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Rabbits in Forest

#1822 Most AskedMedium

Rabbits in Forest #1822 Most Asked

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Topics:

ArraysGreedy Algorithms

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 i^th 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 <= 1000
0 <= answers[i] < 1000

Try coding it in LeetCode

Rabbits in Forest

#1822 Most AskedMedium

Rabbits in Forest #1822 Most Asked

Medium

Google

9 views

Topics:

ArraysGreedy Algorithms

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 <= 1000
0 <= answers[i] < 1000

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of possible values for the number of rabbits reporting the same color?
Can the input array `answers` be empty or null?
If the input `answers` leads to multiple valid solutions, can I return any one of them?
Can the same number appear multiple times in the `answers` array, and if so, does each occurrence represent a separate group of rabbits or are they all from the same group?
Does the `answers` array contain only non-negative integers?

Brute Force Solution

Approach

The brute-force method in this problem is like trying every possible combination of rabbit groups to see which one requires the fewest number of rabbits in total. We explore all the options until we find the one that works according to the rules of the problem.

Here's how the algorithm would work step-by-step:

Consider each number reported by a rabbit as the size of a group of rabbits.
Start by assuming that each rabbit who said '0' represents a separate group of one rabbit. Count these rabbits.
For all other numbers, try treating them as though they came from completely separate groups. For example, if two rabbits say '1', assume they are in separate groups of two.
Count the total number of rabbits this would require.
Then, consider the case where some of the rabbits reporting the same number belong to the same group. For example, if two rabbits said '1', now assume they belong to the same group of two.
Make sure if multiple rabbits claim to be in the same group of a certain size, the number of those rabbits does not exceed that group size.
Calculate the total number of rabbits required in each grouping scenario.
Keep track of the smallest total number of rabbits you find that fits the description reported by all the rabbits.

Code Implementation

def rabbits_in_forest(answers):
    minimum_number_of_rabbits = float('inf')

    # Consider each number reported by a rabbit as the size of a group
    unique_answers = set(answers)

    for answer in unique_answers:
        total_rabbits = 0

        # Handle the case where rabbits say '0' separately
        zero_count = answers.count(0)
        if 0 in answers:
            total_rabbits += zero_count

        non_zero_answers = [a for a in answers if a != 0]
        counts = {}
        for a in non_zero_answers:
            counts[a] = counts.get(a, 0) + 1

        # Iterate through the counts to calculate the min number of rabbits.
        for reported_number, number_of_rabbits_reporting in counts.items():

            # Calculate number of groups based on reports. 
            number_of_groups = (number_of_rabbits_reporting + reported_number) // (reported_number + 1)

            # Add rabbits in the group to the total.
            total_rabbits += number_of_groups * (reported_number + 1)

        minimum_number_of_rabbits = min(minimum_number_of_rabbits, total_rabbits)

    return minimum_number_of_rabbits

Big(O) Analysis

Time Complexity

O(n!) – The described brute-force approach explores every possible grouping of rabbits, which is equivalent to generating all partitions of a set. Consider 'n' rabbits reporting different numbers. For each number reported, the algorithm attempts every possible combination of treating them as separate groups or merging them. The number of possible combinations grows factorially with 'n' because it needs to consider all possible ways of partitioning 'n' elements into different groups. Therefore, the runtime is proportional to the number of possible groupings, which can be approximated as O(n!).

Space Complexity

O(N) – The described brute-force approach potentially explores different grouping scenarios. To efficiently keep track of the counts of rabbits reporting each number, a hash map or an array of size N (where N is the maximum possible reported number, as each reported number indicates a group size) can be used. This hash map stores the counts of each reported number which contributes to the space complexity. Thus, auxiliary space usage scales linearly with the maximum possible reported number. Therefore, the space complexity is O(N).

Optimal Solution

Approach

To count the minimum number of rabbits, we need to avoid overcounting. The key idea is to realize that rabbits claiming the same number of same-color siblings must belong to the same group. By grouping these rabbits intelligently, we can minimize the total rabbit count.

Here's how the algorithm would work step-by-step:

Imagine you interview each rabbit and ask how many other rabbits have the same color as them.
Create a way to remember the number of siblings each rabbit reports.
If a rabbit says it has zero siblings of the same color, that means it's the only rabbit of that color, so count one rabbit for that color.
If a rabbit says it has 'n' siblings of the same color, it means that entire group has 'n + 1' rabbits in it.
If you see multiple rabbits reporting the same number of siblings, say 'n', you know that there is a group of 'n + 1' rabbits of that color.
If you see more than 'n + 1' rabbits reporting 'n' siblings, it means there are multiple groups of that color. Calculate how many full groups of 'n+1' rabbits exist. The amount of extra rabbits that exist forms an entirely new group of size 'n+1'
Add up the sizes of all of these groups of rabbits to get the total number of rabbits in the forest.

