Computer Science
Updated 2 days ago | GitHub

Rabbits In The Forest

Problem Highlights

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

  • Established a set (2-3) of test cases to verify their own solution later.
  • Established a set (1-2) of edge cases to verify their solution handles complexities.
  • Have fully understood the problem and have no clarifying questions.
  • Have you verified any Time/Space Constraints for this problem?
  • Can I see what color the rabbits are?
    • No
  • What are the constraints?
    • 1 <= answers.length <= 1000
    • 0 <= answers[i] < 1000
  • If there are fewer than i + 1 answers of i, they could be the same color
    • There could be a group of 2 rabbits who say “1”
    • There could be a group of 3 rabbits who say “2”
  • If a rabbit says:
    • 0: add 1 to our count every time
    • 1: add 2 to our count every two rabbits
    • 2: add 3 to our count every 3 rabbits
    • n: add n+1 to our count every n rabbits
HAPPY CASE
Input: answers = [1,1,2]
Output: 5

EDGE CASE
Input: answers = [10,10,10]
Output: 11

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

  • Greedy Algorithm: Greedily put as many rabbits into the same color as possible, until it is not possible to fit, create a new color. If rabbits share the same answer x, we can put at most x+1 rabbits in this color group. Because the result is min of rabbits count, so put as many rabbits in same color group as possible.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea:

1. Create a hashtable counts
2. Set sum = 0
3. For each answer in answers
If answer == 0, sum += 1
If answer not in counts:
counts[answer] = 1
sum += answer + 1
If counts[answer]  < answer + 1
counts[answer] ++
else
counts[answer] += 1
sum += answer + 1
4. Return sum

⚠️ Common Mistakes

  • Treating each rabbit as if it lived in its own color group instead of noticing that up to x + 1 rabbits that all answer x can share one color.
  • Naively adding count[x] * (x + 1) for every distinct answer x. When more rabbits answer x than fit in one group, they must be split across multiple groups — the count needs to be rounded up to the next multiple of x + 1, which is what -count[x] % (x + 1) + count[x] computes.
  • Overlooking that rabbits answering 0 say they are alone in their color, so each of them contributes exactly one rabbit rather than being pooled together.

To minimize the total number of rabbits, we can divide the y rabbits who answered x into groups of size x + 1, all sharing a color. If y % (x + 1) > 0, then the last y % (x + 1) rabbits must share a color with additional rabbits that did not answer the poll; those pad the group up to x + 1 (otherwise, the claim that x other rabbits share the color would be false).

4: I-mplement

Implement the code to solve the algorithm.

class Solution(object):
    def numRabbits(self, answers):
        count = collections.Counter(answers)
        return sum(-v % (k+1) + v for k, v in count.items())

class Solution {
    public int numRabbits(int[] answers) {
        int[] count = new int[1000];
        for (int x: answers) count[x]++;

        int ans = 0;
        for (int k = 0; k < 1000; ++k)
            ans += Math.floorMod(-count[k], k+1) + count[k];
        return ans;
    }
}

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code’s variables along the way.

  • Trace through your code with an input to check for the expected output
  • Catch possible edge cases and off-by-one errors

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

  • Time Complexity: O(N), where N is the number of rabbits that answered
  • Space Complexity: O(N)