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1719. Number Of Ways To Reconstruct A Tree

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

  • 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;
    }
};

  • 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


  • 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
}

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