Question

1719. Number Of Ways To Reconstruct A Tree

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Description

You are given an array pairs, where pairs[i] = [xi, yi], and:

• There are no duplicates.
• xi < yi

Let ways be the number of rooted trees that satisfy the following conditions:

• The tree consists of nodes whose values appeared in pairs.
• A pair [xi, yi] exists in pairs if and only if xi is an ancestor of yi or yi is an ancestor of xi.
• Note: the tree does not have to be a binary tree.

Two ways are considered to be different if there is at least one node that has different parents in both ways.

Return:

• 0 if ways == 0
• 1 if ways == 1
• 2 if ways > 1

A rooted tree is a tree that has a single root node, and all edges are oriented to be outgoing from the root.

An ancestor of a node is any node on the path from the root to that node (excluding the node itself). The root has no ancestors.

 

Example 1:

Input: pairs = [[1,2],[2,3]]
Output: 1
Explanation: There is exactly one valid rooted tree, which is shown in the above figure.

Example 2:

Input: pairs = [[1,2],[2,3],[1,3]]
Output: 2
Explanation: There are multiple valid rooted trees. Three of them are shown in the above figures.

Example 3:

Input: pairs = [[1,2],[2,3],[2,4],[1,5]]
Output: 0
Explanation: There are no valid rooted trees.

 

Constraints:

• 1 <= pairs.length <= 105
• 1 <= xi < yi <= 500
• The elements in pairs are unique.

Solutions

Solution 1

class Solution:
  def checkWays(self, pairs: List[List[int]]) -> int:
    g = [[False] * 510 for _ in range(510)]
    v = defaultdict(list)
    for x, y in pairs:
      g[x][y] = g[y][x] = True
      v[x].append(y)
      v[y].append(x)
    nodes = []
    for i in range(510):
      if v[i]:
        nodes.append(i)
        g[i][i] = True
    nodes.sort(key=lambda x: len(v[x]))
    equal = False
    root = 0
    for i, x in enumerate(nodes):
      j = i + 1
      while j < len(nodes) and not g[x][nodes[j]]:
        j += 1
      if j < len(nodes):
        y = nodes[j]
        if len(v[x]) == len(v[y]):
          equal = True
        for z in v[x]:
          if not g[y][z]:
            return 0
      else:
        root += 1
    if root > 1:
      return 0
    return 2 if equal else 1

class Solution {
  public int checkWays(int[][] pairs) {
    boolean[][] g = new boolean[510][510];
    List<Integer>[] v = new List[510];
    Arrays.setAll(v, k -> new ArrayList<>());
    for (int[] p : pairs) {
      int x = p[0], y = p[1];
      g[x][y] = true;
      g[y][x] = true;
      v[x].add(y);
      v[y].add(x);
    }
    List<Integer> nodes = new ArrayList<>();
    for (int i = 0; i < 510; ++i) {
      if (!v[i].isEmpty()) {
        nodes.add(i);
        g[i][i] = true;
      }
    }
    nodes.sort(Comparator.comparingInt(a -> v[a].size()));
    boolean equal = false;
    int root = 0;
    for (int i = 0; i < nodes.size(); ++i) {
      int x = nodes.get(i);
      int j = i + 1;
      for (; j < nodes.size() && !g[x][nodes.get(j)]; ++j)
        ;
      if (j < nodes.size()) {
        int y = nodes.get(j);
        if (v[x].size() == v[y].size()) {
          equal = true;
        }
        for (int z : v[x]) {
          if (!g[y][z]) {
            return 0;
          }
        }
      } else {
        ++root;
      }
    }
    if (root > 1) {
      return 0;
    }
    return equal ? 2 : 1;
  }
}

class Solution {
public:
  int checkWays(vector<vector<int>>& pairs) {
    vector<vector<bool>> g(510, vector<bool>(510));
    vector<vector<int>> v(510);
    for (auto& p : pairs) {
      int x = p[0], y = p[1];
      g[x][y] = g[y][x] = 1;
      v[x].push_back(y);
      v[y].push_back(x);
    }
    vector<int> nodes;
    for (int i = 1; i <= 500; ++i) {
      if (v[i].size()) {
        nodes.push_back(i);
        g[i][i] = 1;
      }
    }
    sort(nodes.begin(), nodes.end(), [&](int x, int y) -> bool { return v[x].size() < v[y].size(); });
    bool equal = 0;
    int root = 0;
    for (int i = 0; i < nodes.size(); ++i) {
      int x = nodes[i];
      int j = i + 1;
      for (; j < nodes.size() && !g[x][nodes[j]]; ++j)
        ;
      if (j < nodes.size()) {
        int y = nodes[j];
        if (v[x].size() == v[y].size()) equal = 1;
        for (int z : v[x])
          if (!g[y][z])
            return 0;
      } else
        ++root;
    }
    if (root > 1) return 0;
    if (equal) return 2;
    return 1;
  }
};

func checkWays(pairs [][]int) int {
  g := make([][]bool, 510)
  v := make([][]int, 510)
  for i := range g {
    g[i] = make([]bool, 510)
  }
  for _, p := range pairs {
    x, y := p[0], p[1]
    g[x][y] = true
    g[y][x] = true
    v[x] = append(v[x], y)
    v[y] = append(v[y], x)
  }
  var nodes []int
  for i := 1; i <= 500; i++ {
    if len(v[i]) > 0 {
      nodes = append(nodes, i)
      g[i][i] = true
    }
  }
  sort.Slice(nodes, func(i, j int) bool {
    return len(v[nodes[i]]) < len(v[nodes[j]])
  })
  equal := false
  root := 0
  for i, x := range nodes {
    j := i + 1
    for ; j < len(nodes) && !g[x][nodes[j]]; j++ {
    }
    if j < len(nodes) {
      y := nodes[j]
      if len(v[x]) == len(v[y]) {
        equal = true
      }
      for _, z := range v[x] {
        if !g[y][z] {
          return 0
        }
      }
    } else {
      root++
    }
  }
  if root > 1 {
    return 0
  }
  if equal {
    return 2
  }
  return 1
}