README_EN.md

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

Python3

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

Java

class Solution {
    public boolean hasGroupsSizeX(int[] deck) {
        int[] cnt = new int[10000];
        for (int v : deck) {
            ++cnt[v];
        }
        int g = -1;
        for (int v : cnt) {
            if (v > 0) {
                g = g == -1 ? v : gcd(g, v);
            }
        }
        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) {
        int cnt[10000] = {0};
        for (int& v : deck) ++cnt[v];
        int g = -1;
        for (int& v : cnt) {
            if (v) {
                g = g == -1 ? v : __gcd(g, v);
            }
        }
        return g >= 2;
    }
};

Go

func hasGroupsSizeX(deck []int) bool {
	cnt := make([]int, 10000)
	for _, v := range deck {
		cnt[v]++
	}
	g := -1
	for _, v := range cnt {
		if v > 0 {
			if g == -1 {
				g = v
			} else {
				g = gcd(g, v)
			}
		}
	}
	return g >= 2
}

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

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