LeetCode-in-Java

3244. Shortest Distance After Road Addition Queries II

Hard

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

public class Solution {
    public int[] shortestDistanceAfterQueries(int n, int[][] queries) {
        int[] flag = new int[n];
        int[] res = new int[queries.length];
        for (int i = 0; i < n; i++) {
            flag[i] = i + 1;
        }
        for (int k = 0; k < queries.length; k++) {
            int[] query = queries[k];
            int preRes = k == 0 ? (n - 1) : res[k - 1];
            if (flag[query[0]] >= query[1]) {
                res[k] = preRes;
                continue;
            }
            int subDis = 0;
            int curr = query[0];
            while (curr < query[1]) {
                int next = flag[curr];
                subDis += 1;
                flag[curr] = query[1];
                curr = next;
            }
            res[k] = preRes + 1 - subDis;
        }
        return res;
    }
}

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