Today I have done leetcode's February LeetCoding Challenge with cpp.

February LeetCoding Challenge 11

Description

Shortest Path with Alternating Colors

You are given an integer n, the number of nodes in a directed graph where the nodes are labeled from 0 to n - 1. Each edge is red or blue in this graph, and there could be self-edges and parallel edges.

You are given two arrays redEdges and blueEdges where:

• redEdges[i] = [ai, bi] indicates that there is a directed red edge from node ai to node bi in the graph, and
• blueEdges[j] = [uj, vj] indicates that there is a directed blue edge from node uj to node vj in the graph.

Return an array answer of length n, where each answer[x] is the length of the shortest path from node 0 to node x such that the edge colors alternate along the path, or -1 if such a path does not exist.

Example 1:

Input: n = 3, redEdges = [[0,1],[1,2]], blueEdges = []
Output: [0,1,-1]

Example 2:

Input: n = 3, redEdges = [[0,1]], blueEdges = [[2,1]]
Output: [0,1,-1]

Constraints:

• 1 <= n <= 100
• 0 <= redEdges.length, blueEdges.length <= 400
• redEdges[i].length == blueEdges[j].length == 2
• 0 <= ai, bi, uj, vj < n

Solution

class Solution {
public:
  vector<int> shortestAlternatingPaths(int n, vector<vector<int>>& redEdges, vector<vector<int>>& blueEdges) {
    vector<vector<int>> neighbors[2];
    neighbors[0].resize(n);
    neighbors[1].resize(n);
    for(const auto &redEdge : redEdges) {
      neighbors[0][redEdge[0]].push_back(redEdge[1]);
    }
    for(const auto &blueEdge : blueEdges) {
      neighbors[1][blueEdge[0]].push_back(blueEdge[1]);
    }
    vector<int> distance[2];
    distance[0].resize(n, INT_MAX);
    distance[1].resize(n, INT_MAX);
    queue<pair<int, int>> q;
    q.push({0, 0});
    q.push({0, 1});
    distance[0][0] = 0;
    distance[1][0] = 0;

    int current = 0;
    while(q.size()) {
      int sz = q.size();
      for(int _ = 0; _ < sz; ++_) {
        auto [pos, color] = q.front();
        q.pop();
        for(auto next : neighbors[!color][pos]) {
          if(distance[!color][next] != INT_MAX) continue;
          distance[!color][next] = current + 1;
          q.push({next, !color});
        }
      }
      current += 1;
    }

    for(int i = 0; i < n; ++i) {
      distance[0][i] = min(distance[0][i], distance[1][i]);
      if(distance[0][i] == INT_MAX) {
        distance[0][i] = -1;
      }
    }

    return distance[0];
  }
};

// Accepted
// 90/90 cases passed (15 ms)
// Your runtime beats 95.27 % of cpp submissions
// Your memory usage beats 89.64 % of cpp submissions (14.7 MB)