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In a forest, each distinct color rabbit that you see might claim to have seen some number of other rabbits of the same color. These claims are arranged in an array. The array is such that answers[i] is the number of other rabbits that the ith rabbit claims to have seen of the same color.

Your task is to determine the minimum number of rabbits that could be in the forest based on the answers given.

Here is the function signature in Python:

def numRabbits(answers: List[int]) -> int:

Strategy:

To solve this problem, follow these steps:

  1. Group Rabbits by Answer:
    • Use a dictionary to count the occurrences of each distinct answer.
  2. Calculate Minimum Rabbits:
    • For each distinct answer x:
      • If x rabbits say they see x other rabbits, then there are x+1 rabbits in total of that color.
      • If we have count rabbits giving the same answer x, then:
        • We need to group these rabbits into groups of size x+1.
        • Each group of x+1 rabbits requires at least one more rabbit to be complete.
        • Therefore, the total groups required are ceil(count / (x+1)).
  3. Sum Up the Total Rabbits:
    • Add up the total number of rabbits for each color group calculated using the above rules.

Clarifying Questions:

Before proceeding, you might want to ask:

  1. Is there always at least one rabbit?
  2. Are there any constraints on the size of the answers array?
  3. Can answers[i] be zero, and if so, what does it mean?

Implementation:

Here’s the Python code to solve the problem:

from math import ceil
from collections import Counter
from typing import List

def numRabbits(answers: List[int]) -> int:
    count = Counter(answers)
    total_rabbits = 0
    
    for answer, num_rabbits in count.items():
        groups = ceil(num_rabbits / (answer + 1))
        total_rabbits += groups * (answer + 1)
        
    return total_rabbits

Time Complexity:

Overall, the time complexity is O(n), where n is the length of the answers list. The space complexity is also O(n) due to the space needed to store the count dictionary.

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