LeetCode #2858 — HARD

Minimum Edge Reversals So Every Node Is Reachable

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

There is a simple directed graph with n nodes labeled from 0 to n - 1. The graph would form a tree if its edges were bi-directional.

You are given an integer n and a 2D integer array edges, where edges[i] = [u_i, v_i] represents a directed edge going from node u_i to node v_i.

An edge reversal changes the direction of an edge, i.e., a directed edge going from node u_i to node v_i becomes a directed edge going from node v_i to node u_i.

For every node i in the range [0, n - 1], your task is to independently calculate the minimum number of edge reversals required so it is possible to reach any other node starting from node i through a sequence of directed edges.

Return an integer array answer, where answer[i] is the minimum number of edge reversals required so it is possible to reach any other node starting from node i through a sequence of directed edges.

Example 1:

Input: n = 4, edges = [[2,0],[2,1],[1,3]]
Output: [1,1,0,2]
Explanation: The image above shows the graph formed by the edges.
For node 0: after reversing the edge [2,0], it is possible to reach any other node starting from node 0.
So, answer[0] = 1.
For node 1: after reversing the edge [2,1], it is possible to reach any other node starting from node 1.
So, answer[1] = 1.
For node 2: it is already possible to reach any other node starting from node 2.
So, answer[2] = 0.
For node 3: after reversing the edges [1,3] and [2,1], it is possible to reach any other node starting from node 3.
So, answer[3] = 2.

Example 2:

Input: n = 3, edges = [[1,2],[2,0]]
Output: [2,0,1]
Explanation: The image above shows the graph formed by the edges.
For node 0: after reversing the edges [2,0] and [1,2], it is possible to reach any other node starting from node 0.
So, answer[0] = 2.
For node 1: it is already possible to reach any other node starting from node 1.
So, answer[1] = 0.
For node 2: after reversing the edge [1, 2], it is possible to reach any other node starting from node 2.
So, answer[2] = 1.

Constraints:

2 <= n <= 10⁵
edges.length == n - 1
edges[i].length == 2
0 <= u_i == edges[i][0] < n
0 <= v_i == edges[i][1] < n
u_i != v_i
The input is generated such that if the edges were bi-directional, the graph would be a tree.

Step 01

Brute Force Baseline

Problem summary: There is a simple directed graph with n nodes labeled from 0 to n - 1. The graph would form a tree if its edges were bi-directional. You are given an integer n and a 2D integer array edges, where edges[i] = [ui, vi] represents a directed edge going from node ui to node vi. An edge reversal changes the direction of an edge, i.e., a directed edge going from node ui to node vi becomes a directed edge going from node vi to node ui. For every node i in the range [0, n - 1], your task is to independently calculate the minimum number of edge reversals required so it is possible to reach any other node starting from node i through a sequence of directed edges. Return an integer array answer, where answer[i] is the minimum number of edge reversals required so it is possible to reach any other node starting from node i through a sequence of directed edges.

Baseline thinking

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

Pattern signal: Dynamic Programming

Example 1

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

Example 2

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

Core Insight

What unlocks the optimal approach

The problem can be solved using tree DP.
Using node <code>0</code> as the root, let <code>dp[x]</code> be the minimum number of edge reversals so node <code>x</code> can reach every node in its subtree.
Using a DFS traversing the edges bidirectionally, we can compute <code>dp</code>.<br /> <code>dp[x] = dp[y] +</code> (<code>1</code> if the edge between <code>x</code> and <code>y</code> is going from <code>y</code> to <code>x</code>; <code>0</code> otherwise), where <code>x</code> is the parent of <code>y</code>.
Let <code>answer[x]</code> be the minimum number of edge reversals so it is possible to reach any other node starting from node <code>x</code>.
Using another DFS starting from node <code>0</code> and traversing the edges bidirectionally, we can compute <code>answer</code>.<br /> <code>answer[0] = dp[0]</code><br /> <code>answer[y] = answer[x] +</code> (<code>1</code> if the edge between <code>x</code> and <code>y</code> is going from <code>x</code> to <code>y</code>; <code>-1</code> otherwise), where <code>x</code> is the parent of <code>y</code>.

