3243 - Shortest Distance After Road Addition Queries I

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.

C++

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;
    }
};

PYTHON

from collections import deque

class Solution:
    def shortestDistanceAfterQueries(self, n, queries):
        g = [[] for _ in range(n)]
        for i in range(n - 1):
            g[i].append(i + 1)
        
        def bfs(i):
            q = deque([i])
            vis = [False] * n
            vis[i] = True
            d = 0
            while q:
                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
        
        ans = []
        for q in queries:
            g[q[0]].append(q[1])
            ans.append(bfs(0))
        
        return ans

JAVA

• 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);
                      }
                  }
              }
          }
      }
  }