Question

2858. Minimum Edge Reversals So Every Node Is Reachable

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Description

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._

 

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 <= ui == edges[i][0] < n
• 0 <= vi == edges[i][1] < n
• ui != vi
• The input is generated such that if the edges were bi-directional, the graph would be a tree.

Solutions

Solution 1

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

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);
      }
    }
  }
}

class Solution {
public:
  vector<int> minEdgeReversals(int n, vector<vector<int>>& edges) {
    vector<pair<int, int>> g[n];
    vector<int> ans(n);
    for (auto& e : edges) {
      int x = e[0], y = e[1];
      g[x].emplace_back(y, 1);
      g[y].emplace_back(x, -1);
    }
    function<void(int, int)> dfs = [&](int i, int fa) {
      for (auto& [j, k] : g[i]) {
        if (j != fa) {
          ans[0] += k < 0;
          dfs(j, i);
        }
      }
    };
    function<void(int, int)> dfs2 = [&](int i, int fa) {
      for (auto& [j, k] : g[i]) {
        if (j != fa) {
          ans[j] = ans[i] + k;
          dfs2(j, i);
        }
      }
    };
    dfs(0, -1);
    dfs2(0, -1);
    return ans;
  }
};

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
}

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;
}