n 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]

Input: [1,1,1,2,2,2,3,3]
Output: false
Explanation: No possible partition.

Input: [1]
Output: false
Explanation: No possible partition.

Input: [1,1]
Output: true
Explanation: Possible partition [1,1]

Input: [1,1,2,2,2,2]
Output: true
Explanation: Possible partition [1,1],[2,2],[2,2]

// Author: Huahua, 16 ms
class Solution {
public:
  bool hasGroupsSizeX(vector<int>& deck) {
    unordered_map<int, int> counts;
    for (int card : deck) ++counts[card];
    for (int i = 2; i <= deck.size(); ++i) {
      if (deck.size() % i) continue;
      bool ok = true;
      for (const auto& pair : counts)
        if (pair.second % i) {
          ok = false;
          break;
        }
      if (ok) return true;
    }
    return false;
  }
};

// Author: Huahua, 16 ms
class Solution {
public:
  bool hasGroupsSizeX(vector<int>& deck) {
    unordered_map<int, int> counts;
    for (int card : deck) ++counts[card];
    int X = deck.size();
    for (const auto& p : counts) 
      X = __gcd(X, p.second);
    return X >= 2;
  }
};

// Author: Somebody
class Solution {
    public boolean hasGroupsSizeX(int[] deck) {
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        for (int i=0; i<deck.length; i++) {
            int curr = deck[i];
            if (map.containsKey(curr)) {
                map.put(curr, map.get(curr) + 1);
            } else {
                map.put(curr, 1);
            }
        }
        
        // O(nLogn)
        int gcd = 0;
        for(Integer key: map.keySet()) {
            gcd = findGCDOfTwoNumbers(map.get(key), gcd);
        }
        
        return gcd >= 2;      
    }
    
    // O(Log n)
    public static int findGCDOfTwoNumbers (int a, int b) {
        if (b == 0) {
            return a;
        }
        return findGCDOfTwoNumbers(b, a%b);
    }
}

The post 花花酱 LeetCode 914. X of a Kind in a Deck of Cards appeared first on Huahua's Tech Road.

Problem

Given a string S of digits, such as S = "123456579", we can split it into a Fibonacci-like sequence [123, 456, 579].

Formally, a Fibonacci-like sequence is a list F of non-negative integers such that:

• 0 <= F[i] <= 2^31 - 1, (that is, each integer fits a 32-bit signed integer type);
• F.length >= 3;
• and F[i] + F[i+1] = F[i+2] for all 0 <= i < F.length - 2.

Also, note that when splitting the string into pieces, each piece must not have extra leading zeroes, except if the piece is the number 0 itself.

Return any Fibonacci-like sequence split from S, or return [] if it cannot be done.

Example 1:

Input: "123456579"
Output: [123,456,579]

Example 2:

Input: "11235813"
Output: [1,1,2,3,5,8,13]

Example 3:

Input: "112358130"
Output: []
Explanation: The task is impossible.

Example 4:

Input: "0123"
Output: []
Explanation: Leading zeroes are not allowed, so "01", "2", "3" is not valid.

Example 5:

Input: "1101111"
Output: [110, 1, 111]
Explanation: The output [11, 0, 11, 11] would also be accepted.

Note:

1. 1 <= S.length <= 200
2. S contains only digits.

Solution: DFS

Time complexity: O(2^n)

Space complexity: O(n)

C++

// Author: Huahua
// Running time: 0 ms 
class Solution {
public:
  vector splitIntoFibonacci(string s) {
    const int n = s.length();
    vector nums;
    function dfs = [&](int pos) {
      if (pos == n) return nums.size() >= 3;
      int max_len = s[pos] == '0' ? 1 : 10;
      long num = 0;
      for (int i = pos; i < min(pos + max_len, n); ++i) {
        num = num * 10 + (s[i] - '0');
        if (num > INT_MAX) break;
        if (nums.size() >= 2) {
          long sum = nums.rbegin()[0];
          sum += nums.rbegin()[1];
          if (num > sum) break;
          else if (num < sum) continue;
          // num must equaals to sum.
        }
        nums.push_back(num);
        if (dfs(i + 1)) return true;
        nums.pop_back();
      }
      return false;
    };
    dfs(0);
    return nums;
  }
};

The post 花花酱 LeetCode 842. Split Array into Fibonacci Sequence appeared first on Huahua's Tech Road.