You are given a positive integer n representing n cities numbered from 1 to n. You are also given a 2D array roads where roads[i] = [ai, bi, distancei] indicates that there is a bidirectional road between cities ai and bi with a distance equal to distancei. The cities graph is not necessarily connected.

The score of a path between two cities is defined as the minimum distance of a road in this path.

Return the minimum possible score of a path between cities 1 and n.

Note:

• A path is a sequence of roads between two cities.
• It is allowed for a path to contain the same road multiple times, and you can visit cities 1 and n multiple times along the path.
• The test cases are generated such that there is at least one path between 1 and n.

Example

Input: n = 4, roads = [[1,2,9],[2,3,6],[2,4,5],[1,4,7]]
Output: 5
Explanation: The path from city 1 to 4 with the minimum score is: 1 -> 2 -> 4. The score of this path is min(9,5) = 5.
It can be shown that no other path has less score.

Solution

Great solution here

/**
 * @param {number} n
 * @param {number[][]} roads
 * @return {number}
 */
var minScore = function (n, roads) {
  //create a graph
  const graph = new Array(n + 1).fill().map(() => []);
  //create a set to keep track of visited nodes
  const visited = new Set([]);

  //add the roads to the graph
  for (const [v1, v2, distance] of roads) {
    //add the distance to the graph
    graph[v1].push([v2, distance]);
    //add the distance to the graph
    graph[v2].push([v1, distance]);
  }
  //start at node 1
  const queue = [1];
  //add node 1 to the visited set
  visited.add(1);
  let ans = Infinity;

  while (queue.length > 0) {
    //remove the first node from the queue
    const node = queue.shift();
    // For each neighbor of the current node
    for (const [next, distance] of graph[node]) {
      //if the next node is the last node, return the answer
      ans = Math.min(ans, distance);
      //if the next node is the last node, return the answer
      if (visited.has(next)) continue;
      //add the next node to the visited set
      visited.add(next);
      //add the next node to the queue
      queue.push(next);
    }
  }
  return ans;
};