1466. Reorder Routes to Make All Paths Lead to The City Zero - Explanation

Description

There are n cities labeled 0 through n - 1. The pairs in connections describe one-way roads: [a, b] lets you travel from city a to city b. Ignoring their directions, the roads form a tree.

You may reverse the direction of any road. Return the minimum number of reversals needed so that every city has a directed path to city 0.

Reversing a road changes only its direction; its endpoints remain the same.

Example 1:

Reorder Routes to Make All Paths Lead to the City Zero, example 1: illustration of the example

Input: n = 4
connections = [[0,1],[2,1],[2,3]]

Output: 2

Explanation: Reverse 0 → 1 and 2 → 3. The resulting roads are 1 → 0, 2 → 1, and 3 → 2, so every city can reach 0. Both reversals are necessary.

Example 2:

Reorder Routes to Make All Paths Lead to the City Zero, example 2: illustration of the example

Input: n = 4
connections = [[1,0],[2,0],[3,0]]

Output: 0

Explanation: Each of cities 1, 2, and 3 already has a road directly to city 0. No road needs to change direction.

Example 3:

Reorder Routes to Make All Paths Lead to the City Zero, example 3: illustration of the example

Input: n = 5
connections = [[0,1],[0,2],[0,3],[0,4]]

Output: 4

Explanation: All four roads point away from city 0. Each leaf has only one road, so all four roads must be reversed.

Constraints:

2 <= n <= 50000
connections.length == n - 1
connections[i].length == 2
Each road joins two distinct city labels between 0 and n - 1.
Ignoring road directions, the connections form a tree.

Topics

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Prerequisites

Before attempting this problem, you should be comfortable with:

Graph Representation - Building adjacency lists from edge lists to represent graph connections
Depth First Search (DFS) - Recursively traversing all nodes in a graph while tracking visited nodes
Breadth First Search (BFS) - Level-by-level graph traversal using a queue
Hash Set - Efficiently storing and checking edge directions in O(1) time

1. Depth First Search - I

Intuition

We need all cities to reach city 0, so we traverse outward from city 0 and check edge directions. Build an undirected neighbors graph for traversal, but store original edges in edges set to check direction. When moving from city A to neighbor B, if the original edge goes from A to B (away from 0), it needs to be reversed. dfs ensures we visit every city exactly once.

Algorithm

Store all original directed edges in a set called edges as (a, b) pairs.
Build an adjacency list called neighbors treating edges as undirected.
Start dfs from city 0, marking visited cities in visit set.
For each neighbor:
- If not visited, check if (neighbor, city) exists in the edges set.
- If not, the edge points away from 0 and needs reversal; increment changes.
- Recursively visit the neighbor.
Return the total count of edges to reverse.

class Solution:
    def minReorder(self, n: int, connections: List[List[int]]) -> int:
        edges = {(a, b) for a, b in connections}
        neighbors = {city: [] for city in range(n)}
        visit = set()
        changes = 0

        for a, b in connections:
            neighbors[a].append(b)
            neighbors[b].append(a)

        def dfs(city):
            nonlocal changes
            visit.add(city)
            for neighbor in neighbors[city]:
                if neighbor in visit:
                    continue
                if (neighbor, city) not in edges:
                    changes += 1
                dfs(neighbor)

        dfs(0)
        return changes

public class Solution {
    private Map<Integer, List<Integer>> neighbors;
    private boolean[] visit;
    private Set<String> edges;

    public int minReorder(int n, int[][] connections) {
        edges = new HashSet<>();
        neighbors = new HashMap<>();
        visit = new boolean[n];
        int[] changes = {0};

        for (int[] conn : connections) {
            int a = conn[0], b = conn[1];
            edges.add(a + "," + b);
            neighbors.computeIfAbsent(a, k -> new ArrayList<>()).add(b);
            neighbors.computeIfAbsent(b, k -> new ArrayList<>()).add(a);
        }

