LeetCode #781 — MEDIUM

Rabbits in Forest

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Math · Greedy

Example 1

[1,1,2]

Example 2

[10,10,10]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #781: Rabbits in Forest
class Solution {
    public int numRabbits(int[] answers) {
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int x : answers) {
            cnt.merge(x, 1, Integer::sum);
        }
        int ans = 0;
        for (var e : cnt.entrySet()) {
            int group = e.getKey() + 1;
            ans += (e.getValue() + group - 1) / group * group;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #781: Rabbits in Forest
func numRabbits(answers []int) (ans int) {
	cnt := map[int]int{}
	for _, x := range answers {
		cnt[x]++
	}
	for x, v := range cnt {
		group := x + 1
		ans += (v + group - 1) / group * group
	}
	return
}

# Accepted solution for LeetCode #781: Rabbits in Forest
class Solution:
    def numRabbits(self, answers: List[int]) -> int:
        cnt = Counter(answers)
        ans = 0
        for x, v in cnt.items():
            group = x + 1
            ans += (v + group - 1) // group * group
        return ans

// Accepted solution for LeetCode #781: Rabbits in Forest
struct Solution;
use std::collections::HashMap;

impl Solution {
    fn num_rabbits(answers: Vec<i32>) -> i32 {
        let mut hm: HashMap<i32, i32> = HashMap::new();
        for x in answers {
            *hm.entry(x).or_default() += 1;
        }
        let mut res = 0;
        for (k, v) in hm {
            res += (v + k) / (k + 1) * (k + 1);
        }
        res
    }
}

#[test]
fn test() {
    let answers = vec![1, 1, 2];
    let res = 5;
    assert_eq!(Solution::num_rabbits(answers), res);
    let answers = vec![10, 10, 10];
    let res = 11;
    assert_eq!(Solution::num_rabbits(answers), res);
    let answers = vec![];
    let res = 0;
    assert_eq!(Solution::num_rabbits(answers), res);
}

// Accepted solution for LeetCode #781: Rabbits in Forest
function numRabbits(answers: number[]): number {
    const cnt = new Map<number, number>();
    for (const x of answers) {
        cnt.set(x, (cnt.get(x) || 0) + 1);
    }
    let ans = 0;
    for (const [x, v] of cnt.entries()) {
        const group = x + 1;
        ans += Math.floor((v + group - 1) / group) * group;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.