LeetCode #1466 — MEDIUM

Reorder Routes to Make All Paths Lead to the City Zero

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

The Problem

Problem Statement

There are n cities numbered from 0 to n - 1 and n - 1 roads such that there is only one way to travel between two different cities (this network form a tree). Last year, The ministry of transport decided to orient the roads in one direction because they are too narrow.

Roads are represented by connections where connections[i] = [a_i, b_i] represents a road from city a_i to city b_i.

This year, there will be a big event in the capital (city 0), and many people want to travel to this city.

Your task consists of reorienting some roads such that each city can visit the city 0. Return the minimum number of edges changed.

It's guaranteed that each city can reach city 0 after reorder.

Example 1:

Input: n = 6, connections = [[0,1],[1,3],[2,3],[4,0],[4,5]]
Output: 3
Explanation: Change the direction of edges show in red such that each node can reach the node 0 (capital).

Example 2:

Input: n = 5, connections = [[1,0],[1,2],[3,2],[3,4]]
Output: 2
Explanation: Change the direction of edges show in red such that each node can reach the node 0 (capital).

Example 3:

Input: n = 3, connections = [[1,0],[2,0]]
Output: 0

Constraints:

2 <= n <= 5 * 10⁴
connections.length == n - 1
connections[i].length == 2
0 <= a_i, b_i <= n - 1
a_i != b_i

Step 01

Brute Force Baseline

Problem summary: There are n cities numbered from 0 to n - 1 and n - 1 roads such that there is only one way to travel between two different cities (this network form a tree). Last year, The ministry of transport decided to orient the roads in one direction because they are too narrow. Roads are represented by connections where connections[i] = [ai, bi] represents a road from city ai to city bi. This year, there will be a big event in the capital (city 0), and many people want to travel to this city. Your task consists of reorienting some roads such that each city can visit the city 0. Return the minimum number of edges changed. It's guaranteed that each city can reach city 0 after reorder.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

6
[[0,1],[1,3],[2,3],[4,0],[4,5]]

Example 2

5
[[1,0],[1,2],[3,2],[3,4]]

Example 3

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

Core Insight

What unlocks the optimal approach

Treat the graph as undirected. Start a dfs from the root, if you come across an edge in the forward direction, you need to reverse the edge.

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 #1466: Reorder Routes to Make All Paths Lead to the City Zero
class Solution {
    private List<int[]>[] g;

    public int minReorder(int n, int[][] connections) {
        g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : connections) {
            int a = e[0], b = e[1];
            g[a].add(new int[] {b, 1});
            g[b].add(new int[] {a, 0});
        }
        return dfs(0, -1);
    }

    private int dfs(int a, int fa) {
        int ans = 0;
        for (var e : g[a]) {
            int b = e[0], c = e[1];
            if (b != fa) {
                ans += c + dfs(b, a);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1466: Reorder Routes to Make All Paths Lead to the City Zero
func minReorder(n int, connections [][]int) int {
	g := make([][][2]int, n)
	for _, e := range connections {
		a, b := e[0], e[1]
		g[a] = append(g[a], [2]int{b, 1})
		g[b] = append(g[b], [2]int{a, 0})
	}
	var dfs func(int, int) int
	dfs = func(a, fa int) (ans int) {
		for _, e := range g[a] {
			if b, c := e[0], e[1]; b != fa {
				ans += c + dfs(b, a)
			}
		}
		return
	}
	return dfs(0, -1)
}

# Accepted solution for LeetCode #1466: Reorder Routes to Make All Paths Lead to the City Zero
class Solution:
    def minReorder(self, n: int, connections: List[List[int]]) -> int:
        def dfs(a: int, fa: int) -> int:
            return sum(c + dfs(b, a) for b, c in g[a] if b != fa)

        g = [[] for _ in range(n)]
        for a, b in connections:
            g[a].append((b, 1))
            g[b].append((a, 0))
        return dfs(0, -1)

// Accepted solution for LeetCode #1466: Reorder Routes to Make All Paths Lead to the City Zero
impl Solution {
    pub fn min_reorder(n: i32, connections: Vec<Vec<i32>>) -> i32 {
        let mut g: Vec<Vec<(i32, i32)>> = vec![vec![]; n as usize];
        for e in connections.iter() {
            let a = e[0] as usize;
            let b = e[1] as usize;
            g[a].push((b as i32, 1));
            g[b].push((a as i32, 0));
        }
        fn dfs(a: usize, fa: i32, g: &Vec<Vec<(i32, i32)>>) -> i32 {
            let mut ans = 0;
            for &(b, c) in g[a].iter() {
                if b != fa {
                    ans += c + dfs(b as usize, a as i32, g);
                }
            }
            ans
        }
        dfs(0, -1, &g)
    }
}

// Accepted solution for LeetCode #1466: Reorder Routes to Make All Paths Lead to the City Zero
function minReorder(n: number, connections: number[][]): number {
    const g: [number, number][][] = Array.from({ length: n }, () => []);
    for (const [a, b] of connections) {
        g[a].push([b, 1]);
        g[b].push([a, 0]);
    }
    const dfs = (a: number, fa: number): number => {
        let ans = 0;
        for (const [b, c] of g[a]) {
            if (b !== fa) {
                ans += c + dfs(b, a);
            }
        }
        return ans;
    };
    return dfs(0, -1);
}

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)

Space

O(n)

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.