Question

914. X of a Kind in a Deck of Cards

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Description

You are given an integer array deck where deck[i] represents the number written on the ith card.

Partition the cards into one or more groups such that:

• Each group has exactly x cards where x > 1, and
• All the cards in one group have the same integer written on them.

Return true_ if such partition is possible, or _false_ otherwise_.

 

Example 1:

Input: deck = [1,2,3,4,4,3,2,1]
Output: true
Explanation: Possible partition [1,1],[2,2],[3,3],[4,4].

Example 2:

Input: deck = [1,1,1,2,2,2,3,3]
Output: false
Explanation: No possible partition.

 

Constraints:

• 1 <= deck.length <= 104
• 0 <= deck[i] < 104

Solutions

Solution 1

class Solution:
  def hasGroupsSizeX(self, deck: List[int]) -> bool:
    vals = Counter(deck).values()
    return reduce(gcd, vals) >= 2

class Solution {
  public boolean hasGroupsSizeX(int[] deck) {
    int[] cnt = new int[10000];
    for (int v : deck) {
      ++cnt[v];
    }
    int g = -1;
    for (int v : cnt) {
      if (v > 0) {
        g = g == -1 ? v : gcd(g, v);
      }
    }
    return g >= 2;
  }

  private int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a % b);
  }
}

class Solution {
public:
  bool hasGroupsSizeX(vector<int>& deck) {
    int cnt[10000] = {0};
    for (int& v : deck) ++cnt[v];
    int g = -1;
    for (int& v : cnt) {
      if (v) {
        g = g == -1 ? v : __gcd(g, v);
      }
    }
    return g >= 2;
  }
};

func hasGroupsSizeX(deck []int) bool {
  cnt := make([]int, 10000)
  for _, v := range deck {
    cnt[v]++
  }
  g := -1
  for _, v := range cnt {
    if v > 0 {
      if g == -1 {
        g = v
      } else {
        g = gcd(g, v)
      }
    }
  }
  return g >= 2
}

func gcd(a, b int) int {
  if b == 0 {
    return a
  }
  return gcd(b, a%b)
}