Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/781.html

781. Rabbits in Forest (Medium)

In a forest, each rabbit has some color. Some subset of rabbits (possibly all of them) tell you how many other rabbits have the same color as them. Those answers are placed in an array.

Return the minimum number of rabbits that could be in the forest.

Examples:
Input: answers = [1, 1, 2]
Output: 5
Explanation:
The two rabbits that answered "1" could both be the same color, say red.
The rabbit than answered "2" can't be red or the answers would be inconsistent.
Say the rabbit that answered "2" was blue.
Then there should be 2 other blue rabbits in the forest that didn't answer into the array.
The smallest possible number of rabbits in the forest is therefore 5: 3 that answered plus 2 that didn't.

Input: answers = [10, 10, 10]
Output: 11

Input: answers = []
Output: 0

Note:

  1. answers will have length at most 1000.
  2. Each answers[i] will be an integer in the range [0, 999].

Related Topics:
Hash Table, Math

Solution 1.

  • class Solution {
    public int numRabbits(int[] answers) {
        int minRabbits = 0;
        Arrays.sort(answers);
        int length = answers.length;
        int startIndex = -1;
        for (int i = 0; i < length; i++) {
            if (answers[i] == 0)
                minRabbits++;
            else {
                startIndex = i;
                break;
            }
        }
        if (startIndex < 0)
            return minRabbits;
        int prevAnswer = -1;
        int count = 0;
        for (int i = startIndex; i < length; i++) {
            int answer = answers[i];
            if (prevAnswer < 0)
                count = 1;
            else if (answer == prevAnswer)
                count++;
            else {
                int prevRabbits = minRabbitsPossible(prevAnswer, count);
                minRabbits += prevRabbits;
                count = 1;
            }
            prevAnswer = answer;
        }
        int prevRabbits = minRabbitsPossible(prevAnswer, count);
        minRabbits += prevRabbits;
        return minRabbits;
    }

    public int minRabbitsPossible(int answer, int count) {
        int groupSize = answer + 1;
        int groupsCount = (int) Math.ceil(1.0 * count / groupSize);
        return groupSize * groupsCount;
    }
}

############

class Solution {
    public int numRabbits(int[] answers) {
        Map<Integer, Integer> counter = new HashMap<>();
        for (int e : answers) {
            counter.put(e, counter.getOrDefault(e, 0) + 1);
        }
        int res = 0;
        for (Map.Entry<Integer, Integer> entry : counter.entrySet()) {
            int answer = entry.getKey(), count = entry.getValue();
            res += (int) Math.ceil(count / ((answer + 1) * 1.0)) * (answer + 1);
        }
        return res;
    }
}

  • // OJ: https://leetcode.com/problems/rabbits-in-forest/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int numRabbits(vector<int>& A) {
        unordered_map<int, int> m;
        for (int n : A) ++m[n];
        int ans = 0;
        for (auto &[n, cnt] : m) ans += (cnt + n) / (n + 1) * (n + 1);
        return ans;
    }
};

  • class Solution:
    def numRabbits(self, answers: List[int]) -> int:
        counter = Counter(answers)
        return sum([math.ceil(v / (k + 1)) * (k + 1) for k, v in counter.items()])

############

class Solution(object):
    def numRabbits(self, answers):
        """
        :type answers: List[int]
        :rtype: int
        """
        count = collections.Counter(answers)
        print count
        return sum((count[x] + x) / (x + 1) * (x + 1) for x in count)

All Problems

All Solutions