LeetCode 781 - Rabbits in Forest

LeetCode Problem 781

Difficulty: 🟡 Medium
Topics: Array, Hash Table, Math, Greedy

Solution

Problem Understanding

In this problem, every rabbit tells us how many other rabbits share its color. The input array answers contains these responses. If a rabbit says x, that means there are exactly x + 1 rabbits of that color group in total, including itself.

The challenge is that we do not know which rabbits belong to the same color group. Multiple rabbits may be describing the same group, or they may belong to different groups that coincidentally give the same answer. Our task is to compute the smallest total number of rabbits that could exist while still making all answers consistent.

For example, if two rabbits both answer 1, they could belong to the same color group of size 2. However, if three rabbits answer 1, they cannot all belong to the same group because a group of rabbits answering 1 can contain only two rabbits total. Therefore, the third rabbit must start another group of size 2.

The constraints are relatively small, with at most 1000 answers, but the problem is primarily about reasoning correctly rather than handling huge input sizes. Since answers are bounded below 1000, we can comfortably use a hash map to count frequencies.

Several edge cases are important:

• Rabbits answering 0 are special because they claim no other rabbit shares their color, meaning each such rabbit forms a unique group of size 1.
• Multiple rabbits giving the same answer may require splitting into multiple groups.
• Frequencies that are not exact multiples of the group size are tricky because partially filled groups still require allocating the entire group size.

The key difficulty is understanding that reported counts describe group sizes, not direct rabbit counts.

Approaches

Brute Force Approach

A brute force solution would attempt to explicitly assign rabbits into possible color groups while ensuring all answers remain consistent. For every rabbit, we would try to place it into an existing compatible group or create a new group.

This approach eventually finds the minimum valid arrangement because it explores all possible groupings. However, the number of possible assignments grows combinatorially. Even for modest input sizes, the search space becomes enormous because rabbits with identical answers can be partitioned in many different ways.

This makes brute force impractical.

Optimal Greedy Counting Approach

The important observation is that a rabbit answering x implies a group size of exactly x + 1.

Suppose:

• x = 2
• Five rabbits answered 2

Each valid color group can contain at most 3 rabbits that answered 2, because the total group size must be 3.

Therefore:

• First group uses 3 rabbits
• Remaining 2 rabbits require another full group of size 3

Total rabbits contributed become 6.

This means we only need frequency counting.

For each distinct answer x:

• Group size is x + 1
• If frequency is freq
• Number of required groups is:

$$\left\lceil \frac{freq}{x+1} \right\rceil$$

Each group contributes x + 1 rabbits to the total.

This greedy grouping is optimal because combining as many rabbits as possible into the same valid group minimizes the total number of groups.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	Exponential	Exponential	Tries all possible group assignments
Optimal	O(n)	O(n)	Uses frequency counting and greedy grouping

Algorithm Walkthrough

1. Create a hash map to count how many times each answer appears.

If three rabbits answer 2, we store:

2 -> 3

The hash map allows us to process each distinct answer independently. 2. Iterate through each (answer, frequency) pair.

For a rabbit answering x, the implied group size is:

$$x + 1$$ 3. Determine how many groups are needed.

If frequency exceeds the group size, multiple groups are required.

The number of groups is:

$$\left\lceil \frac{frequency}{groupSize} \right\rceil$$

In code, this can be computed efficiently as:

(frequency + groupSize - 1) // groupSize

4. Add the total rabbits contributed by those groups.

Each group contributes exactly groupSize rabbits, even if not all rabbits answered.

Therefore:

total += groupsNeeded * groupSize

5. Return the accumulated total.

Why it works

The algorithm works because rabbits giving the same answer can only belong to groups of one fixed size. To minimize the total rabbit count, we should pack as many rabbits as possible into each valid group before creating another one. This greedy grouping strategy minimizes the number of groups, and therefore minimizes the total rabbits counted.

Python Solution

from collections import Counter
from typing import List

class Solution:
    def numRabbits(self, answers: List[int]) -> int:
        frequency_map = Counter(answers)

        total_rabbits = 0

        for answer, frequency in frequency_map.items():
            group_size = answer + 1

            groups_needed = (frequency + group_size - 1) // group_size

            total_rabbits += groups_needed * group_size

        return total_rabbits

The implementation begins by building a frequency map using Counter. This lets us process identical answers together rather than handling rabbits individually.

For every distinct answer:

• group_size is computed as answer + 1
• groups_needed uses ceiling division to determine how many valid color groups are required
• The total rabbits contributed by those groups are added to total_rabbits

The algorithm never explicitly constructs color groups. Instead, it mathematically computes the minimum number of groups necessary.

