What is the main algorithmic approach to solve 'X of a Kind in a Deck of Cards'?

The main approach is to use hash tables for frequency counting, followed by computing the GCD of the frequencies to check if they can be divided evenly into groups.

How do I calculate the GCD of multiple numbers in this problem?

You can use the Euclidean algorithm to compute the GCD of the card frequencies, or use a built-in function to simplify this process.

What should I do if the GCD is 1?

If the GCD is 1, it's impossible to partition the deck into groups with the same number of identical cards, so you should return `false`.

How do I handle large input sizes for this problem?

For large inputs, the algorithm needs to efficiently count card frequencies and calculate the GCD without unnecessary recomputation. Using hash tables or dictionaries ensures this step is performed efficiently.

What edge cases should I consider when solving 'X of a Kind in a Deck of Cards'?

Consider cases where there is only one type of card, or where the cards appear in very uneven frequencies, as these cases can test the correctness of the solution.

#914

Easy

auto_awesome数组·哈希·扫描

LeetCode 题解工作台

卡牌分组

给定一副牌，每张牌上都写着一个整数。此时，你需要选定一个数字 X ，使我们可以将整副牌按下述规则分成 1 组或更多组：每组都有 X 张牌。组内所有的牌上都写着相同的整数。仅当你可选的 X >= 2 时返回 true 。示例 1：输入： deck = [1,2,3,4,4,3,2,1] 输…

数组哈希表数学计数数论

题目描述

给定一副牌，每张牌上都写着一个整数。

此时，你需要选定一个数字 X，使我们可以将整副牌按下述规则分成 1 组或更多组：

每组都有 X 张牌。
组内所有的牌上都写着相同的整数。

仅当你可选的 X >= 2 时返回 true。

示例 1：

输入：deck = [1,2,3,4,4,3,2,1]
输出：true
解释：可行的分组是 [1,1]，[2,2]，[3,3]，[4,4]

示例 2：

输入：deck = [1,1,1,2,2,2,3,3]
输出：false
解释：没有满足要求的分组。

提示：

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

lightbulb

解题思路

方法一：最大公约数

我们先用数组或哈希表 cnt 统计每个数字出现的次数，只有当 $X$ 是所有数字出现次数的约数时，即 $X$ 是所有 cnt[i] 的最大公约数的约数时，才能满足题意。

因此，我们求出所有数字出现次数的最大公约数 $g$ ，然后判断其是否大于等于 $2$ 即可。

时间复杂度 $O(n \times \log M)$ ，空间复杂度 $O(n + \log M)$ 。其中 $n$ 和 $M$ 分别是数组 deck 的长度和数组 deck 中的最大值。

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

speed

复杂度分析

指标	值
时间	O(N \log^2 N)
空间	O(N)

psychology

面试官常问的追问

外企场景

question_mark
Check if the candidate understands how to utilize hash tables for counting elements efficiently.
question_mark
Observe if they can correctly identify the need for computing the GCD to ensure divisibility.
question_mark
Pay attention to how the candidate approaches the verification of group divisibility and their understanding of the problem constraints.

warning

常见陷阱

外企场景

error
Misunderstanding the need for GCD: Candidates might incorrectly assume that the card counts should simply be equal without considering GCD as a condition for partitioning.
error
Failure to handle edge cases: Candidates should be cautious of scenarios where some cards have very few occurrences or all cards are identical.
error
Inefficient frequency counting: If a candidate does not choose an efficient way to count card frequencies, such as using a hash table or dictionary, the solution could become too slow for larger inputs.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Consider variations where the deck contains a very large number of cards, requiring an optimized approach for counting and GCD computation.
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What if some card values are missing or are spread out very unevenly? Discuss how the algorithm scales under these conditions.
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Consider a constraint where the deck has a restricted number of card types or unique values, which could lead to different optimization strategies.

help

常见问题

外企场景

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