LeetCode #1129 — MEDIUM

Shortest Path with Alternating Colors

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

3
[[0,1],[1,2]]
[]

Example 2

3
[[0,1]]
[[2,1]]

Step 02

Core Insight

What unlocks the optimal approach

Do a breadth-first search, where the "nodes" are actually (Node, color of last edge taken).

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1129: Shortest Path with Alternating Colors
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;
    }
}

// Accepted solution for LeetCode #1129: Shortest Path with Alternating Colors
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
}

# Accepted solution for LeetCode #1129: Shortest Path with Alternating Colors
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

// Accepted solution for LeetCode #1129: Shortest Path with Alternating Colors
struct Solution;
use std::collections::VecDeque;

impl Solution {
    fn shortest_alternating_paths(
        n: i32,
        red_edges: Vec<Vec<i32>>,
        blue_edges: Vec<Vec<i32>>,
    ) -> Vec<i32> {
        let n = n as usize;
        let mut graph: Vec<Vec<Vec<usize>>> = vec![vec![vec![]; n]; 2];
        let mut visited: Vec<Vec<bool>> = vec![vec![false; n]; 2];
        let mut res: Vec<i32> = vec![-1; n];
        for e in red_edges {
            let u = e[0] as usize;
            let v = e[1] as usize;
            graph[0][u].push(v);
        }
        for e in blue_edges {
            let u = e[0] as usize;
            let v = e[1] as usize;
            graph[1][u].push(v);
        }
        let mut queue: VecDeque<(usize, usize, usize)> = VecDeque::new();
        queue.push_back((0, 0, 0));
        queue.push_back((0, 1, 0));
        visited[0][0] = true;
        visited[1][0] = true;
        while let Some((u, c, d)) = queue.pop_front() {
            if res[u] == -1 {
                res[u] = d as i32;
            }
            for &v in &graph[c][u] {
                if !visited[1 - c][v] {
                    visited[1 - c][v] = true;
                    queue.push_back((v, 1 - c, d + 1));
                }
            }
        }
        res
    }
}

#[test]
fn test() {
    let n = 3;
    let red_edges = vec_vec_i32![[0, 1], [1, 2]];
    let blue_edges = vec_vec_i32![];
    let res = vec![0, 1, -1];
    assert_eq!(
        Solution::shortest_alternating_paths(n, red_edges, blue_edges),
        res
    );
    let n = 3;
    let red_edges = vec_vec_i32![[0, 1]];
    let blue_edges = vec_vec_i32![[2, 1]];
    let res = vec![0, 1, -1];
    assert_eq!(
        Solution::shortest_alternating_paths(n, red_edges, blue_edges),
        res
    );
    let n = 3;
    let red_edges = vec_vec_i32![[1, 0]];
    let blue_edges = vec_vec_i32![[2, 1]];
    let res = vec![0, -1, -1];
    assert_eq!(
        Solution::shortest_alternating_paths(n, red_edges, blue_edges),
        res
    );
    let n = 3;
    let red_edges = vec_vec_i32![[0, 1]];
    let blue_edges = vec_vec_i32![[1, 2]];
    let res = vec![0, 1, 2];
    assert_eq!(
        Solution::shortest_alternating_paths(n, red_edges, blue_edges),
        res
    );
    let n = 3;
    let red_edges = vec_vec_i32![[0, 1], [0, 2]];
    let blue_edges = vec_vec_i32![[1, 0]];
    let res = vec![0, 1, 1];
    assert_eq!(
        Solution::shortest_alternating_paths(n, red_edges, blue_edges),
        res
    );
}

// Accepted solution for LeetCode #1129: Shortest Path with Alternating Colors
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];

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n + m)

Space

O(n + m)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.