Go Solution

func numRabbits(answers []int) int {
    frequencyMap := make(map[int]int)

    for _, answer := range answers {
        frequencyMap[answer]++
    }

    totalRabbits := 0

    for answer, frequency := range frequencyMap {
        groupSize := answer + 1

        groupsNeeded := (frequency + groupSize - 1) / groupSize

        totalRabbits += groupsNeeded * groupSize
    }

    return totalRabbits
}

The Go implementation follows the same logic as the Python version.

A map[int]int is used instead of Counter to track frequencies. Integer division in Go naturally truncates toward zero, so the standard ceiling division formula still works correctly.

There are no overflow concerns because the maximum possible answer count is very small under the given constraints.

Worked Examples

Example 1

answers = [1,1,2]

First build the frequency map.

Answer	Frequency
1	2
2	1

Now process each entry.

Processing answer = 1

Variable	Value
answer	1
frequency	2
group_size	2
groups_needed	1
rabbits_added	2

Current total:

total = 2

Processing answer = 2

Variable	Value
answer	2
frequency	1
group_size	3
groups_needed	1
rabbits_added	3

Final total:

total = 5

Answer:

5

Example 2

answers = [10,10,10]

Frequency map:

Answer	Frequency
10	3

Processing:

Variable	Value
answer	10
frequency	3
group_size	11
groups_needed	1
rabbits_added	11

Final total:

11

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each rabbit is counted once
Space	O(n)	Hash map may store up to n distinct answers

The time complexity is linear because we first count all answers and then iterate through distinct values. The space complexity is also linear in the worst case when every rabbit gives a unique answer.

Test Cases

from typing import List

class Solution:
    def numRabbits(self, answers: List[int]) -> int:
        from collections import Counter

        frequency_map = Counter(answers)

        total_rabbits = 0

        for answer, frequency in frequency_map.items():
            group_size = answer + 1
            groups_needed = (frequency + group_size - 1) // group_size
            total_rabbits += groups_needed * group_size

        return total_rabbits

solution = Solution()

assert solution.numRabbits([1, 1, 2]) == 5  # Provided example
assert solution.numRabbits([10, 10, 10]) == 11  # Large group size
assert solution.numRabbits([0]) == 1  # Single rabbit with unique color
assert solution.numRabbits([0, 0, 0]) == 3  # Every rabbit unique
assert solution.numRabbits([1, 1]) == 2  # Perfectly filled group
assert solution.numRabbits([1, 1, 1]) == 4  # Requires two groups
assert solution.numRabbits([2, 2, 2]) == 3  # Exact full group
assert solution.numRabbits([2, 2, 2, 2]) == 6  # Overflow into second group
assert solution.numRabbits([3, 3, 3, 3]) == 4  # One complete group
assert solution.numRabbits([3, 3, 3, 3, 3]) == 8  # Two required groups
assert solution.numRabbits([0, 1, 0, 1, 1]) == 5  # Mixed answers
assert solution.numRabbits([5]) == 6  # One rabbit implies unseen rabbits

Test	Why
[1,1,2]	Validates provided example
[10,10,10]	Tests large implied group size
[0]	Smallest non-empty input
[0,0,0]	Every rabbit has unique color
[1,1]	Exact group fit
[1,1,1]	Requires partially filled second group
[2,2,2]	Exact size-3 grouping
[2,2,2,2]	Spillover into additional group
[3,3,3,3]	One fully populated group
[3,3,3,3,3]	Multiple groups required
[0,1,0,1,1]	Mixed frequencies and group sizes
[5]	Single rabbit requiring unseen rabbits

Edge Cases

One important edge case occurs when rabbits answer 0. A rabbit answering 0 claims that no other rabbit shares its color. Therefore, every such rabbit must belong to a unique single-rabbit group. A common bug is accidentally grouping multiple 0 answers together, which would violate the condition. The implementation handles this naturally because the computed group size becomes 1.

Another tricky case occurs when the number of rabbits does not perfectly divide into group sizes. For example, three rabbits answering 1 require two groups of size 2, resulting in four rabbits total. A naive implementation might incorrectly count only the rabbits seen in the input. The ceiling division ensures partially filled groups still allocate the full required group size.

A third important edge case is when only one rabbit gives a very large answer, such as [999]. Even though we only observed one rabbit, that rabbit claims there are 999 other rabbits of the same color. The implementation correctly creates one full group of size 1000, ensuring consistency with the rabbit's statement.