Question

1617. Count Subtrees With Max Distance Between Cities

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Description

There are n cities numbered from 1 to n. You are given an array edges of size n-1, where edges[i] = [ui, vi] represents a bidirectional edge between cities ui and vi. 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 _dth_ __element (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 <= ui, vi <= n
• All pairs (ui, vi) are distinct.

Solutions

Solution 1

class Solution:
  def countSubgraphsForEachDiameter(
    self, n: int, edges: List[List[int]]
  ) -> List[int]:
    def dfs(u: int, d: int = 0):
      nonlocal mx, nxt, msk
      if mx < d:
        mx, nxt = d, u
      msk ^= 1 << u
      for v in g[u]:
        if msk >> v & 1:
          dfs(v, d + 1)

    g = defaultdict(list)
    for u, v in edges:
      u, v = u - 1, v - 1
      g[u].append(v)
      g[v].append(u)
    ans = [0] * (n - 1)
    nxt = mx = 0
    for mask in range(1, 1 << n):
      if mask & (mask - 1) == 0:
        continue
      msk, mx = mask, 0
      cur = msk.bit_length() - 1
      dfs(cur)
      if msk == 0:
        msk, mx = mask, 0
        dfs(nxt)
        ans[mx - 1] += 1
    return ans

class Solution {
  private List<Integer>[] g;
  private int msk;
  private int nxt;
  private int mx;

  public int[] countSubgraphsForEachDiameter(int n, int[][] edges) {
    g = new List[n];
    Arrays.setAll(g, k -> new ArrayList<>());
    for (int[] e : edges) {
      int u = e[0] - 1, v = e[1] - 1;
      g[u].add(v);
      g[v].add(u);
    }
    int[] ans = new int[n - 1];
    for (int mask = 1; mask < 1 << n; ++mask) {
      if ((mask & (mask - 1)) == 0) {
        continue;
      }
      msk = mask;
      mx = 0;
      int cur = 31 - Integer.numberOfLeadingZeros(msk);
      dfs(cur, 0);
      if (msk == 0) {
        msk = mask;
        mx = 0;
        dfs(nxt, 0);
        ++ans[mx - 1];
      }
    }
    return ans;
  }

  private void dfs(int u, int d) {
    msk ^= 1 << u;
    if (mx < d) {
      mx = d;
      nxt = u;
    }
    for (int v : g[u]) {
      if ((msk >> v & 1) == 1) {
        dfs(v, d + 1);
      }
    }
  }
}

class Solution {
public:
  vector<int> countSubgraphsForEachDiameter(int n, vector<vector<int>>& edges) {
    vector<vector<int>> g(n);
    for (auto& e : edges) {
      int u = e[0] - 1, v = e[1] - 1;
      g[u].emplace_back(v);
      g[v].emplace_back(u);
    }
    vector<int> ans(n - 1);
    int nxt = 0, msk = 0, mx = 0;
    function<void(int, int)> dfs = [&](int u, int d) {
      msk ^= 1 << u;
      if (mx < d) {
        mx = d;
        nxt = u;
      }
      for (int& v : g[u]) {
        if (msk >> v & 1) {
          dfs(v, d + 1);
        }
      }
    };
    for (int mask = 1; mask < 1 << n; ++mask) {
      if ((mask & (mask - 1)) == 0) {
        continue;
      }
      msk = mask;
      mx = 0;
      int cur = 31 - __builtin_clz(msk);
      dfs(cur, 0);
      if (msk == 0) {
        msk = mask;
        mx = 0;
        dfs(nxt, 0);
        ++ans[mx - 1];
      }
    }
    return ans;
  }
};

func countSubgraphsForEachDiameter(n int, edges [][]int) []int {
  g := make([][]int, n)
  for _, e := range edges {
    u, v := e[0]-1, e[1]-1
    g[u] = append(g[u], v)
    g[v] = append(g[v], u)
  }
  ans := make([]int, n-1)
  var msk, nxt, mx int
  var dfs func(int, int)
  dfs = func(u, d int) {
    msk ^= 1 << u
    if mx < d {
      mx, nxt = d, u
    }
    for _, v := range g[u] {
      if msk>>v&1 == 1 {
        dfs(v, d+1)
      }
    }
  }
  for mask := 1; mask < 1<<n; mask++ {
    if mask&(mask-1) == 0 {
      continue
    }
    msk, mx = mask, 0
    cur := bits.Len(uint(msk)) - 1
    dfs(cur, 0)
    if msk == 0 {
      msk, mx = mask, 0
      dfs(nxt, 0)
      ans[mx-1]++
    }
  }
  return ans
}

