Formatted question description: https://leetcode.ca/all/914.html

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]

Note:

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

Companies:
Google

Related Topics:
Array, Math

Solution 1.

// OJ: https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/
// Time: O(N^2loglogN)
// Space: O(N)
class Solution {
public:
    bool hasGroupsSizeX(vector<int>& deck) {
        unordered_map<int, int> cnt;
        for (int n : deck) cnt[n]++;
        int minCnt = INT_MAX;
        for (auto p : cnt) minCnt = min(minCnt, p.second);
        if (minCnt == 1) return false;
        for (int x = 2; x <= minCnt; ++x) {
            bool found = true;
            for (auto p : cnt) {
                if (p.second % x) {
                    found = false;
                    break;
                }
            }
            if (found) return true;
        }
        return false;
    }
};

Java

class Solution {
    public boolean hasGroupsSizeX(int[] deck) {
        Map<Integer, Integer> countMap = new HashMap<Integer, Integer>();
        for (int num : deck) {
            int count = countMap.getOrDefault(num, 0);
            count++;
            countMap.put(num, count);
        }
        int gcd = 0;
        Set<Integer> keySet = countMap.keySet();
        for (int num : keySet) {
            int count = countMap.get(num);
            gcd = gcd(gcd, count);
        }
        return gcd >= 2;
    }

    public int gcd(int num1, int num2) {
        if (num1 == 0 && num2 == 0)
            return 1;
        while (num1 != 0 && num2 != 0) {
            if (num1 > num2) {
                int temp = num1;
                num1 = num2;
                num2 = temp;
            }
            num2 %= num1;
        }
        return num1 == 0 ? num2 : num1;
    }
}

// OJ: https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/
// Time: O(N^2loglogN)
// Space: O(N)
class Solution {
public:
    bool hasGroupsSizeX(vector<int>& deck) {
        unordered_map<int, int> cnt;
        for (int n : deck) cnt[n]++;
        int minCnt = INT_MAX;
        for (auto p : cnt) minCnt = min(minCnt, p.second);
        if (minCnt == 1) return false;
        for (int x = 2; x <= minCnt; ++x) {
            bool found = true;
            for (auto p : cnt) {
                if (p.second % x) {
                    found = false;
                    break;
                }
            }
            if (found) return true;
        }
        return false;
    }
};

class Solution(object):
    def hasGroupsSizeX(self, deck):
        """
        :type deck: List[int]
        :rtype: bool
        """
        count = collections.Counter(deck)
        X = min(count.values())
        for x in range(2, X + 1):
            if all(v % x == 0 for v in count.values()):
                return True
        return False

914 - X of a Kind in a Deck of Cards

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

Solution 1.

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