1129. Shortest Path with Alternating Colors

Description

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] = [a_i, b_i] indicates that there is a directed red edge from node a_i to node b_i in the graph, and
blueEdges[j] = [u_j, v_j] indicates that there is a directed blue edge from node u_j to node v_j 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 <= a_i, b_i, u_j, v_j < n

Solutions

Solution 1: BFS

The problem is essentially a shortest path problem, which we can consider solving using BFS.

First, we preprocess all the edges, categorizing all the edges by color and storing them in a multi-dimensional array $g$. Where $g[0]$ stores all red edges, and $g[1]$ stores all blue edges.

Next, we define the following data structures or variables:

Queue $q$: used to store the currently searched node and the color of the current edge;
Set $vis$: used to store the nodes that have been searched and the color of the current edge;
Variable $d$: used to represent the current search level, i.e., the distance from the currently searched node to the starting point;
Array $ans$: used to store the shortest distance from each node to the starting point. Initially, we initialize all elements in the $ans$ array to $-1$, indicating that the distance from all nodes to the starting point is unknown.

We first enqueue the starting point $0$ and the color of the starting edge $0$ or $1$, indicating that we start from the starting point and the current edge is red or blue.

Next, we start the BFS search. Each time we take out a node $(i, c)$ from the queue, if the answer of the current node has not been updated, then we update the answer of the current node to the current level $d$, i.e., $ans[i] = d$. Then, we flip the color of the current edge $c$, i.e., if the current edge is red, we change it to blue, and vice versa. We take out all edges corresponding to the color, if the other end node $j$ of the edge has not been searched, then we enqueue it.

After the search is over, return the answer array.

The time complexity is $O(n + m)$, and the space complexity is $O(n + m)$. Here, $n$ and $m$ are the number of nodes and edges, respectively.

Python3JavaC++GoTypeScript

class Solution:
    def shortestAlternatingPaths(
        self, n: int, redEdges: List[List[int]], blueEdges: List[List[int]]
    ) -> List[int]:
        g = [defaultdict(list), defaultdict(list)]
        for i, j in redEdges:
            g[0][i].append(j)
        for i, j in blueEdges:
            g[1][i].append(j)
        ans = [-1] * n
        vis = set()
        q = deque([(0, 0), (0, 1)])
        d = 0
        while q:
            for _ in range(len(q)):
                i, c = q.popleft()
                if ans[i] == -1:
                    ans[i] = d
                vis.add((i, c))
                c ^= 1
                for j in g[c][i]:
                    if (j, c) not in vis:
                        q.append((j, c))
            d += 1
        return ans

class Solution {
    public int[] shortestAlternatingPaths(int n, int[][] redEdges, int[][] blueEdges) {
        List<Integer>[][] g = new List[2][n];
        for (var f : g) {
            Arrays.setAll(f, k -> new ArrayList<>());
        }
        for (var e : redEdges) {
            g[0][e[0]].add(e[1]);
        }
        for (var e : blueEdges) {
            g[1][e[0]].add(e[1]);
        }
        Deque<int[]> q = new ArrayDeque<>();
        q.offer(new int[] {0, 0});
        q.offer(new int[] {0, 1});
        boolean[][] vis = new boolean[n][2];
        int[] ans = new int[n];
        Arrays.fill(ans, -1);
        int d = 0;
        while (!q.isEmpty()) {
            for (int k = q.size(); k > 0; --k) {
                var p = q.poll();
                int i = p[0], c = p[1];
                if (ans[i] == -1) {
                    ans[i] = d;
                }
                vis[i][c] = true;
                c ^= 1;
                for (int j : g[c][i]) {
                    if (!vis[j][c]) {
                        q.offer(new int[] {j, c});
                    }
                }
            }
            ++d;
        }
        return ans;
    }
}

class Solution {
public:
    vector<int> shortestAlternatingPaths(int n, vector<vector<int>>& redEdges, vector<vector<int>>& blueEdges) {
        vector<vector<vector<int>>> g(2, vector<vector<int>>(n));
        for (auto& e : redEdges) {
            g[0][e[0]].push_back(e[1]);
        }
        for (auto& e : blueEdges) {
            g[1][e[0]].push_back(e[1]);
        }
        queue<pair<int, int>> q;
        q.emplace(0, 0);
        q.emplace(0, 1);
        bool vis[n][2];
        memset(vis, false, sizeof vis);
        vector<int> ans(n, -1);
        int d = 0;
        while (!q.empty()) {
            for (int k = q.size(); k; --k) {
                auto [i, c] = q.front();
                q.pop();
                if (ans[i] == -1) {
                    ans[i] = d;
                }
                vis[i][c] = true;
                c ^= 1;
                for (int& j : g[c][i]) {
                    if (!vis[j][c]) {
                        q.emplace(j, c);
                    }
                }
            }
            ++d;
        }
        return ans;
    }
};

func shortestAlternatingPaths(n int, redEdges [][]int, blueEdges [][]int) []int {
    g := [2][][]int{}
    for i := range g {
        g[i] = make([][]int, n)
    }
    for _, e := range redEdges {
        g[0][e[0]] = append(g[0][e[0]], e[1])
    }
    for _, e := range blueEdges {
        g[1][e[0]] = append(g[1][e[0]], e[1])
    }
    type pair struct{ i, c int }
    q := []pair{pair{0, 0}, pair{0, 1}}
    ans := make([]int, n)
    vis := make([][2]bool, n)
    for i := range ans {
        ans[i] = -1
    }
    d := 0
    for len(q) > 0 {
        for k := len(q); k > 0; k-- {
            p := q[0]
            q = q[1:]
            i, c := p.i, p.c
            if ans[i] == -1 {
                ans[i] = d
            }
            vis[i][c] = true
            c ^= 1
            for _, j := range g[c][i] {
                if !vis[j][c] {
                    q = append(q, pair{j, c})
                }
            }
        }
        d++
    }
    return ans
}

function shortestAlternatingPaths(
    n: number,
    redEdges: number[][],
    blueEdges: number[][],
): number[] {
    const g: [Graph, Graph] = [{}, {}];
    const ans = Array(n).fill(-1);
    const vis = Array.from({ length: n }, () => Array.from({ length: 2 }, () => false));
    let q: Vertex[] = [
        [0, 0],
        [0, 1],
    ];
    vis[0][0] = vis[0][1] = true;
    let d = 0;
    for (const [i, j] of redEdges) {
        (g[0][i] ??= []).push(j);
    }
    for (const [i, j] of blueEdges) {
        (g[1][i] ??= []).push(j);
    }
    while (q.length) {
        const qNext: Vertex[] = [];
        for (let [i, color] of q) {
            if (ans[i] === -1) {
                ans[i] = d;
            }
            color ^= 1;
            for (const j of g[color][i] ?? []) {
                if (!vis[j][color]) {
                    vis[j][color] = true;
                    qNext.push([j, color as Color]);
                }
            }
        }
        q = qNext;
        d++;
    }
    return ans;
}

type Graph = Record<number, number[]>;
type Color = 0 | 1;
type Vertex = [number, Color];

1129. Shortest Path with Alternating Colors

Description

Solutions

Solution 1: BFS

Comments