Code Implementation

def rabbits_in_forest(answers):    sibling_counts = {}
    for answer in answers:
        sibling_counts[answer] = sibling_counts.get(answer, 0) + 1

    total_rabbits = 0
    for number_of_same_color_siblings, count in sibling_counts.items():
        # If a rabbit says 0, it's a group of 1
        if number_of_same_color_siblings == 0:
            total_rabbits += count
        else:
            group_size = number_of_same_color_siblings + 1
            # Need to account for full groups and potentially a partial group
            number_of_full_groups = count // group_size

            if count % group_size == 0:
                total_rabbits += number_of_full_groups * group_size
            else:
                # Remainder forms a new group
                total_rabbits += (number_of_full_groups + 1) * group_size

    return total_rabbits

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n (representing the rabbit answers) once to count the frequency of each answer. The most computationally expensive operation inside the loop is incrementing the count of a particular answer in a hashmap or similar data structure, which takes O(1) time on average. Subsequently, we iterate through the unique rabbit answers, but this number is at most n, so this takes O(n) time. Therefore, the overall time complexity is dominated by the single loop through the input array, resulting in O(n) time complexity, where n is the number of rabbits.

Space Complexity

O(N) – The algorithm creates a data structure (presumably a hash map or array) to remember the number of siblings each rabbit reports. The size of this data structure is proportional to the number of unique answers given by the rabbits, which in the worst case could be all N rabbits giving different answers. Therefore, the auxiliary space required is O(N), where N is the number of rabbits (size of the input). Other variables use constant space.

Edge Cases

Case	How to Handle
Empty answers array	If the input array is empty, return 0 because no rabbits exist.
All answers are 0	The solution should return the length of the answers array, because each rabbit that answers 0 could be the only rabbit in its group.
Answers contain only 1s	The solution should return the length of the answers array, since each rabbit that answers 1 could be in a group of 2.
Answers contain a large number (close to the max integer)	The solution handles this by checking for positive values of x and ensures valid groups are counted.
Answers contain duplicate numbers, but not enough to form full groups based on those counts.	The solution must appropriately round up the number of groups formed.
Input array has only one element	The solution should create one group of size answers[0] + 1.
Input with duplicate counts that would overfill groups.	When count is greater than expected, start a new group for those excess rabbits
Integer overflow when calculating group size (answer[i] + 1)	Handle integer overflow by checking the result or by constraining the inputs

Empty answers array

How to Handle:

If the input array is empty, return 0 because no rabbits exist.

All answers are 0

How to Handle:

The solution should return the length of the answers array, because each rabbit that answers 0 could be the only rabbit in its group.

Answers contain only 1s

How to Handle:

The solution should return the length of the answers array, since each rabbit that answers 1 could be in a group of 2.

Answers contain a large number (close to the max integer)

How to Handle:

The solution handles this by checking for positive values of x and ensures valid groups are counted.

Answers contain duplicate numbers, but not enough to form full groups based on those counts.

How to Handle:

The solution must appropriately round up the number of groups formed.

Input array has only one element

How to Handle:

The solution should create one group of size answers[0] + 1.

Input with duplicate counts that would overfill groups.

How to Handle:

When count is greater than expected, start a new group for those excess rabbits

Integer overflow when calculating group size (answer[i] + 1)

How to Handle:

Handle integer overflow by checking the result or by constraining the inputs

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of possible values for the number of rabbits reporting the same color?
Can the input array `answers` be empty or null?
If the input `answers` leads to multiple valid solutions, can I return any one of them?
Can the same number appear multiple times in the `answers` array, and if so, does each occurrence represent a separate group of rabbits or are they all from the same group?
Does the `answers` array contain only non-negative integers?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Consider each number reported by a rabbit as the size of a group of rabbits.
Start by assuming that each rabbit who said '0' represents a separate group of one rabbit. Count these rabbits.
For all other numbers, try treating them as though they came from completely separate groups. For example, if two rabbits say '1', assume they are in separate groups of two.
Count the total number of rabbits this would require.
Then, consider the case where some of the rabbits reporting the same number belong to the same group. For example, if two rabbits said '1', now assume they belong to the same group of two.
Make sure if multiple rabbits claim to be in the same group of a certain size, the number of those rabbits does not exceed that group size.
Calculate the total number of rabbits required in each grouping scenario.
Keep track of the smallest total number of rabbits you find that fits the description reported by all the rabbits.

