花花酱 LeetCode 1617. Count Subtrees With Max Distance Between Cities

zxi — Sun, 11 Oct 2020 20:08:05 +0000

There are n cities numbered from 1 to n. You are given an array edges of size n-1, where edges[i] = [u_i, v_i] represents a bidirectional edge between cities u_i and v_i. There exists a unique path between each pair of cities. In other words, the cities form a tree.

A subtree is a subset of cities where every city is reachable from every other city in the subset, where the path between each pair passes through only the cities from the subset. Two subtrees are different if there is a city in one subtree that is not present in the other.

For each d from 1 to n-1, find the number of subtrees in which the maximum distance between any two cities in the subtree is equal to d.

Return an array of size n-1 where the d^thelement (1-indexed) is the number of subtrees in which the maximum distance between any two cities is equal to d.

Notice that the distance between the two cities is the number of edges in the path between them.

Example 1:

Input: n = 4, edges = [[1,2],[2,3],[2,4]]
Output: [3,4,0]
Explanation:
The subtrees with subsets {1,2}, {2,3} and {2,4} have a max distance of 1.
The subtrees with subsets {1,2,3}, {1,2,4}, {2,3,4} and {1,2,3,4} have a max distance of 2.
No subtree has two nodes where the max distance between them is 3.

Example 2:

Input: n = 2, edges = [[1,2]]
Output: [1]

Example 3:

Input: n = 3, edges = [[1,2],[2,3]]
Output: [2,1]

Constraints:

2 <= n <= 15
edges.length == n-1
edges[i].length == 2
1 <= u_i, v_i <= n
All pairs (u_i, v_i) are distinct.

Solution1: Brute Force + Diameter of tree

Try all subtrees and find the diameter of that subtree (longest distance between any node)

Time complexity: O(2^n * n)
Space complexity: O(n)

C++

// Author: Huahua
class Solution {
public:
  vector countSubgraphsForEachDiameter(int n, vector>& edges, int last = -1) {
    vector> g(n);
    for (const auto& e : edges)
      g[e[0] - 1].push_back(e[1] - 1), g[e[1] - 1].push_back(e[0] - 1);
    
    vector seen(n, -1), seen2;
    queue q;
    
    auto bfs = [&](int start, vector& seen) -> pair {
      q.push(start);
      seen[start] = 1;
      int count = 0, dist = -1;
      while (!q.empty()) {
        int s = q.size();        
        while (s--) {
          last = q.front(); q.pop();
          ++count;
          for (int v : g[last])
            if (seen[v] != -1 && !seen[v]++) q.push(v);
        }
        ++dist;
      }
      return {dist, count};
    };
    
    vector ans(n - 1);
    for (int s = 0; s < 1 << n; ++s) {
      if (__builtin_popcount(s) <= 1) continue;
      fill(begin(seen), end(seen), -1);
      for (int i = 0; i < n; ++i) if (s & (1 << i)) seen[i] = 0;
      seen2 = seen;
      const int start = 31 - __builtin_clz(s & (s - 1));
      if (bfs(start, seen).second != __builtin_popcount(s)) continue;      
      ++ans[bfs(last, seen2).first - 1];
    }
    return ans;
  }
};

Solution 2: DP on Trees

dp[i][k][d] := # of subtrees rooted at i with tree diameter of d and the distance from i to the farthest node is k.

