914. X of a Kind in a Deck of Cards （easy）

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: [1,2,3,4,4,3,2,1]
Output: true
Explanation: Possible partition [1,1],[2,2],[3,3],[4,4]

Example 2:

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

Example 3:

Input: [1]
Output: false
Explanation: No possible partition.
 

Example 4:

Input: [1,1]
Output: true
Explanation: Possible partition [1,1]

Example 5:

Input: [1,1,2,2,2,2]
Output: true
Explanation: Possible partition [1,1],[2,2],[2,2]

1 暴力破解(其實very fast Or2

Runtime: 44 ms, faster than 99.67% of Python3

class Solution(object):
    def hasGroupsSizeX(self, deck):
        count = collections.Counter(deck)
        N = len(deck)
        for X in range(2, N+1):
            if N % X == 0:
                if all(v % X == 0 for v in count.values()):
                    return True
        return False
 

2 awkard & stupid me

思路：用存在公倍數求解

e.g.[1,2,3,4,4,3,2,1] ->set = [1,2,3,4] 每個數在deck的count都是2個，公倍數都是2 -》true

e.g.[1,1,1,2,2,2,3,3] -》set = [1,2,3] 每個數在deck的count分別為3，3，2，不存在相同的公倍數 -》false

我的笨辦法求相同的公倍數，先求set裡第一個數的count（即deck.count(set[0])）的公倍數list(所有公倍數)：gbs

遍歷set的其他數，如果gbs裡的數不是遍歷到的這個數count的公倍數 則移除

最後剩下的gbs是所有數count的共同公倍數

class Solution:
    def hasGroupsSizeX(self, deck):
        """
        :type deck: List[int]
        :rtype: bool
        """

       
        setdeck = list(set(deck))
        if len(deck) < 2 or deck.count(setdeck[0]) == 1:return False
        gbs = []
        for i in range(2,deck.count(setdeck[0])+1):
            if deck.count(setdeck[0]) % i == 0:
                gbs.append(i)
        for x in setdeck:
            num = deck.count(x)
            aa = gbs.copy()
            for gb in aa:
                if num% gb != 0:gbs.remove(gb)
        return True if len(gbs) >= 1 else False

1528ms 0.99% Or2

3 結合改進（fast

每一個迴圈都計算count很耗時，發現collections.count先把所有數字出現的次數計算了，return一個字典dict{數字：次數}

class Solution:
    def hasGroupsSizeX(self, deck):
        """
        :type deck: List[int]
        :rtype: bool
        """
        count = collections.Counter(deck)
        setdeck = list(set(deck))
        if len(deck) < 2 or deck.count(setdeck[0]) == 1:return False
        gbs = []
        for i in range(2,count[setdeck[0]]+1):
            if count[setdeck[0]]% i == 0:gbs.append(i)
        for x in setdeck:
            aa = gbs.copy()
            for a in aa:
                if count[x]%a !=0:gbs.remove(a)
        return True if len(gbs)>=1 else False

Runtime: 44 ms, faster than 99.67% of Python3