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] = [ai, bi] indicates that there is a directed red edge from node ai to node bi in the graph, andblueEdges[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 <= 1000 <= redEdges.length, blueEdges.length <= 400redEdges[i].length == blueEdges[j].length == 20 <= ai, bi, uj, vj < 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.
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| 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
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| 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;
}
}
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| 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;
}
};
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| 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
}
|
