Rabbits in Forest
MedPrompt
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] < 1000
Approaches
3 approaches with complexity analysis and trade-offs.
This approach first sorts the input array answers. By sorting, all rabbits who gave the same answer are grouped together. We can then iterate through the sorted array, processing each group of identical answers to calculate the number of rabbits required for that group.
Algorithm
- Sort the
answersarray in non-decreasing order. - Initialize
totalRabbits = 0and an indexi = 0. - Loop while
iis less than the length of the array: a. Get the current answerx = answers[i]. b. The size of a color group for this answer isgroupSize = x + 1. c. Count the number of rabbitskthat gave the same answerx. This can be done by finding the next indexjwhereanswers[j] != x. The countkwill bej - i. d. Calculate the number of groups needed using ceiling division:numGroups = (k + groupSize - 1) / groupSize. e. Add the total rabbits for this color to the result:totalRabbits += numGroups * groupSize. f. Move the indexitojto start processing the next distinct answer. - Return
totalRabbits.
Walkthrough
The fundamental logic is that if a rabbit answers x, it belongs to a color group of size x + 1. All rabbits that give the same answer x can potentially belong to the same color group. By sorting the answers array, we place all identical answers adjacent to each other, making them easy to process as a single block. We iterate through the sorted array, and for each block of identical answers x of size k, we calculate how many groups of size x + 1 are needed. Since each group can contain at most x + 1 such rabbits, we need ceil(k / (x + 1)) groups. The total number of rabbits for this color is then ceil(k / (x + 1)) * (x + 1). We sum this value for each distinct answer to find the overall minimum number of rabbits.
import java.util.Arrays; class Solution { public int numRabbits(int[] answers) { if (answers == null || answers.length == 0) { return 0; } Arrays.sort(answers); int totalRabbits = 0; int i = 0; while (i < answers.length) { int currentAnswer = answers[i]; int groupSize = currentAnswer + 1; int count = 0; int j = i; while (j < answers.length && answers[j] == currentAnswer) { count++; j++; } // Using ceiling division: (numerator + denominator - 1) / denominator int numGroups = (count + groupSize - 1) / groupSize; totalRabbits += numGroups * groupSize; i = j; } return totalRabbits; }}Complexity
Time
O(N log N), where N is the number of elements in `answers`. The dominant operation is sorting the array. The subsequent scan of the array takes O(N) time.
Space
O(log N) or O(N), depending on the implementation of the sorting algorithm. In Java, `Arrays.sort` for primitives uses a variant of Quicksort which has an average space complexity of O(log N).
Trade-offs
Pros
The approach is relatively simple to conceptualize.
It does not require any auxiliary data structures like a HashMap, modifying the input array in-place (if allowed) or using the space provided by the sorting algorithm.
Cons
The sorting step has a time complexity of O(N log N), which is less efficient than linear time solutions.
A more efficient approach is to count the frequency of each answer without sorting. A HashMap is a perfect data structure for this task. We can iterate through the answers array once to populate the HashMap with counts, and then iterate through the HashMap's entries to calculate the total number of rabbits based on these frequencies.
Algorithm
- Create a
HashMap<Integer, Integer>calledcountsto store the frequency of each answer. - Iterate through the
answersarray. For eachanswer, update its frequency in thecountsmap. - Initialize
totalRabbits = 0. - Iterate through each entry
(answer, count)in thecountsmap: a. The size of a color group isgroupSize = answer + 1. b. Calculate the number of groups needed for this answer:numGroups = (count + groupSize - 1) / groupSize. c. Add the total rabbits for this color to the result:totalRabbits += numGroups * groupSize. - Return
totalRabbits.
Walkthrough
This approach avoids the O(N log N) sorting step by using a hash map to count frequencies in linear time. We first populate a map where keys are the answers given by rabbits and values are the number of rabbits that gave that answer. After counting, we iterate through the map. For each answer x with a count of k, we apply the same logic as before: these k rabbits belong to groups of size x + 1. We calculate the minimum number of groups required, which is ceil(k / (x + 1)), and multiply it by the group size x + 1 to find the total rabbits for that color. Summing these results across all unique answers gives the final minimum total.
import java.util.HashMap;import java.util.Map; class Solution { public int numRabbits(int[] answers) { if (answers == null || answers.length == 0) { return 0; } Map<Integer, Integer> counts = new HashMap<>(); for (int answer : answers) { counts.put(answer, counts.getOrDefault(answer, 0) + 1); } int totalRabbits = 0; for (Map.Entry<Integer, Integer> entry : counts.entrySet()) { int answer = entry.getKey(); int count = entry.getValue(); int groupSize = answer + 1; int numGroups = (count + groupSize - 1) / groupSize; totalRabbits += numGroups * groupSize; } return totalRabbits; }}Complexity
Time
O(N), where N is the length of the `answers` array. This is because we iterate through the array once to build the map and then iterate through the unique answers (at most N).
