LeetCode 2086 - Minimum Number of Food Buckets to Feed the Hamsters

The problem is asking us to place the minimum number of food buckets on empty spaces in a linear arrangement of cells so that all hamsters are fed. Each cell in the string hamsters represents either a hamster ('H') or an empty spot ('.').

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LeetCode Problem 2086

Difficulty: 🟡 Medium
Topics: String, Dynamic Programming, Greedy

Solution

Problem Understanding

The problem is asking us to place the minimum number of food buckets on empty spaces in a linear arrangement of cells so that all hamsters are fed. Each cell in the string hamsters represents either a hamster ('H') or an empty spot ('.'). A hamster can eat if there is at least one food bucket directly adjacent to it, either to its left or to its right. We are asked to determine the minimum number of buckets needed or return -1 if it is impossible to feed all hamsters.

The input string length can be up to 10^5, so any approach that is worse than linear time will likely be too slow. The key challenge is deciding where to place the buckets optimally, because placing them greedily without thought may result in an unfeedable hamster.

Edge cases include strings with consecutive hamsters, strings with no empty spaces, single hamsters at the edges, and strings with all empty spots.

Approaches

The brute-force approach would try all combinations of bucket placements. For each empty cell, we could consider placing a bucket or not, then check whether all hamsters are fed. While this guarantees correctness, it is computationally infeasible because the number of combinations grows exponentially with the number of empty spots.

The optimal approach relies on greedy placement. The observation is that placing a bucket to the right of a hamster is usually better because it can feed both the current hamster and the next one if they are consecutive. If the right cell is not empty, we place the bucket to the left. If neither side is empty, feeding is impossible. This greedy strategy ensures minimal buckets and linear traversal of the string.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n) O(n) Tries all combinations of bucket placements, infeasible for n up to 10^5
Optimal Greedy O(n) O(1) Linear scan placing buckets optimally to feed hamsters, minimal extra space

Algorithm Walkthrough

  1. Initialize a counter buckets to 0 to track the number of buckets placed. Convert the string into a list for mutability if needed.
  2. Traverse the string from left to right using an index i.
  3. When encountering a hamster at position i, check if it is already adjacent to a bucket (we can mark buckets with 'B' or track positions). If yes, continue to the next index.
  4. If the hamster is not yet fed, try to place a bucket to its right (i+1) if that index exists and is empty. Mark the bucket and increment buckets. Skip i+1 in the next iteration because that cell now contains a bucket.
  5. If placing to the right is not possible, attempt to place a bucket to the left (i-1) if it is empty. Mark the bucket and increment buckets.
  6. If neither side is empty or valid, return -1 because the hamster cannot be fed.
  7. Continue until the end of the string. Return buckets.

Why it works: The greedy choice always places a bucket where it can feed as many consecutive hamsters as possible, ensuring a minimal total count. Each hamster is considered only once, and we place buckets only when needed.

Python Solution

class Solution:
    def minimumBuckets(self, hamsters: str) -> int:
        buckets = 0
        hamsters = list(hamsters)
        n = len(hamsters)
        i = 0
        
        while i < n:
            if hamsters[i] == 'H':
                if i + 1 < n and hamsters[i + 1] == '.':
                    buckets += 1
                    hamsters[i + 1] = 'B'
                    i += 2
                elif i - 1 >= 0 and hamsters[i - 1] == '.':
                    buckets += 1
                    hamsters[i - 1] = 'B'
                    i += 1
                else:
                    return -1
            else:
                i += 1
        
        return buckets

The implementation converts the string to a list to allow marking placed buckets with 'B'. We scan left to right and place buckets to the right if possible, otherwise to the left. If neither is possible, we return -1. Skipping indices after placing a bucket ensures we do not double-count.

Go Solution

func minimumBuckets(hamsters string) int {
    buckets := 0
    n := len(hamsters)
    hamstersArr := []rune(hamsters)
    
    for i := 0; i < n; i++ {
        if hamstersArr[i] == 'H' {
            if i+1 < n && hamstersArr[i+1] == '.' {
                buckets++
                hamstersArr[i+1] = 'B'
                i++ 
            } else if i-1 >= 0 && hamstersArr[i-1] == '.' {
                buckets++
                hamstersArr[i-1] = 'B'
            } else {
                return -1
            }
        }
    }
    
    return buckets
}

The Go version mirrors the Python solution, using a rune slice to allow mutation. Index handling and greedy placement logic are identical.

Worked Examples

Example 1: "H..H"

i hamsters[i] Action Buckets Notes
0 H Place bucket at i+1 1 Cell 1 marked 'B'
2 . Skip 1 Nothing to do
3 H Place bucket at i-1 2 Cell 2 marked 'B'

Result: 2 buckets

Example 2: ".H.H."

i hamsters[i] Action Buckets Notes
1 H Place bucket at i+1 1 Cell 2 marked 'B'
3 H Already fed from left 1 Skip

Result: 1 bucket

Example 3: ".HHH."

i hamsters[i] Action Buckets Notes
1 H Place bucket at i+1 1 Cell 2 marked 'B'
2 H Already fed 1 Skip
3 H Cannot place bucket left or right -1 Return -1

Complexity Analysis

Measure Complexity Explanation
Time O(n) Single pass through the string, checking neighbors
Space O(n) Mutable copy of string to mark buckets

The algorithm is linear in time because each index is visited at most twice. Space is linear due to the copy of the string for marking bucket placements.

Test Cases

# Basic examples
assert Solution().minimumBuckets("H..H") == 2  # Standard placement at empty cells
assert Solution().minimumBuckets(".H.H.") == 1  # Greedy bucket feeds multiple hamsters
assert Solution().minimumBuckets(".HHH.") == -1  # Impossible to feed all

# Edge cases
assert Solution().minimumBuckets("H") == -1  # Single hamster cannot be fed
assert Solution().minimumBuckets(".") == 0  # No hamsters, no buckets needed
assert Solution().minimumBuckets("HH") == -1  # Consecutive hamsters with no empty space
assert Solution().minimumBuckets("H.H.H.H") == 3  # Alternating hamsters
assert Solution().minimumBuckets("H..H..H") == 3  # Multiple groups with gaps
Test Why
"H..H" Standard scenario with gaps between hamsters
".H.H." Tests minimal bucket sharing
".HHH." Impossible case with consecutive hamsters
"H" Single hamster, no space to feed
"." Empty string, no buckets needed
"HH" Consecutive hamsters with no empty spots
"H.H.H.H" Alternating hamsters, greedy choice works
"H..H..H" Multiple isolated groups, each needs buckets

Edge Cases

The first edge case is a single hamster with no adjacent empty cells. The algorithm correctly returns -1 because neither left nor right bucket placement is possible.

The second edge case is consecutive hamsters. If there are two or more hamsters in a row with no empty space between them, the algorithm detects impossibility because a single bucket cannot feed both at once.

The third edge case is hamsters at the string boundaries. For a hamster at index 0, only the right side can be used for bucket placement. At index n-1, only the left side is available. The implementation correctly checks array bounds before placing buckets.