Code Implementation

def rabbits_in_forest(answers):
    minimum_number_of_rabbits = float('inf')

    # Consider each number reported by a rabbit as the size of a group
    unique_answers = set(answers)

    for answer in unique_answers:
        total_rabbits = 0

        # Handle the case where rabbits say '0' separately
        zero_count = answers.count(0)
        if 0 in answers:
            total_rabbits += zero_count

        non_zero_answers = [a for a in answers if a != 0]
        counts = {}
        for a in non_zero_answers:
            counts[a] = counts.get(a, 0) + 1

        # Iterate through the counts to calculate the min number of rabbits.
        for reported_number, number_of_rabbits_reporting in counts.items():

            # Calculate number of groups based on reports. 
            number_of_groups = (number_of_rabbits_reporting + reported_number) // (reported_number + 1)

            # Add rabbits in the group to the total.
            total_rabbits += number_of_groups * (reported_number + 1)

        minimum_number_of_rabbits = min(minimum_number_of_rabbits, total_rabbits)

    return minimum_number_of_rabbits

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Imagine you interview each rabbit and ask how many other rabbits have the same color as them.
Create a way to remember the number of siblings each rabbit reports.
If a rabbit says it has zero siblings of the same color, that means it's the only rabbit of that color, so count one rabbit for that color.
If a rabbit says it has 'n' siblings of the same color, it means that entire group has 'n + 1' rabbits in it.
If you see multiple rabbits reporting the same number of siblings, say 'n', you know that there is a group of 'n + 1' rabbits of that color.
If you see more than 'n + 1' rabbits reporting 'n' siblings, it means there are multiple groups of that color. Calculate how many full groups of 'n+1' rabbits exist. The amount of extra rabbits that exist forms an entirely new group of size 'n+1'
Add up the sizes of all of these groups of rabbits to get the total number of rabbits in the forest.

Code Implementation

def rabbits_in_forest(answers):    sibling_counts = {}
    for answer in answers:
        sibling_counts[answer] = sibling_counts.get(answer, 0) + 1

    total_rabbits = 0
    for number_of_same_color_siblings, count in sibling_counts.items():
        # If a rabbit says 0, it's a group of 1
        if number_of_same_color_siblings == 0:
            total_rabbits += count
        else:
            group_size = number_of_same_color_siblings + 1
            # Need to account for full groups and potentially a partial group
            number_of_full_groups = count // group_size

            if count % group_size == 0:
                total_rabbits += number_of_full_groups * group_size
            else:
                # Remainder forms a new group
                total_rabbits += (number_of_full_groups + 1) * group_size

    return total_rabbits

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty answers array	If the input array is empty, return 0 because no rabbits exist.
All answers are 0	The solution should return the length of the answers array, because each rabbit that answers 0 could be the only rabbit in its group.
Answers contain only 1s	The solution should return the length of the answers array, since each rabbit that answers 1 could be in a group of 2.
Answers contain a large number (close to the max integer)	The solution handles this by checking for positive values of x and ensures valid groups are counted.
Answers contain duplicate numbers, but not enough to form full groups based on those counts.	The solution must appropriately round up the number of groups formed.
Input array has only one element	The solution should create one group of size answers[0] + 1.
Input with duplicate counts that would overfill groups.	When count is greater than expected, start a new group for those excess rabbits
Integer overflow when calculating group size (answer[i] + 1)	Handle integer overflow by checking the result or by constraining the inputs

Empty answers array

How to Handle:

If the input array is empty, return 0 because no rabbits exist.

All answers are 0

How to Handle:

The solution should return the length of the answers array, because each rabbit that answers 0 could be the only rabbit in its group.

Answers contain only 1s

How to Handle:

The solution should return the length of the answers array, since each rabbit that answers 1 could be in a group of 2.

Answers contain a large number (close to the max integer)

How to Handle:

The solution handles this by checking for positive values of x and ensures valid groups are counted.

Answers contain duplicate numbers, but not enough to form full groups based on those counts.

How to Handle:

The solution must appropriately round up the number of groups formed.

Input array has only one element

How to Handle:

The solution should create one group of size answers[0] + 1.

Input with duplicate counts that would overfill groups.

How to Handle:

When count is greater than expected, start a new group for those excess rabbits

Integer overflow when calculating group size (answer[i] + 1)

How to Handle:

Handle integer overflow by checking the result or by constraining the inputs

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