        dfs(0, changes);
        return changes[0];
    }

    private void dfs(int city, int[] changes) {
        visit[city] = true;
        for (int neighbor : neighbors.get(city)) {
            if (visit[neighbor]) continue;
            if (!edges.contains(neighbor + "," + city)) {
                changes[0]++;
            }
            dfs(neighbor, changes);
        }
    }
}

class Solution {
public:
    int minReorder(int n, vector<vector<int>>& connections) {
        unordered_set<string> edges;
        unordered_map<int, vector<int>> neighbors;
        vector<bool> visit(n, false);
        int changes = 0;

        for (auto& c : connections) {
            edges.insert(to_string(c[0]) + "," + to_string(c[1]));
            neighbors[c[0]].push_back(c[1]);
            neighbors[c[1]].push_back(c[0]);
        }

        function<void(int)> dfs = [&](int city) {
            visit[city] = true;
            for (int neighbor : neighbors[city]) {
                if (visit[neighbor]) continue;
                if (edges.find(to_string(neighbor) + "," + to_string(city)) == edges.end()) {
                    changes++;
                }
                dfs(neighbor);
            }
        };

        dfs(0);
        return changes;
    }
};

class Solution {
    /**
     * @param {number} n
     * @param {number[][]} connections
     * @return {number}
     */
    minReorder(n, connections) {
        const edges = new Set();
        const neighbors = Array.from({ length: n }, () => []);
        const visit = new Array(n).fill(false);
        let changes = 0;

        for (const [a, b] of connections) {
            edges.add(`${a},${b}`);
            neighbors[a].push(b);
            neighbors[b].push(a);
        }

        const dfs = (city) => {
            visit[city] = true;
            for (const neighbor of neighbors[city]) {
                if (visit[neighbor]) continue;
                if (!edges.has(`${neighbor},${city}`)) changes++;
                dfs(neighbor);
            }
        };

        dfs(0);
        return changes;
    }
}

public class Solution {
    private Dictionary<int, List<int>> neighbors;
    private bool[] visit;
    private HashSet<string> edges;

    public int MinReorder(int n, int[][] connections) {
        edges = new HashSet<string>();
        neighbors = new Dictionary<int, List<int>>();
        visit = new bool[n];
        int changes = 0;

        for (int i = 0; i < n; i++) {
            neighbors[i] = new List<int>();
        }

        foreach (var conn in connections) {
            int a = conn[0], b = conn[1];
            edges.Add($"{a},{b}");
            neighbors[a].Add(b);
            neighbors[b].Add(a);
        }

        void Dfs(int city) {
            visit[city] = true;
            foreach (int neighbor in neighbors[city]) {
                if (visit[neighbor]) continue;
                if (!edges.Contains($"{neighbor},{city}")) changes++;
                Dfs(neighbor);
            }
        }

        Dfs(0);
        return changes;
    }
}

func minReorder(n int, connections [][]int) int {
    edges := make(map[string]bool)
    neighbors := make([][]int, n)
    visit := make([]bool, n)
    changes := 0

    for i := 0; i < n; i++ {
        neighbors[i] = []int{}
    }

    for _, conn := range connections {
        a, b := conn[0], conn[1]
        edges[fmt.Sprintf("%d,%d", a, b)] = true
        neighbors[a] = append(neighbors[a], b)
        neighbors[b] = append(neighbors[b], a)
    }

    var dfs func(city int)
    dfs = func(city int) {
        visit[city] = true
        for _, neighbor := range neighbors[city] {
            if visit[neighbor] {
                continue
            }
            if !edges[fmt.Sprintf("%d,%d", neighbor, city)] {
                changes++
            }
            dfs(neighbor)
        }
    }

    dfs(0)
    return changes
}

class Solution {
    fun minReorder(n: Int, connections: Array<IntArray>): Int {
        val edges = HashSet<String>()
        val neighbors = Array(n) { mutableListOf<Int>() }
        val visit = BooleanArray(n)
        var changes = 0

