# Minimum Edge Reversals So Every Node Is Reachable LeetCode Solution

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### We Are Discuss About LeetCode Solution

### Minimum Edge Reversals So Every Node Is Reachable LeetCode Solution

## Problem

### Minimum Edge Reversals So Every Node Is Reachable

**User Accepted:**0**User Tried:**3**Total Accepted:**0**Total Submissions:**4**Difficulty:**Hard

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`

represents a _{i}, v_{i}]**directed edge** going from node `u`

to node _{i}`v`

._{i}

An **edge reversal** changes the direction of an edge, i.e., a directed edge going from node `u`

to node _{i}`v`

becomes a directed edge going from node _{i}`v`

to node _{i}`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 <= 10`

^{5}`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.

6 digit code appear after 69 seconds.

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