function countSubgraphsForEachDiameter(n: number, edges: number[][]): number[] {
  const g = Array.from({ length: n }, () => []);
  for (const [u, v] of edges) {
    g[u - 1].push(v - 1);
    g[v - 1].push(u - 1);
  }
  const ans: number[] = new Array(n - 1).fill(0);
  let [mx, msk, nxt] = [0, 0, 0];
  const dfs = (u: number, d: number) => {
    if (mx < d) {
      mx = d;
      nxt = u;
    }
    msk ^= 1 << u;
    for (const v of g[u]) {
      if ((msk >> v) & 1) {
        dfs(v, d + 1);
      }
    }
  };
  for (let mask = 1; mask < 1 << n; ++mask) {
    if ((mask & (mask - 1)) === 0) {
      continue;
    }
    msk = mask;
    mx = 0;
    const cur = 31 - numberOfLeadingZeros(msk);
    dfs(cur, 0);
    if (msk === 0) {
      msk = mask;
      mx = 0;
      dfs(nxt, 0);
      ++ans[mx - 1];
    }
  }
  return ans;
}

function numberOfLeadingZeros(i: number): number {
  if (i == 0) return 32;
  let n = 1;
  if (i >>> 16 == 0) {
    n += 16;
    i <<= 16;
  }
  if (i >>> 24 == 0) {
    n += 8;
    i <<= 8;
  }
  if (i >>> 28 == 0) {
    n += 4;
    i <<= 4;
  }
  if (i >>> 30 == 0) {
    n += 2;
    i <<= 2;
  }
  n -= i >>> 31;
  return n;
}

Solution 2

class Solution:
  def countSubgraphsForEachDiameter(
    self, n: int, edges: List[List[int]]
  ) -> List[int]:
    def bfs(u: int) -> int:
      d = -1
      q = deque([u])
      nonlocal msk, nxt
      msk ^= 1 << u
      while q:
        d += 1
        for _ in range(len(q)):
          nxt = u = q.popleft()
          for v in g[u]:
            if msk >> v & 1:
              msk ^= 1 << v
              q.append(v)
      return d

    g = defaultdict(list)
    for u, v in edges:
      u, v = u - 1, v - 1
      g[u].append(v)
      g[v].append(u)
    ans = [0] * (n - 1)
    nxt = 0
    for mask in range(1, 1 << n):
      if mask & (mask - 1) == 0:
        continue
      msk = mask
      cur = msk.bit_length() - 1
      bfs(cur)
      if msk == 0:
        msk = mask
        mx = bfs(nxt)
        ans[mx - 1] += 1
    return ans

class Solution {
  private List<Integer>[] g;
  private int msk;
  private int nxt;

  public int[] countSubgraphsForEachDiameter(int n, int[][] edges) {
    g = new List[n];
    Arrays.setAll(g, k -> new ArrayList<>());
    for (int[] e : edges) {
      int u = e[0] - 1, v = e[1] - 1;
      g[u].add(v);
      g[v].add(u);
    }
    int[] ans = new int[n - 1];
    for (int mask = 1; mask < 1 << n; ++mask) {
      if ((mask & (mask - 1)) == 0) {
        continue;
      }
      msk = mask;
      int cur = 31 - Integer.numberOfLeadingZeros(msk);
      bfs(cur);
      if (msk == 0) {
        msk = mask;
        int mx = bfs(nxt);
        ++ans[mx - 1];
      }
    }
    return ans;
  }

  private int bfs(int u) {
    int d = -1;
    Deque<Integer> q = new ArrayDeque<>();
    q.offer(u);
    msk ^= 1 << u;
    while (!q.isEmpty()) {
      ++d;
      for (int k = q.size(); k > 0; --k) {
        u = q.poll();
        nxt = u;
        for (int v : g[u]) {
          if ((msk >> v & 1) == 1) {
            msk ^= 1 << v;
            q.offer(v);
          }
        }
      }
    }
    return d;
  }
}

