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

## 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)
}

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);
}

}

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