README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0900-0999/0914.X%20of%20a%20Kind%20in%20a%20Deck%20of%20Cards/README_EN.md	Array Hash Table Math Counting Number Theory

914. X of a Kind in a Deck of Cards

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Description

You are given an integer array deck where deck[i] represents the number written on the ith card.

Partition the cards into one or more groups such that:

• Each group has exactly x cards where x > 1, and
• All the cards in one group have the same integer written on them.

Return true if such partition is possible, or false otherwise.

 

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.

 

Constraints:

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

Solutions

Solution 1: Greatest Common Divisor

First, we use an array or hash table cnt to count the occurrence of each number. Only when $X$ is a divisor of the greatest common divisor of all cnt[i], can it satisfy the problem's requirement.

Therefore, we find the greatest common divisor $g$ of the occurrence of all numbers, and then check whether it is greater than or equal to $2$.

The time complexity is $O(n \times \log M)$, and the space complexity is $O(n + \log M)$. Where $n$ and $M$ are the length of the array deck and the maximum value in the array deck, respectively.

Python3

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

Java

class Solution {
    public boolean hasGroupsSizeX(int[] deck) {
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int x : deck) {
            cnt.merge(x, 1, Integer::sum);
        }
        int g = cnt.get(deck[0]);
        for (int x : cnt.values()) {
            g = gcd(g, x);
        }
        return g >= 2;
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}

C++

class Solution {
public:
    bool hasGroupsSizeX(vector<int>& deck) {
        unordered_map<int, int> cnt;
        for (int x : deck) {
            ++cnt[x];
        }
        int g = cnt[deck[0]];
        for (auto& [_, x] : cnt) {
            g = gcd(g, x);
        }
        return g >= 2;
    }
};

Go

func hasGroupsSizeX(deck []int) bool {
	cnt := map[int]int{}
	for _, x := range deck {
		cnt[x]++
	}
	g := cnt[deck[0]]
	for _, x := range cnt {
		g = gcd(g, x)
	}
	return g >= 2
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}

TypeScript

function hasGroupsSizeX(deck: number[]): boolean {
    const cnt: Record<number, number> = {};
    for (const x of deck) {
        cnt[x] = (cnt[x] || 0) + 1;
    }
    const gcd = (a: number, b: number): number => (b === 0 ? a : gcd(b, a % b));
    let g = cnt[deck[0]];
    for (const [_, x] of Object.entries(cnt)) {
        g = gcd(g, x);
    }
    return g >= 2;
}