        for (conn in connections) {
            val (a, b) = conn[0] to conn[1]
            edges.add("$a,$b")
            neighbors[a].add(b)
            neighbors[b].add(a)
        }

        fun dfs(city: Int) {
            visit[city] = true
            for (neighbor in neighbors[city]) {
                if (visit[neighbor]) continue
                if (!edges.contains("$neighbor,$city")) changes++
                dfs(neighbor)
            }
        }

        dfs(0)
        return changes
    }
}

class Solution {
    func minReorder(_ n: Int, _ connections: [[Int]]) -> Int {
        var edges = Set<String>()
        var neighbors = [[Int]](repeating: [], count: n)
        var visit = [Bool](repeating: false, count: n)
        var changes = 0

        for conn in connections {
            let a = conn[0], b = conn[1]
            edges.insert("\(a),\(b)")
            neighbors[a].append(b)
            neighbors[b].append(a)
        }

        func dfs(_ city: Int) {
            visit[city] = true
            for neighbor in neighbors[city] {
                if visit[neighbor] { continue }
                if !edges.contains("\(neighbor),\(city)") { changes += 1 }
                dfs(neighbor)
            }
        }

        dfs(0)
        return changes
    }
}

impl Solution {
    pub fn min_reorder(n: i32, connections: Vec<Vec<i32>>) -> i32 {
        let n = n as usize;
        let mut edges = HashSet::new();
        let mut neighbors = vec![vec![]; n];
        let mut visit = vec![false; n];
        let mut changes = 0;

        for conn in &connections {
            let (a, b) = (conn[0] as usize, conn[1] as usize);
            edges.insert((a, b));
            neighbors[a].push(b);
            neighbors[b].push(a);
        }

        fn dfs(
            city: usize,
            visit: &mut Vec<bool>,
            neighbors: &Vec<Vec<usize>>,
            edges: &HashSet<(usize, usize)>,
            changes: &mut i32,
        ) {
            visit[city] = true;
            for &neighbor in &neighbors[city] {
                if visit[neighbor] {
                    continue;
                }
                if !edges.contains(&(neighbor, city)) {
                    *changes += 1;
                }
                dfs(neighbor, visit, neighbors, edges, changes);
            }
        }

        dfs(0, &mut visit, &neighbors, &edges, &mut changes);
        changes
    }
}

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(n)$

2. Depth First Search - II

Intuition

Instead of using a separate set to track edge directions, we can encode direction information directly in the adjacency list. When adding edges, we store the original direction as a positive value and the reverse as negative. During dfs, positive nei values indicate edges pointing away from city 0, which need reversal.

Algorithm

Build an adjacency list called adj where each edge (u, v) adds v to adj[u] and -u to adj[v].
Start dfs from city 0 with parent = -1.
For each nei (neighbor):
- Skip if abs(nei) equals the parent (to avoid going back).
- Add 1 to the count if nei > 0 (edge needs reversal).
- Recursively process abs(nei) as the next node.
Return the total count.

class Solution:
    def minReorder(self, n: int, connections: List[List[int]]) -> int:
        adj = [[] for _ in range(n)]
        for u, v in connections:
            adj[u].append(v)
            adj[v].append(-u)

        def dfs(node, parent):
            changes = 0
            for nei in adj[node]:
                if abs(nei) == parent:
                    continue
                changes += dfs(abs(nei), node) + (nei > 0)
            return changes

        return dfs(0, -1)

public class Solution {
    public int minReorder(int n, int[][] connections) {
        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) adj.add(new ArrayList<>());

        for (int[] conn : connections) {
            adj.get(conn[0]).add(conn[1]);
            adj.get(conn[1]).add(-conn[0]);
        }