class Solution {
public:
  vector<int> countSubgraphsForEachDiameter(int n, vector<vector<int>>& edges) {
    vector<vector<int>> g(n);
    for (auto& e : edges) {
      int u = e[0] - 1, v = e[1] - 1;
      g[u].emplace_back(v);
      g[v].emplace_back(u);
    }
    vector<int> ans(n - 1);
    int nxt = 0, msk = 0;
    auto bfs = [&](int u) -> int {
      int d = -1;
      msk ^= 1 << u;
      queue<int> q{{u}};
      while (!q.empty()) {
        ++d;
        for (int k = q.size(); k; --k) {
          u = q.front();
          nxt = u;
          q.pop();
          for (int& v : g[u]) {
            if (msk >> v & 1) {
              msk ^= 1 << v;
              q.push(v);
            }
          }
        }
      }
      return d;
    };
    for (int mask = 1; mask < 1 << n; ++mask) {
      if ((mask & (mask - 1)) == 0) {
        continue;
      }
      msk = mask;
      int cur = 31 - __builtin_clz(msk);
      bfs(cur);
      if (msk == 0) {
        msk = mask;
        int mx = bfs(nxt);
        ++ans[mx - 1];
      }
    }
    return ans;
  }
};

func countSubgraphsForEachDiameter(n int, edges [][]int) []int {
  g := make([][]int, n)
  for _, e := range edges {
    u, v := e[0]-1, e[1]-1
    g[u] = append(g[u], v)
    g[v] = append(g[v], u)
  }
  ans := make([]int, n-1)
  var msk, nxt int
  bfs := func(u int) int {
    d := -1
    q := []int{u}
    msk ^= 1 << u
    for len(q) > 0 {
      d++
      for k := len(q); k > 0; k-- {
        u = q[0]
        q = q[1:]
        nxt = u
        for _, v := range g[u] {
          if msk>>v&1 == 1 {
            msk ^= 1 << v
            q = append(q, v)
          }
        }
      }
    }
    return d
  }
  for mask := 1; mask < 1<<n; mask++ {
    if mask&(mask-1) == 0 {
      continue
    }
    msk = mask
    cur := bits.Len(uint(msk)) - 1
    bfs(cur)
    if msk == 0 {
      msk = mask
      mx := bfs(nxt)
      ans[mx-1]++
    }
  }
  return ans
}

function countSubgraphsForEachDiameter(n: number, edges: number[][]): number[] {
  const g = Array.from({ length: n }, () => []);
  for (const [u, v] of edges) {
    g[u - 1].push(v - 1);
    g[v - 1].push(u - 1);
  }
  const ans: number[] = new Array(n - 1).fill(0);
  let [msk, nxt] = [0, 0];
  const bfs = (u: number) => {
    let d = -1;
    const q = [u];
    msk ^= 1 << u;
    while (q.length) {
      ++d;
      for (let k = q.length; k; --k) {
        u = q.shift()!;
        nxt = u;
        for (const v of g[u]) {
          if ((msk >> v) & 1) {
            msk ^= 1 << v;
            q.push(v);
          }
        }
      }
    }
    return d;
  };
  for (let mask = 1; mask < 1 << n; ++mask) {
    if ((mask & (mask - 1)) === 0) {
      continue;
    }
    msk = mask;
    const cur = 31 - numberOfLeadingZeros(msk);
    bfs(cur);
    if (msk === 0) {
      msk = mask;
      const mx = bfs(nxt);
      ++ans[mx - 1];
    }
  }
  return ans;
}

function numberOfLeadingZeros(i: number): number {
  if (i == 0) return 32;
  let n = 1;
  if (i >>> 16 == 0) {
    n += 16;
    i <<= 16;
  }
  if (i >>> 24 == 0) {
    n += 8;
    i <<= 8;
  }
  if (i >>> 28 == 0) {
    n += 4;
    i <<= 4;
  }
  if (i >>> 30 == 0) {
    n += 2;
    i <<= 2;
  }
  n -= i >>> 31;
  return n;
}