1129. Shortest Path with Alternating Colors

import java.util.*;
public class LC {
    static Scanner sc=new Scanner(System.in);
    public static void main(String[] args) {
int n = 3, redEdges[][] = {{0,1},{1,2}}, blueEdges[][] = {};
int res[]=shortestAlternatingPaths(n,redEdges,blueEdges);
for(int x:res)
    System.out.println(x);

    }
    public static int[] shortestAlternatingPaths(int n, int[][] redEdges, int[][] blueEdges) {
        Map<Integer, List<List<Integer>>> adj = new HashMap<>();
        for (int[] red : redEdges) {
            adj.computeIfAbsent(red[0], k -> new ArrayList<>()).add(
                    Arrays.asList(red[1], 0));//[dest , 0:as red color]
        }

        for (int[] blue : blueEdges) {
            adj.computeIfAbsent(blue[0], k -> new ArrayList<>()).add(
                    Arrays.asList(blue[1], 1));//[dest , 1 as blue ]
        }

        int[] ans = new int[n];
        Arrays.fill(ans, -1);
        boolean[][] visit = new boolean[n][2];//node , color
        Queue<int[]> q = new LinkedList<>();

        // Start with node 0, with number of steps as 0 and undefined color -1.
        q.offer(new int[] { 0, 0, -1 });//node , length, previous edge color
        ans[0] = 0;
        visit[0][1] = visit[0][0] = true;

        while (!q.isEmpty()) {
            int[] cur = q.poll();
            int node = cur[0], length = cur[1], prevColor = cur[2];

            if (!adj.containsKey(node)) {
                continue;
            }
//here we do not store minimum as which path visited first , then its corresponding length is going to minimum
            for (List<Integer> nei : adj.get(node)) {
                int neighbor = nei.get(0);
                int color = nei.get(1);
                if (!visit[neighbor][color] && color != prevColor) {
                    if (ans[neighbor] == -1)
                        ans[neighbor] = 1 + length;
                    visit[neighbor][color] = true;
                    q.offer(new int[] { neighbor, 1 + length, color });
                }
            }
        }
        return ans;
    }

}