Time complexity: O(n^5)
Space complexity: O(n^3)

C++

// Author: Huahua, 4ms, 8.8MB
class Solution {
public:
  vector countSubgraphsForEachDiameter(int n, vector>& edges) {    
    vector> g(n);
    for (const auto& e : edges) {
      g[e[0] - 1].push_back(e[1] - 1);
      g[e[1] - 1].push_back(e[0] - 1);
    }
    vector>> dp(n);    
    vector sizes(n);
    function dfs = [&](int u, int p) {
      if (!dp[u].empty()) return;      
      dp[u] = vector>(n, vector(n));
      dp[u][0][0] = 1;
      sizes[u] = 1;
      for (int v : g[u]) {
        if (v == p) continue;
        dfs(v, u);
        vector> dpu(dp[u]);
        for (int d1 = 0; d1 < sizes[u]; ++d1)
          for (int k1 = 0; k1 <= d1; ++k1) {
            if (!dp[u][k1][d1]) continue;
            for (int d2 = 0; d2 < sizes[v]; ++d2)
              for (int k2 = 0; k2 <= d2; ++k2) {
                const int d = max({d1, d2, k1 + k2 + 1});
                const int k = max(k1, k2 + 1);
                dpu[k][d] += dp[u][k1][d1] * dp[v][k2][d2];
              }
          }
        swap(dpu, dp[u]);
        sizes[u] += sizes[v];
      }
    };
    vector ans(n - 1);
    dfs(0, -1);
    for (int i = 0; i < n; ++i) 
      for (int k = 0; k < n; ++k)
        for (int d = 1; d < n; ++d)
          ans[d - 1] += dp[i][k][d];
    return ans;
  }
};

The post 花花酱 LeetCode 1617. Count Subtrees With Max Distance Between Cities appeared first on Huahua's Tech Road.

花花酱 LeetCode 1372. Longest ZigZag Path in a Binary Tree

zxi — Sun, 08 Mar 2020 08:53:52 +0000

Given a binary tree root, a ZigZag path for a binary tree is defined as follow:

Choose any node in the binary tree and a direction (right or left).
If the current direction is right then move to the right child of the current node otherwise move to the left child.
Change the direction from right to left or right to left.
Repeat the second and third step until you can’t move in the tree.

Zigzag length is defined as the number of nodes visited – 1. (A single node has a length of 0).

Return the longest ZigZag path contained in that tree.

Example 1:

Input: root = [1,null,1,1,1,null,null,1,1,null,1,null,null,null,1,null,1]
Output: 3
Explanation: Longest ZigZag path in blue nodes (right -> left -> right).

Example 2:

Input: root = [1,1,1,null,1,null,null,1,1,null,1]
Output: 4
Explanation: Longest ZigZag path in blue nodes (left -> right -> left -> right).

Example 3:

Input: root = [1]
Output: 0

Constraints:

Each tree has at most 50000 nodes..
Each node’s value is between [1, 100].

Solution: Recursion

For each node return
1. max ZigZag length if go left
2. max ZigZag length if go right
3. maz ZigZag length within the subtree

ZigZag(root):
ll, lr, lm = ZigZag(root.left)
rl, rr, rm = ZigZag(root.right)
return (lr+1, rl + 1, max(lr+1, rl+1, lm, rm))

Time complexity: O(n)
Space complexity: O(h)

C++

// Author: Huahua
class Solution {
public:
  int longestZigZag(TreeNode* root) {    
    return get<2>(ZigZag(root));
  }
    
  // Returns {left, right, max}
  tuple ZigZag(TreeNode* root) {
    if (!root) return {-1, -1, -1};
    auto [ll, lr, lm] = ZigZag(root->left);
    auto [rl, rr, rm] = ZigZag(root->right);
    int l = lr + 1;
    int r = rl + 1;
    return {l, r, max({l, r, lm, rm})};
  }
};

Python3

# Author: Huahua
class Solution:
  def longestZigZag(self, root: TreeNode) -> int:
    def ZigZag(node):
      if not node: return (-1, -1, -1)
      ll, lr, lm = ZigZag(node.left)
      rl, rr, rm = ZigZag(node.right)
      return (lr + 1, rl + 1, max(lr + 1, rl + 1, lm, rm))
    return ZigZag(root)[2]

The post 花花酱 LeetCode 1372. Longest ZigZag Path in a Binary Tree appeared first on Huahua's Tech Road.

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