914.X-of-a-Kind-in-a-Deck-of-Cards.md

914. X of a Kind in a Deck of Cards

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题目地址

https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/

题目描述

In a deck of cards, each card has an integer written on it.

Return true if and only if you can choose X >= 2 such that it is possible to split the entire deck into 1 or more groups of cards, where:

Each group has exactly X cards.
All the cards in each group have the same integer.
 
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.

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

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

Example 5:
Input: deck = [1,1,2,2,2,2]
Output: true
Explanation: Possible partition [1,1],[2,2],[2,2].
 
Constraints:
1 <= deck.length <= 10^4
0 <= deck[i] < 10^4

代码

Approach #1 Brute Force

Time: O(N^2 * long log N) && Space: O(N)

the number of divisors of N is bounded by O(N log log N)

class Solution {
  public boolean hasGroupsSizeX(int[] deck) {
		int N = deck.length;
    int[] count = new int[10000];
    for (int c: deck)
      count[c]++;
    
    List<Integer> values = new ArrayList();
    for (int i = 0; i < 10000; i++) {
      if (count[i] > 0)
        values.add(count[i]);
    }
    
    search: for (int x = 2; x <= N; x++) {
      if (N % x == 0) {
        for (int v: values) {
          if (v % x != 0)
            continue search;
        }
        return true;
      }
    }
  }
}

Approach #2 Greatest Common Divisor

Time Complexity: O(N log^2 N) & Space Complexity: O(N)

class Solution {
  public boolean hasGroupsSizeX(int[] deck) {
    int[] count = new int[10000];
    for (int c: deck)
      count[c]++;
    
    int g = -1;
    for (int i = 0; i < 10000; i++) {
      if (count[i] > 0) {
        if (g == -1) {
          g = count[i];
        } else  {
        	g = gcd(g, count[i]);
      	}
      }
    }
    return g >= 2;
  }
  
  public int gcd(int x, int y) {
    return x == 0 ? y : gcd(y % x, x);
  }
  
}