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3243. Shortest Distance After Road Addition Queries I

MediumOpen on LeetCodeProblem statement

Problem Statement

3243. Shortest Distance After Road Addition Queries I

Medium

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:

Java

Source file
class Solution {
    public int[] shortestDistanceAfterQueries(int n, int[][] queries) {
        int[] answer = new int[queries.length];
        List<Integer>[] adj = new List[n];
        for (int i = 0; i < n; i++) {
            adj[i] = new ArrayList<>();
            adj[i].add(i + 1);
        }
        adj[n - 1].clear();
        for (int i = 0; i < queries.length; i++) {
            int[] q = queries[i];
            adj[q[0]].add(q[1]);
            answer[i] = bfs(n, adj);
        }
        return answer;
    }

    private int bfs(int n, List<Integer>[] adj) {
        Queue<Integer> q = new LinkedList<>();
        boolean vis[] = new boolean[n];
        q.add(0);
        int nodesAtLevel = 1, pathLen = 0;
        while (!q.isEmpty()) {
            if (nodesAtLevel == 0) {
                nodesAtLevel = q.size();
                pathLen++;
            }
            int node = q.poll();
            if (node == n - 1)
                return pathLen;
            nodesAtLevel--;
            for (int nbr : adj[node]) {
                if (vis[nbr])
                    continue;
                q.offer(nbr);
                vis[nbr] = true;
            }
        }
        return -1;
    }
}