LeetCode 914 - X of a Kind in a Deck of Cards

https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/

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.

Approach 2: Greatest Common Divisor

Again, say there are C_i cards of number i. These must be broken down into piles of X cards each, ie. C_i % X == 0 for all i.

Thus, X must divide the greatest common divisor of C_i. If this greatest common divisor g is greater than 1, then X = g will satisfy. Otherwise, it won't.

    public boolean hasGroupsSizeX(int[] deck) {

        int[] count = new int[10000];

        for (int c: deck)

            count[c]++;

        int g = -1;

        for (int i = 0; i < 10000; ++i)

            if (count[i] > 0) {

                if (g == -1)

                    g = count[i];

                else

                    g = gcd(g, count[i]);

            }

        return g >= 2;

    }

    public int gcd(int x, int y) {

        return x == 0 ? y : gcd(y%x, x);

    }

Time Complexity: $O(N \log^2 N)$, where $N$ is the number of votes. If there are $C_i$ cards with number $i$, then each gcd operation is naively $O(\log^2 C_i)$. Better bounds exist, but are outside the scope of this article to develop.
https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/discuss/175845/C%2B%2BJavaPython-Greatest-Common-Divisor

    public boolean hasGroupsSizeX(int[] deck) {
        Map<Integer, Integer> count = new HashMap<>();
        int res = 0;
        for (int i : deck) count.put(i, count.getOrDefault(i, 0) + 1);
        for (int i : count.values()) res = gcd(i, res);
        return res > 1;
    }

    public int gcd(int a, int b) {
        return b > 0 ? gcd(b, a % b) : a;
    }

X.

We can try every possible X.

Algorithm

Since we divide the deck of N cards into say, K piles of X cards each, we must have N % X == 0.

Then, say the deck has C_i copies of cards with number i. Each group with number i has X copies, so we must have C_i % X == 0. These are necessary and sufficient conditions.

Time Complexity: $O(N^2 \log \log N)$, where $N$ is the number of cards. It is outside the scope of this article to prove that the number of divisors of $N$ is bounded by $O(N \log \log N)$.

    public boolean hasGroupsSizeX(int[] deck) {

        int N = deck.length;

        int[] count = new int[10000];

        for (int c: deck)

            count[c]++;

        List<Integer> values = new ArrayList();

        for (int i = 0; i < 10000; ++i)

            if (count[i] > 0)

                values.add(count[i]);

        search: for (int X = 2; X <= N; ++X)

            if (N % X == 0) {

                for (int v: values)

                    if (v % X != 0)

                        continue search;

                return true;

            }

        return false;

    }