Question

2709. Greatest Common Divisor Traversal

中文文档

Description

You are given a 0-indexed integer array nums, and you are allowed to traverse between its indices. You can traverse between index i and index j, i != j, if and only if gcd(nums[i], nums[j]) > 1, where gcd is the greatest common divisor.

Your task is to determine if for every pair of indices i and j in nums, where i < j, there exists a sequence of traversals that can take us from i to j.

Return true_ if it is possible to traverse between all such pairs of indices,__ or _false_ otherwise._

 

Example 1:

Input: nums = [2,3,6]
Output: true
Explanation: In this example, there are 3 possible pairs of indices: (0, 1), (0, 2), and (1, 2).
To go from index 0 to index 1, we can use the sequence of traversals 0 -> 2 -> 1, where we move from index 0 to index 2 because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 > 1, and then move from index 2 to index 1 because gcd(nums[2], nums[1]) = gcd(6, 3) = 3 > 1.
To go from index 0 to index 2, we can just go directly because gcd(nums[0], nums[2]) = gcd(2, 6) = 2 > 1. Likewise, to go from index 1 to index 2, we can just go directly because gcd(nums[1], nums[2]) = gcd(3, 6) = 3 > 1.

Example 2:

Input: nums = [3,9,5]
Output: false
Explanation: No sequence of traversals can take us from index 0 to index 2 in this example. So, we return false.

Example 3:

Input: nums = [4,3,12,8]
Output: true
Explanation: There are 6 possible pairs of indices to traverse between: (0, 1), (0, 2), (0, 3), (1, 2), (1, 3), and (2, 3). A valid sequence of traversals exists for each pair, so we return true.

 

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 105

Solutions

Solution 1

class UnionFind:
  def __init__(self, n):
    self.p = list(range(n))
    self.size = [1] * n

  def find(self, x):
    if self.p[x] != x:
      self.p[x] = self.find(self.p[x])
    return self.p[x]

  def union(self, a, b):
    pa, pb = self.find(a), self.find(b)
    if pa == pb:
      return False
    if self.size[pa] > self.size[pb]:
      self.p[pb] = pa
      self.size[pa] += self.size[pb]
    else:
      self.p[pa] = pb
      self.size[pb] += self.size[pa]
    return True


mx = 100010
p = defaultdict(list)
for x in range(1, mx + 1):
  v = x
  i = 2
  while i <= v // i:
    if v % i == 0:
      p[x].append(i)
      while v % i == 0:
        v //= i
    i += 1
  if v > 1:
    p[x].append(v)


class Solution:
  def canTraverseAllPairs(self, nums: List[int]) -> bool:
    n = len(nums)
    m = max(nums)
    uf = UnionFind(n + m + 1)
    for i, x in enumerate(nums):
      for j in p[x]:
        uf.union(i, j + n)
    return len(set(uf.find(i) for i in range(n))) == 1

class UnionFind {
  private int[] p;
  private int[] size;

  public UnionFind(int n) {
    p = new int[n];
    size = new int[n];
    for (int i = 0; i < n; ++i) {
      p[i] = i;
      size[i] = 1;
    }
  }

  public int find(int x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }

  public boolean union(int a, int b) {
    int pa = find(a), pb = find(b);
    if (pa == pb) {
      return false;
    }
    if (size[pa] > size[pb]) {
      p[pb] = pa;
      size[pa] += size[pb];
    } else {
      p[pa] = pb;
      size[pb] += size[pa];
    }
    return true;
  }
}

class Solution {
  private static final int MX = 100010;
  private static final List<Integer>[] P = new List[MX];

  static {
    Arrays.setAll(P, k -> new ArrayList<>());
    for (int x = 1; x < MX; ++x) {
      int v = x;
      int i = 2;
      while (i <= v / i) {
        if (v % i == 0) {
          P[x].add(i);
          while (v % i == 0) {
            v /= i;
          }
        }
        ++i;
      }
      if (v > 1) {
        P[x].add(v);
      }
    }
  }

  public boolean canTraverseAllPairs(int[] nums) {
    int m = Arrays.stream(nums).max().getAsInt();
    int n = nums.length;
    UnionFind uf = new UnionFind(n + m + 1);
    for (int i = 0; i < n; ++i) {
      for (int j : P[nums[i]]) {
        uf.union(i, j + n);
      }
    }
    Set<Integer> s = new HashSet<>();
    for (int i = 0; i < n; ++i) {
      s.add(uf.find(i));
    }
    return s.size() == 1;
  }
}

int MX = 100010;
vector<int> P[100010];

int init = []() {
  for (int x = 1; x < MX; ++x) {
    int v = x;
    int i = 2;
    while (i <= v / i) {
      if (v % i == 0) {
        P[x].push_back(i);
        while (v % i == 0) {
          v /= i;
        }
      }
      ++i;
    }
    if (v > 1) {
      P[x].push_back(v);
    }
  }
  return 0;
}();

class UnionFind {
public:
  UnionFind(int n) {
    p = vector<int>(n);
    size = vector<int>(n, 1);
    iota(p.begin(), p.end(), 0);
  }

  bool unite(int a, int b) {
    int pa = find(a), pb = find(b);
    if (pa == pb) {
      return false;
    }
    if (size[pa] > size[pb]) {
      p[pb] = pa;
      size[pa] += size[pb];
    } else {
      p[pa] = pb;
      size[pb] += size[pa];
    }
    return true;
  }

  int find(int x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }

private:
  vector<int> p, size;
};

class Solution {
public:
  bool canTraverseAllPairs(vector<int>& nums) {
    int m = *max_element(nums.begin(), nums.end());
    int n = nums.size();
    UnionFind uf(m + n + 1);
    for (int i = 0; i < n; ++i) {
      for (int j : P[nums[i]]) {
        uf.unite(i, j + n);
      }
    }
    unordered_set<int> s;
    for (int i = 0; i < n; ++i) {
      s.insert(uf.find(i));
    }
    return s.size() == 1;
  }
};

const mx = 100010

var p = make([][]int, mx)

func init() {
  for x := 1; x < mx; x++ {
    v := x
    i := 2
    for i <= v/i {
      if v%i == 0 {
        p[x] = append(p[x], i)
        for v%i == 0 {
          v /= i
        }
      }
      i++
    }
    if v > 1 {
      p[x] = append(p[x], v)
    }
  }
}

type unionFind struct {
  p, size []int
}

func newUnionFind(n int) *unionFind {
  p := make([]int, n)
  size := make([]int, n)
  for i := range p {
    p[i] = i
    size[i] = 1
  }
  return &unionFind{p, size}
}

func (uf *unionFind) find(x int) int {
  if uf.p[x] != x {
    uf.p[x] = uf.find(uf.p[x])
  }
  return uf.p[x]
}

func (uf *unionFind) union(a, b int) bool {
  pa, pb := uf.find(a), uf.find(b)
  if pa == pb {
    return false
  }
  if uf.size[pa] > uf.size[pb] {
    uf.p[pb] = pa
    uf.size[pa] += uf.size[pb]
  } else {
    uf.p[pa] = pb
    uf.size[pb] += uf.size[pa]
  }
  return true
}

func canTraverseAllPairs(nums []int) bool {
  m := slices.Max(nums)
  n := len(nums)
  uf := newUnionFind(m + n + 1)
  for i, x := range nums {
    for _, j := range p[x] {
      uf.union(i, j+n)
    }
  }
  s := map[int]bool{}
  for i := 0; i < n; i++ {
    s[uf.find(i)] = true
  }
  return len(s) == 1
}