        return dfs(0, -1, adj);
    }

    private int dfs(int node, int parent, List<List<Integer>> adj) {
        int changes = 0;
        for (int nei : adj.get(node)) {
            if (Math.abs(nei) == parent) continue;
            changes += dfs(Math.abs(nei), node, adj) + (nei > 0 ? 1 : 0);
        }
        return changes;
    }
}

class Solution {
public:
    int minReorder(int n, vector<vector<int>>& connections) {
        vector<vector<int>> adj(n);
        for (auto& conn : connections) {
            int u = conn[0], v = conn[1];
            adj[u].push_back(v);
            adj[v].push_back(-u);
        }

        return dfs(0, -1, adj);
    }

private:
    int dfs(int node, int parent, vector<vector<int>>& adj) {
        int changes = 0;
        for (int nei : adj[node]) {
            if (abs(nei) == parent) continue;
            changes += dfs(abs(nei), node, adj) + (nei > 0);
        }
        return changes;
    }
};

class Solution {
    /**
     * @param {number} n
     * @param {number[][]} connections
     * @return {number}
     */
    minReorder(n, connections) {
        const adj = Array.from({ length: n }, () => []);
        for (const [u, v] of connections) {
            adj[u].push(v);
            adj[v].push(-u);
        }

        const dfs = (node, parent) => {
            let changes = 0;
            for (const nei of adj[node]) {
                if (Math.abs(nei) === parent) continue;
                changes += dfs(Math.abs(nei), node) + (nei > 0 ? 1 : 0);
            }
            return changes;
        };

        return dfs(0, -1);
    }
}

func minReorder(n int, connections [][]int) int {
    adj := make([][]int, n)
    for i := 0; i < n; i++ {
        adj[i] = []int{}
    }

    for _, conn := range connections {
        u, v := conn[0], conn[1]
        adj[u] = append(adj[u], v)
        adj[v] = append(adj[v], -u)
    }

    var dfs func(node, parent int) int
    dfs = func(node, parent int) int {
        changes := 0
        for _, nei := range adj[node] {
            absNei := nei
            if absNei < 0 {
                absNei = -absNei
            }
            if absNei == parent {
                continue
            }
            add := 0
            if nei > 0 {
                add = 1
            }
            changes += dfs(absNei, node) + add
        }
        return changes
    }

    return dfs(0, -1)
}

class Solution {
    fun minReorder(n: Int, connections: Array<IntArray>): Int {
        val adj = Array(n) { mutableListOf<Int>() }

        for (conn in connections) {
            adj[conn[0]].add(conn[1])
            adj[conn[1]].add(-conn[0])
        }

        fun dfs(node: Int, parent: Int): Int {
            var changes = 0
            for (nei in adj[node]) {
                if (kotlin.math.abs(nei) == parent) continue
                changes += dfs(kotlin.math.abs(nei), node) + if (nei > 0) 1 else 0
            }
            return changes
        }

        return dfs(0, -1)
    }
}

class Solution {
    func minReorder(_ n: Int, _ connections: [[Int]]) -> Int {
        var adj = [[Int]](repeating: [], count: n)

        for conn in connections {
            adj[conn[0]].append(conn[1])
            adj[conn[1]].append(-conn[0])
        }

        func dfs(_ node: Int, _ parent: Int) -> Int {
            var changes = 0
            for nei in adj[node] {
                if abs(nei) == parent { continue }
                changes += dfs(abs(nei), node) + (nei > 0 ? 1 : 0)
            }
            return changes
        }

        return dfs(0, -1)
    }
}

impl Solution {
    pub fn min_reorder(n: i32, connections: Vec<Vec<i32>>) -> i32 {
        let n = n as usize;
        let mut adj = vec![vec![]; n];

        for conn in &connections {
            adj[conn[0] as usize].push(conn[1]);
            adj[conn[1] as usize].push(-conn[0]);
        }

        fn dfs(node: usize, parent: i32, adj: &Vec<Vec<i32>>) -> i32 {
            let mut changes = 0;
            for &nei in &adj[node] {
                if nei.abs() == parent {
                    continue;
                }
                changes += dfs(nei.unsigned_abs() as usize, node as i32, adj)
                    + if nei > 0 { 1 } else { 0 };
            }
            changes
        }

        dfs(0, -1, &adj)
    }
}

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(n)$

3. Breadth First Search

Intuition

bfs works equally well since we just need to visit all nodes once from city 0. We store each edge with a flag isForward indicating if it is a forward edge (pointing away from city 0). Processing level by level in queue, whenever we traverse a forward edge, we count it as needing reversal.

