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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] = [ui, vi] represents the addition of a new unidirectional road from city ui to city vi. 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

Thinking

The graph starts as the path \(0\to 1\to\cdots\to n-1\); after each forward edge we want the distance from \(0\) to \(n-1\). \(n,q\le 500\), so a fresh shortest path per query is acceptable.

Edges have weight \(1\), so BFS from \(0\) after each insertion records the distance. The total time is \(O(q(n+q))\).

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.

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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

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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);
                    }
                }
            }
        }
    }
}

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class Solution {
public:
    vector<int> shortestDistanceAfterQueries(int n, vector<vector<int>>& queries) {
        vector<int> g[n];
        for (int i = 0; i < n - 1; ++i) {
            g[i].push_back(i + 1);
        }
        auto bfs = [&](int i) -> int {
            queue<int> q{{i}};
            vector<bool> vis(n);
            vis[i] = true;
            for (int d = 0;; ++d) {
                for (int k = q.size(); k; --k) {
                    int u = q.front();
                    q.pop();
                    if (u == n - 1) {
                        return d;
                    }
                    for (int v : g[u]) {
                        if (!vis[v]) {
                            vis[v] = true;
                            q.push(v);
                        }
                    }
                }
            }
        };
        vector<int> ans;
        for (const auto& q : queries) {
            g[q[0]].push_back(q[1]);
            ans.push_back(bfs(0));
        }
        return ans;
    }
};

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func shortestDistanceAfterQueries(n int, queries [][]int) []int {
    g := make([][]int, n)
    for i := range g {
        g[i] = append(g[i], i+1)
    }
    bfs := func(i int) int {
        q := []int{i}
        vis := make([]bool, n)
        vis[i] = true
        for d := 0; ; d++ {
            for k := len(q); k > 0; k-- {
                u := q[0]
                if u == n-1 {
                    return d
                }
                q = q[1:]
                for _, v := range g[u] {
                    if !vis[v] {
                        vis[v] = true
                        q = append(q, v)
                    }
                }
            }
        }
    }
    ans := make([]int, len(queries))
    for i, q := range queries {
        g[q[0]] = append(g[q[0]], q[1])
        ans[i] = bfs(0)
    }
    return ans
}

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function shortestDistanceAfterQueries(n: number, queries: number[][]): number[] {
    const g: number[][] = Array.from({ length: n }, () => []);
    for (let i = 0; i < n - 1; ++i) {
        g[i].push(i + 1);
    }
    const bfs = (i: number): number => {
        const q: number[] = [i];
        const vis: boolean[] = Array(n).fill(false);
        vis[i] = true;
        for (let d = 0; ; ++d) {
            const nq: number[] = [];
            for (const u of q) {
                if (u === n - 1) {
                    return d;
                }
                for (const v of g[u]) {
                    if (!vis[v]) {
                        vis[v] = true;
                        nq.push(v);
                    }
                }
            }
            q.splice(0, q.length, ...nq);
        }
    };
    const ans: number[] = [];
    for (const [u, v] of queries) {
        g[u].push(v);
        ans.push(bfs(0));
    }
    return ans;
}

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