781. Rabbits in Forest

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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].

有相同值的尽量设置成同一种颜色，可以排序，也可以直接用Map

class Solution:
    def numRabbits(self, answers):
        """
        :type answers: List[int]
        :rtype: int
        """
        answers.sort(reverse=True)
        res = 0
        i = 0
        while 0<=i<len(answers):
            res += answers[i]+1
            for j in range(i+1, i+answers[i]+2):
                if j>=len(answers) or answers[j]!=answers[i]:
                    break
            i = j
        return res
    
s=Solution()
print(s.numRabbits([1,1,2]))
print(s.numRabbits([10,10,10]))
print(s.numRabbits([]))
        

class Solution {
    public int numRabbits(int[] answers) {
        int res = 0;
        Map<Integer,Integer> map = new HashMap<>();
        for(int answer : answers){
            map.put(answer,map.getOrDefault(answer,0)+1);
        }
        for(Integer n : map.keySet()){
            int group = map.get(n)/(n+1);
            res += map.get(n)%(n+1) != 0 ? (group+1)*(n+1) : group*(n+1);
        }
        return res;
    }
}