[Power button - small daily practice] 914. The card packet (python)

914. The card packet

Topic links: https://leetcode-cn.com/problems/x-of-a-kind-in-a-deck-of-cards/
Difficulty: Easy

Subject description:

Given a deck of cards, each card are written on an integer.

At this point, you need to select a number X, so that we can be full decks of cards according to the following rules into 1 组or 更多组:

• Each group has a Xcard.
• Within the group are written the same integer on all the cards.

Only when you optional X >= 2returns true.

Examples

Example 1:
Input: [1,2,3,4,4,3,2,1]
Output: true
interpretation: grouping is possible [1,1], [2,2], [3,3], [ 4,4]

Example 2:
Input: [1,1,1,2,2,2,3,3]
Output: false
interpretation: the packet does not satisfy the requirements.

Example 3:
Input: [1]
Output: false
interpretation: the packet does not satisfy the requirements.

Example 4:
Input: [1,1]
Output: true
interpretation: grouping is possible [1,1]

Example 5:
Input: [1,1,2,2,2,2]
Output: true
interpretation: grouping is possible [1,1], [2,2], [2,2]

prompt：

1 <= deck.length <= 10000
0 <= deck[i] < 10000

Problem-solving ideas

This question I think for a long Leng Shimo made, or road simple questions, self-confidence is rolling, and actually did not expect to use 最大公约数to do, hey, hum

This question is crucial is that we can think of using 最大公约数to calculate.

• 1, the first statistics of the number of each element, consisting of a list

  • Simple way: num = list(collections.Counter(deck).values())one step

• 2, Step 1 list calculated greatest common divisor obtained

  • Simple way: n = reduce(math.gcd, num)one step

• 3, if the divisor than or equal to 2, True is returned, otherwise returns False

Wherein the request nnumber 最大公约数, or 最小公倍数if you want to manually realized, calculated as follows:

• 1> first find the greatest common divisor and least common multiple of the number 2
• 2> and the results of the third number and the least common multiple and the greatest common divisor continued
• 3> The same demand has been, until all numbers are again seeking

The sample code

# 简单方法
import collections
import math
from functools import reduce

class Solution:
    def hasGroupsSizeX(self, deck: List[int]) -> bool:
        num = list(collections.Counter(deck).values())
        n = reduce(math.gcd,num)
    	
        return False if n<2 else True

#该简单方法参考自：
##作者：jutraman
##链接：https://leetcode-cn.com/problems/x-of-a-kind-in-a-deck-of-cards/solution/pythonqia-pai-fen-zu-by-jutraman/

# 复杂一点的，其实就是把上述简单方法调的函数自己手动实现而已
class Solution:
    def hasGroupsSizeX(self, deck: List[int]) -> bool:
        # 第1步：统计，结果保存在num列表中
        # 简单方法：num = list(collections.Counter(deck).values())
        only_ele = [deck[0]]
        num = {deck[0]: 1}
        for i in deck[1:]:
            if i in only_ele:
                n = num[i] + 1
                num[i] = n
            else:
                num[i] = 1
                only_ele.append(i)
        num = list(num.values())
        
        if len(num)==1:
            if num[0] > 1:
                return True
            return False
         
        # 第2步：计算最大公约数n
        # 简单方法：n = reduce(math.gcd, num)
        n = self.gcd(num[0],num[1])
        for i in range(len(num)-2):
            n = gcd(n,num[i+2])
        
        # 第3步：判断最大公约数是否大于等于2
        return True if n>1 else False

    def gcd(self,a,b):# 计算两个数的最大公约数
        if a < b:#让a保持是大的数
            temp = b
            b = a
            a = temp
        n = 1
        i = 1
        while i <= b:            
            if a % i == 0 and b % i == 0:
                n = i
            i += 1
        return n