3244. Shortest Distance After Road Addition Queries II

Problem Link

Description

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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.

There are no two queries such that queries[i][0] < queries[j][0] < queries[i][1] < queries[j][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 <= 105
• 1 <= queries.length <= 105
• 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.
• There are no two queries such that i != j and queries[i][0] < queries[j][0] < queries[i][1] < queries[j][1].

Solution

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C++

class Solution {
public:
    vector<int> shortestDistanceAfterQueries(int n, vector<vector<int>>& queries) {
        set<int> s;
        vector<int> res;
        for (int x = 0; x < n; x++) s.insert(x);
 
        for (auto& query: queries) {
            int u = query[0], v = query[1];
            auto it = s.upper_bound(u);
            auto it2 = s.lower_bound(v);
            s.erase(it, it2);
            res.push_back(s.size() - 1);
        }
 
        return res;
    }
};

Python3

class Solution:
    def shortestDistanceAfterQueries(self, n: int, queries: List[List[int]]) -> List[int]:
        d = {i: i + 1 for i in range(n - 1)}
        res = []
 
        for i, j in queries:
            if i in d and d[i] < j:
                v = i
                while v < j:
                    v = d.pop(v)
                d[i] = j
                
            res.append(len(d))
 
        return res