3243. Shortest Distance After Road Addition Queries I

Description

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.

 

Solutions

Solution: Dynamic Programming

• Time complexity: O(queries.length*n²)
• Space complexity: O(queries.length*n)

 

JavaScript

/**
 * @param {number} n
 * @param {number[][]} queries
 * @return {number[]}
 */
const shortestDistanceAfterQueries = function (n, queries) {
  const graph = Array.from({ length: n }, () => []);
  const dp = Array.from({ length: n }, (_, index) => index);

  return queries.map(([from, to]) => {
    graph[to].push(from);

    for (let node = to; node < n; node++) {
      dp[node] = Math.min(dp[node], dp[node - 1] + 1);

      for (const fromNode of graph[node]) {
        dp[node] = Math.min(dp[node], dp[fromNode] + 1);
      }
    }
    return dp[n - 1];
  });
};