$. Initially, each city $i$ has a one-way road to city $i + 1$.

Then, for each query $[u, v]$, we add $v$ to the list of reachable cities from $u$, and then use BFS to find the shortest path length from city $0$ to city $n - 1$, adding the result to the answer array.

Finally, we return the answer array.

The time complexity is $O(q \times (n + q))$, and the space complexity is $O(n + q)$. Here, $n$ and $q$ are the number of cities and the number of queries, respectively.

Python3JavaC++GoTypeScript

class Solution:
    def shortestDistanceAfterQueries(
        self, n: int, queries: List[List[int]]
    ) -> List[int]:
        def bfs(i: int) -> int:
            q = deque([i])
            vis = [False] * n
            vis[i] = True
            d = 0
            while 1:
                for _ in range(len(q)):
                    u = q.popleft()
                    if u == n - 1:
                        return d
                    for v in g[u]:
                        if not vis[v]:
                            vis[v] = True
                            q.append(v)
                d += 1

        g = [[i + 1] for i in range(n - 1)]
        ans = []
        for u, v in queries:
            g[u].append(v)
            ans.append(bfs(0))
        return ans

class Solution {
    private List<Integer>[] g;
    private int n;

    public int[] shortestDistanceAfterQueries(int n, int[][] queries) {
        this.n = n;
        g = new List[n];
        Arrays.setAll(g, i -> new ArrayList<>());
        for (int i = 0; i < n - 1; ++i) {
            g[i].add(i + 1);
        }
        int m = queries.length;
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            int u = queries[i][0], v = queries[i][1];
            g[u].add(v);
            ans[i] = bfs(0);
        }
        return ans;
    }

    private int bfs(int i) {
        Deque<Integer> q = new ArrayDeque<>();
        q.offer(i);
        boolean[] vis = new boolean[n];
        vis[i] = true;
        for (int d = 0;; ++d) {
            for (int k = q.size(); k > 0; --k) {
                int u = q.poll();
                if (u == n - 1) {
                    return d;
                }
                for (int v : g[u]) {
                    if (!vis[v]) {
                        vis[v] = true;
                        q.offer(v);
                    }
                }
            }
        }
    }
}