README.md

914. 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.

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. 1 <= deck.length <= 10000
2. 0 <= deck[i] < 10000

Solutions (Rust)

1. Brute Force

use std::collections::HashMap;

impl Solution {
    pub fn has_groups_size_x(deck: Vec<i32>) -> bool {
        let mut map = HashMap::new();
        for n in &deck {
            *map.entry(n).or_insert(0) += 1;
        }

        for x in (2..=(deck.len() / map.len())).filter(|x| deck.len() % x == 0) {
            if map.values().all(|v| v % x == 0) {
                return true;
            }
        }

        false
    }
}

2. GCD

use std::collections::HashMap;

impl Solution {
    pub fn has_groups_size_x(deck: Vec<i32>) -> bool {
        let mut map = HashMap::new();
        for n in deck {
            *map.entry(n).or_insert(0) += 1;
        }

        let mut x = *map.values().nth(0).unwrap();
        map.values().fold(x, |x, y| Self::gcd(x, *y)) > 1
    }

    pub fn gcd(mut x: i32, mut y: i32) -> i32 {
        while x % y != 0 {
            let tmp = x;
            x = y;
            y = tmp % y;
        }
        y
    }
}