Algorithm

Build an adjacency list called adj where each edge (u, v) stores (v, 1) in adj[u] and (u, 0) in adj[v].
Initialize a queue with city 0 and mark it visited in visit array.
While the queue is not empty:
- Dequeue a node.
- For each unvisited neighbor, mark it visited, add isForward to the count, and enqueue it.
Return the total count.

class Solution:
    def minReorder(self, n: int, connections: List[List[int]]) -> int:
        adj = [[] for _ in range(n)]
        for u, v in connections:
            adj[u].append((v, 1))
            adj[v].append((u, 0))

        visit = [False] * n
        queue = deque([0])
        visit[0] = True
        changes = 0

        while queue:
            node = queue.popleft()
            for neighbor, isForward in adj[node]:
                if not visit[neighbor]:
                    visit[neighbor] = True
                    changes += isForward
                    queue.append(neighbor)
        return changes

public class Solution {
    public int minReorder(int n, int[][] connections) {
        List<int[]>[] adj = new ArrayList[n];
        for (int i = 0; i < n; i++) adj[i] = new ArrayList<>();

        for (int[] conn : connections) {
            adj[conn[0]].add(new int[]{conn[1], 1});
            adj[conn[1]].add(new int[]{conn[0], 0});
        }

        boolean[] visited = new boolean[n];
        Queue<Integer> queue = new LinkedList<>();
        queue.add(0);
        visited[0] = true;
        int changes = 0;

        while (!queue.isEmpty()) {
            int node = queue.poll();
            for (int[] edge : adj[node]) {
                int neighbor = edge[0], isForward = edge[1];
                if (!visited[neighbor]) {
                    visited[neighbor] = true;
                    changes += isForward;
                    queue.add(neighbor);
                }
            }
        }
        return changes;
    }
}

class Solution {
public:
    int minReorder(int n, vector<vector<int>>& connections) {
        vector<vector<pair<int, int>>> adj(n);
        for (auto& conn : connections) {
            adj[conn[0]].push_back({conn[1], 1});
            adj[conn[1]].push_back({conn[0], 0});
        }

        vector<bool> visit(n, false);
        queue<int> q;
        q.push(0);
        visit[0] = true;
        int changes = 0;

        while (!q.empty()) {
            int node = q.front();
            q.pop();
            for (auto& [neighbor, isForward] : adj[node]) {
                if (!visit[neighbor]) {
                    visit[neighbor] = true;
                    changes += isForward;
                    q.push(neighbor);
                }
            }
        }
        return changes;
    }
};

class Solution {
    /**
     * @param {number} n
     * @param {number[][]} connections
     * @return {number}
     */
    minReorder(n, connections) {
        const adj = Array.from({ length: n }, () => []);
        for (const [u, v] of connections) {
            adj[u].push([v, 1]);
            adj[v].push([u, 0]);
        }

        const visited = Array(n).fill(false);
        const queue = new Queue();
        queue.push(0);
        visited[0] = true;
        let changes = 0;

        while (!queue.isEmpty()) {
            const node = queue.pop();
            for (const [neighbor, isForward] of adj[node]) {
                if (!visited[neighbor]) {
                    visited[neighbor] = true;
                    changes += isForward;
                    queue.push(neighbor);
                }
            }
        }
        return changes;
    }
}

func minReorder(n int, connections [][]int) int {
    adj := make([][][2]int, n)
    for i := 0; i < n; i++ {
        adj[i] = [][2]int{}
    }