Space
O(U), where U is the number of unique answers. In the worst-case scenario where all N answers are distinct, the space complexity is O(N).
Trade-offs
Pros
Achieves optimal O(N) time complexity.
It is flexible and works for any range of integer values in the
answersarray, not just small, bounded ones.
Cons
Uses extra space for the HashMap, which can be up to O(N) in the worst case where all answers are unique.
May have a slight performance overhead due to hash calculations and potential collisions compared to a direct-access array.
Given the problem's constraints that answers are non-negative and less than 1000, we can optimize the frequency counting by using a fixed-size array instead of a HashMap. This approach leverages direct array indexing, which is generally faster than hashing, making it the most performant solution for the given constraints.
Algorithm
- Create an integer array
countsof size 1001 (since0 <= answers[i] < 1000) and initialize it with zeros. - Iterate through the
answersarray. For eachanswer, incrementcounts[answer]. - Initialize
totalRabbits = 0. - Iterate from
i = 0to1000(the range of possible answers): a. Ifcounts[i] == 0, continue to the next iteration. b. Letanswer = iandcount = counts[i]. c. The size of a color group isgroupSize = answer + 1. d. Calculate the number of groups needed:numGroups = (count + groupSize - 1) / groupSize. e. Add the total rabbits for this color to the result:totalRabbits += numGroups * groupSize. - Return
totalRabbits.
Walkthrough
This method is a specialized version of the HashMap approach, tailored to the specific constraints of the problem. Since answers[i] is between 0 and 999, we can use an array of size 1001 as a direct address table or frequency map. counts[i] will store the number of rabbits that answered i. We first pass through the answers array to populate our counts array. Then, we iterate through the counts array from index 0 to 1000. For each index i where counts[i] is greater than zero, we perform the same calculation as in the previous approaches to find the total number of rabbits for that color group and add it to our running total. This avoids the overhead of HashMap's hashing and object creation, leading to better performance.
class Solution { public int numRabbits(int[] answers) { if (answers == null || answers.length == 0) { return 0; } // answers[i] is in the range [0, 999] int[] counts = new int[1000]; for (int answer : answers) { counts[answer]++; } int totalRabbits = 0; // A rabbit answering 0 is in a group of 1 (itself) totalRabbits += counts[0]; for (int i = 1; i < 1000; i++) { if (counts[i] == 0) { continue; } int count = counts[i]; int answer = i; int groupSize = answer + 1; int numGroups = (count + groupSize - 1) / groupSize; totalRabbits += numGroups * groupSize; } return totalRabbits; }}Complexity
Time
O(N + M), where N is the length of `answers` and M is the range of possible answer values. Given M=1000, this is effectively O(N) as N can be up to 1000.
Space
O(M), where M is the maximum possible value for an answer plus one. Given the constraints, this is O(1000), which is constant space.
Trade-offs
Pros
Extremely fast due to direct array access, avoiding hashing overhead.
Maintains an optimal linear time complexity.
Simple to implement.
Cons
The space complexity is dependent on the maximum possible value of an answer, not the number of answers or unique answers. If the range of answers were very large, this approach would be infeasible.
It is less flexible than the HashMap approach if the constraints on the answer values change.
Solutions
Solution
class Solution {public int numRabbits(int[] answers) { Map<Integer, Integer> counter = new HashMap<>(); for (int e : answers) { counter.put(e, counter.getOrDefault(e, 0) + 1); } int res = 0; for (Map.Entry<Integer, Integer> entry : counter.entrySet()) { int answer = entry.getKey(), count = entry.getValue(); res += (int)Math.ceil(count / ((answer + 1) * 1.0)) * (answer + 1); } return res; }}class Solution {public: int numRabbits(vector<int> &answers) { unordered_map<int, int> cnt; for (int x : answers) { ++cnt[x]; } int ans = 0; for (auto &[x, v] : cnt) { int group = x + 1; ans += (v + group - 1) / group * group; } return ans; }};class Solution: def numRabbits(self, answers: List[int]) -> int: counter = Counter(answers) return sum([math . ceil(v / (k + 1)) * (k + 1) for k, v in counter . items()])Video walkthrough
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