Shortest Distance After Road Addition Queries II - Practice Coding | SlaveCode
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3244. Shortest Distance After Road Addition Queries II
Hard
50 Points
Array
Greedy
Graph
Ordered Set
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.
Examples
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].
3244. Shortest Distance After Road Addition Queries II
Hard
50 Points
Array
Greedy
Graph
Ordered Set
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.
Examples
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].