There is a forest with an unknown number of rabbits. We asked n rabbits "How many rabbits have the same color as you?" and collected the answers in an integer array answers where answers[i] is the answer of the ith rabbit.
Given the array answers, return the minimum number of rabbits that could be in the forest.
Example 1:
Input: answers = [1,1,2] Output: 5 Explanation: The two rabbits that answered "1" could both be the same color, say red. The rabbit that 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.
Example 2:
Input: answers = [10,10,10] Output: 11
Constraints:
1 <= answers.length <= 10000 <= answers[i] < 1000This approach involves using a greedy strategy to group rabbits based on their answers. The idea is to minimize the new groups formed by properly counting how many times each answer is reported. For an answer 'x', at most 'x + 1' rabbits can share the same color including the answering rabbit. You should calculate the number of groups required to accommodate rabbits with the same answer.
This Python solution first counts the frequency of each answer using a Counter. For each unique answer 'x', it calculates how many groups of size 'x + 1' are needed to account for all rabbits that gave this answer. It then accumulates this into the total count of rabbits.
C++
Time Complexity: O(n), where n is the length of the answers array. Because we maintain a frequency count with a dictionary and process the results.
Space Complexity: O(n), which is the space required to store the counter dictionary.
In this approach, we use discrete buckets to count the occurrences of each answer. By directly iterating and placing rabbits in their respective buckets based on their answer, we can calculate the minimum number needed. This approach optimizes space and iteration by leveraging mathematical group handling of repeated counts.
This Java implementation also utilizes a hashmap to count the occurrences of each distinct answer. By calculating the minimal groups necessary with a simple ceiling division, the method aggregates the total rabbits needed.
JavaScript
Time Complexity: O(n), iterating through the array and the map.
Space Complexity: O(n), storing frequencies in the hashmap.
| Approach | Complexity |
|---|---|
| Greedy Grouping Approach | Time Complexity: O(n), where n is the length of the answers array. Because we maintain a frequency count with a dictionary and process the results. Space Complexity: O(n), which is the space required to store the counter dictionary. |
| Bucket Counting Approach | Time Complexity: O(n), iterating through the array and the map. Space Complexity: O(n), storing frequencies in the hashmap. |
Rabbits in Forest (Leetcode 781) | Hashmap Interview Question Playlist • Pepcoding • 11,169 views views
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