    for _, conn := range connections {
        adj[conn[0]] = append(adj[conn[0]], [2]int{conn[1], 1})
        adj[conn[1]] = append(adj[conn[1]], [2]int{conn[0], 0})
    }

    visited := make([]bool, n)
    queue := []int{0}
    visited[0] = true
    changes := 0

    for len(queue) > 0 {
        node := queue[0]
        queue = queue[1:]
        for _, edge := range adj[node] {
            neighbor, isForward := edge[0], edge[1]
            if !visited[neighbor] {
                visited[neighbor] = true
                changes += isForward
                queue = append(queue, neighbor)
            }
        }
    }
    return changes
}

class Solution {
    fun minReorder(n: Int, connections: Array<IntArray>): Int {
        val adj = Array(n) { mutableListOf<IntArray>() }

        for (conn in connections) {
            adj[conn[0]].add(intArrayOf(conn[1], 1))
            adj[conn[1]].add(intArrayOf(conn[0], 0))
        }

        val visited = BooleanArray(n)
        val queue = ArrayDeque<Int>()
        queue.add(0)
        visited[0] = true
        var changes = 0

        while (queue.isNotEmpty()) {
            val node = queue.removeFirst()
            for (edge in adj[node]) {
                val (neighbor, isForward) = edge[0] to edge[1]
                if (!visited[neighbor]) {
                    visited[neighbor] = true
                    changes += isForward
                    queue.add(neighbor)
                }
            }
        }
        return changes
    }
}

class Solution {
    func minReorder(_ n: Int, _ connections: [[Int]]) -> Int {
        var adj = [[(Int, Int)]](repeating: [], count: n)

        for conn in connections {
            adj[conn[0]].append((conn[1], 1))
            adj[conn[1]].append((conn[0], 0))
        }

        var visited = [Bool](repeating: false, count: n)
        var queue = [Int]()
        queue.append(0)
        visited[0] = true
        var changes = 0

        while !queue.isEmpty {
            let node = queue.removeFirst()
            for (neighbor, isForward) in adj[node] {
                if !visited[neighbor] {
                    visited[neighbor] = true
                    changes += isForward
                    queue.append(neighbor)
                }
            }
        }
        return changes
    }
}

impl Solution {
    pub fn min_reorder(n: i32, connections: Vec<Vec<i32>>) -> i32 {
        let n = n as usize;
        let mut adj = vec![vec![]; n];

        for conn in &connections {
            adj[conn[0] as usize].push((conn[1] as usize, 1));
            adj[conn[1] as usize].push((conn[0] as usize, 0));
        }

        let mut visited = vec![false; n];
        let mut queue = VecDeque::new();
        queue.push_back(0);
        visited[0] = true;
        let mut changes = 0;

        while let Some(node) = queue.pop_front() {
            for &(neighbor, is_forward) in &adj[node] {
                if !visited[neighbor] {
                    visited[neighbor] = true;
                    changes += is_forward;
                    queue.push_back(neighbor);
                }
            }
        }
        changes
    }
}

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(n)$

Common Pitfalls

Treating the Graph as Directed Only

A common mistake is building only a directed adjacency list based on the given edges. Since we need to traverse the entire tree starting from city 0, we must treat edges as undirected for traversal purposes while separately tracking the original direction to count reversals.

Reversing the Logic of Edge Direction Check

When checking if an edge needs reversal, it is easy to get the condition backwards. Remember that edges pointing away from city 0 need reversal. If traversing from node A to neighbor B and the original edge is (A, B), it points away from 0 and must be counted; if the original edge is (B, A), it already points toward 0.

Not Handling the Sign Encoding for Node 0

In the DFS-II approach that uses negative values to encode direction, node 0 cannot be represented as negative since -0 == 0. This edge case must be handled by using the parent check (abs(nei) == parent) rather than relying solely on sign for all nodes.