1129 Shortest Path with Alternating Colors

Question

My Solution:

class Solution {
    public int[] shortestAlternatingPaths(int n, int[][] redEdges, int[][] blueEdges) {
        int[][] g = new int[n][n];
        buildGraph(g, n, redEdges, blueEdges);
        
        Queue<int[]> q = new LinkedList<>();
        q.offer(new int[]{0, 1});
        q.offer(new int[]{0, -1});
        
        int len = 0;
        int[] res = new int[n];
        Arrays.fill(res, Integer.MAX_VALUE);
        res[0] = 0;
        
        Set<String> visited = new HashSet<>();
        while(!q.isEmpty()){
            int size = q.size();
            len++;
            for(int i = 0; i < size; i++){
                int[] cur = q.poll();
                int node = cur[0];
                int color = cur[1];
                int oppoColor = -color;
                
                for(int j = 1; j<n; j++){
                    if(g[node][j] == oppoColor
                      || g[node][j] == 0){
                        if(!visited.add(j+""+oppoColor)){
                            continue;
                        }
                        q.offer(new int[]{j, oppoColor});
                        res[j] = Math.min(res[j], len);
                    }
                }
            }
        }
        
        for( int i = 1; i < n;i++){
            if(res[i] == Integer.MAX_VALUE){
                res[i] = -1;
            }
        }
        
        return res;
    }
    
    
    // 1 red
    // -1 blue
    // 0 both
    private void buildGraph(int[][] g, int n, int[][] red_edges, int[][] blue_edges){
        for(int i =0; i < n; i++){
            Arrays.fill(g[i], -n); // initial array with n
            
            for(int[] e: red_edges){
                int from = e[0];
                int to = e[1];
                g[from][to] = 1;
            }
            
            for(int[] e: blue_edges){
                int from = e[0];
                int to = e[1];
                
                if(g[from][to] == 1){
                    g[from][to] = 0;
                }else{
                    g[from][to] = -1;
                }
            }
        }
    }
}