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

Largest constraint values

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 #2858: Minimum Edge Reversals So Every Node Is Reachable
class Solution {
    private List<int[]>[] g;
    private int[] ans;

    public int[] minEdgeReversals(int n, int[][] edges) {
        ans = new int[n];
        g = new List[n];
        Arrays.setAll(g, i -> new ArrayList<>());
        for (var e : edges) {
            int x = e[0], y = e[1];
            g[x].add(new int[] {y, 1});
            g[y].add(new int[] {x, -1});
        }
        dfs(0, -1);
        dfs2(0, -1);
        return ans;
    }

    private void dfs(int i, int fa) {
        for (var ne : g[i]) {
            int j = ne[0], k = ne[1];
            if (j != fa) {
                ans[0] += k < 0 ? 1 : 0;
                dfs(j, i);
            }
        }
    }

    private void dfs2(int i, int fa) {
        for (var ne : g[i]) {
            int j = ne[0], k = ne[1];
            if (j != fa) {
                ans[j] = ans[i] + k;
                dfs2(j, i);
            }
        }
    }
}

// Accepted solution for LeetCode #2858: Minimum Edge Reversals So Every Node Is Reachable
func minEdgeReversals(n int, edges [][]int) []int {
	g := make([][][2]int, n)
	for _, e := range edges {
		x, y := e[0], e[1]
		g[x] = append(g[x], [2]int{y, 1})
		g[y] = append(g[y], [2]int{x, -1})
	}
	ans := make([]int, n)
	var dfs func(int, int)
	var dfs2 func(int, int)
	dfs = func(i, fa int) {
		for _, ne := range g[i] {
			j, k := ne[0], ne[1]
			if j != fa {
				if k < 0 {
					ans[0]++
				}
				dfs(j, i)
			}
		}
	}
	dfs2 = func(i, fa int) {
		for _, ne := range g[i] {
			j, k := ne[0], ne[1]
			if j != fa {
				ans[j] = ans[i] + k
				dfs2(j, i)
			}
		}
	}
	dfs(0, -1)
	dfs2(0, -1)
	return ans
}

# Accepted solution for LeetCode #2858: Minimum Edge Reversals So Every Node Is Reachable
class Solution:
    def minEdgeReversals(self, n: int, edges: List[List[int]]) -> List[int]:
        ans = [0] * n
        g = [[] for _ in range(n)]
        for x, y in edges:
            g[x].append((y, 1))
            g[y].append((x, -1))

        def dfs(i: int, fa: int):
            for j, k in g[i]:
                if j != fa:
                    ans[0] += int(k < 0)
                    dfs(j, i)

        dfs(0, -1)

        def dfs2(i: int, fa: int):
            for j, k in g[i]:
                if j != fa:
                    ans[j] = ans[i] + k
                    dfs2(j, i)

        dfs2(0, -1)
        return ans

// Accepted solution for LeetCode #2858: Minimum Edge Reversals So Every Node Is Reachable
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2858: Minimum Edge Reversals So Every Node Is Reachable
// class Solution {
//     private List<int[]>[] g;
//     private int[] ans;
// 
//     public int[] minEdgeReversals(int n, int[][] edges) {
//         ans = new int[n];
//         g = new List[n];
//         Arrays.setAll(g, i -> new ArrayList<>());
//         for (var e : edges) {
//             int x = e[0], y = e[1];
//             g[x].add(new int[] {y, 1});
//             g[y].add(new int[] {x, -1});
//         }
//         dfs(0, -1);
//         dfs2(0, -1);
//         return ans;
//     }
// 
//     private void dfs(int i, int fa) {
//         for (var ne : g[i]) {
//             int j = ne[0], k = ne[1];
//             if (j != fa) {
//                 ans[0] += k < 0 ? 1 : 0;
//                 dfs(j, i);
//             }
//         }
//     }
// 
//     private void dfs2(int i, int fa) {
//         for (var ne : g[i]) {
//             int j = ne[0], k = ne[1];
//             if (j != fa) {
//                 ans[j] = ans[i] + k;
//                 dfs2(j, i);
//             }
//         }
//     }
// }

// Accepted solution for LeetCode #2858: Minimum Edge Reversals So Every Node Is Reachable
function minEdgeReversals(n: number, edges: number[][]): number[] {
    const g: number[][][] = Array.from({ length: n }, () => []);
    for (const [x, y] of edges) {
        g[x].push([y, 1]);
        g[y].push([x, -1]);
    }
    const ans: number[] = Array(n).fill(0);
    const dfs = (i: number, fa: number) => {
        for (const [j, k] of g[i]) {
            if (j !== fa) {
                ans[0] += k < 0 ? 1 : 0;
                dfs(j, i);
            }
        }
    };
    const dfs2 = (i: number, fa: number) => {
        for (const [j, k] of g[i]) {
            if (j !== fa) {
                ans[j] = ans[i] + k;
                dfs2(j, i);
            }
        }
    };
    dfs(0, -1);
    dfs2(0, -1);
    return ans;
}

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

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.