Problem Description

You are given a string hamsters where each character represents either:

• 'H' - a hamster at that position
• '.' - an empty position

Your task is to place food buckets at empty positions (.) to feed all hamsters. A hamster can be fed if there is at least one food bucket either immediately to its left or immediately to its right.

For example, a hamster at index i is fed if there's a food bucket at index i-1 or i+1 (or both).

The goal is to find the minimum number of food buckets needed to feed all hamsters. If it's impossible to feed all hamsters (for instance, if a hamster has no empty positions adjacent to it), return -1.

Example scenarios:

• "H.H" - You can place 1 bucket at index 1, which feeds both hamsters
• "HHH" - Impossible to feed the middle hamster since it has no adjacent empty positions, return -1
• ".H.H." - You need 2 buckets minimum to feed both hamsters

The solution uses a greedy approach: when encountering a hamster, it first tries to place a bucket to its right (if possible), as this bucket might also feed the next hamster. If that's not possible, it checks if a bucket can be placed to its left. If neither position is available, it returns -1.

Intuition

The key insight is that we want to minimize the number of buckets, so we should try to make each bucket feed as many hamsters as possible. A bucket can potentially feed up to two hamsters - one on its left and one on its right.

When we encounter a hamster that needs feeding, we have two choices: place a bucket to its left or to its right. Which should we choose?

Consider this: if we place a bucket to the right of the current hamster, that bucket might also feed the next hamster (if there is one). This gives us a chance to use one bucket for two hamsters. However, if we place a bucket to the left, it can only feed the current hamster (since we've already processed everything to the left).

This leads to a greedy strategy: always try to place the bucket to the right first. This maximizes our chances of feeding multiple hamsters with a single bucket.

The algorithm becomes:

1. Scan the string from left to right
2. When we find a hamster 'H':
   • First check if we can place a bucket to its right (position i+1 must be '.')
   • If yes, place it there and skip the next position (since if it's a hamster, it's already fed)
   • If no, check if we can place a bucket to its left (position i-1 must be '.')
   • If neither is possible, the hamster cannot be fed, return -1

This greedy approach works because once we've decided the optimal placement for feeding earlier hamsters, we never need to reconsider those decisions. Each hamster is handled optimally based on the current state, leading to the global minimum.

Learn more about Greedy and Dynamic Programming patterns.

Solution Approach

The implementation uses a single-pass greedy algorithm with a pointer to traverse the string:

Variables:

• ans: Counter for the number of buckets placed
• i: Current position in the string
• n: Length of the string

Algorithm Steps:

1. Initialize pointer i = 0 and bucket counter ans = 0

2. Iterate through the string using a while loop:

   while i < n:

3. For each position, check if it's a hamster (street[i] == 'H'):

   a. Try placing bucket to the right first:

   if i + 1 < n and street[i + 1] == '.':
       i += 2  # Skip the next position
       ans += 1

   • Check if right position exists and is empty
   • If yes, increment bucket count and skip 2 positions (current hamster + bucket position)
   • This optimization works because if the next position after the bucket is also a hamster, it's already fed

   b. If right placement fails, try left:

   elif i and street[i - 1] == '.':
       ans += 1

   • Check if left position exists (i > 0) and is empty
   • If yes, increment bucket count
   • Note: We don't need to track which positions have buckets since we only check i-1 when we couldn't place at i+1

   c. If both fail, return -1:

   else:
       return -1

   • The hamster cannot be fed

4. Move to next position if current position is not a hamster or after processing:

   i += 1

5. Return the total bucket count after processing all positions

Time Complexity: O(n) - single pass through the string
Space Complexity: O(1) - only using a few variables

The greedy choice of preferring right placement ensures we minimize bucket usage by maximizing the chance of feeding consecutive hamsters with a single bucket.

Example Walkthrough

Let's trace through the string "H.H.H" to see how the greedy algorithm works:

Initial state: i = 0, ans = 0, string = "H.H.H"

Step 1: i = 0

• Current position: 'H' (hamster at index 0)
• Check right (index 1): It's '.' (empty) ✓
• Place bucket at index 1: "H[B]H.H"
• Increment ans = 1
• Skip next position, set i = 2

Step 2: i = 2

• Current position: 'H' (hamster at index 2)
• Check right (index 3): It's '.' (empty) ✓
• Place bucket at index 3: "H[B]H[B]H"
• Increment ans = 2
• Skip next position, set i = 4

Step 3: i = 4

• Current position: 'H' (hamster at index 4)
• Check right (index 5): Out of bounds ✗
• Check left (index 3): It's '.' where we placed a bucket ✓
• Hamster is already fed by the bucket at index 3
• Increment ans remains 2 (bucket already placed)
• Move to next position, set i = 5

Step 4: i = 5

• Out of bounds, loop ends

Result: ans = 2 buckets needed

The buckets at positions 1 and 3 feed all three hamsters:

• Bucket at index 1 feeds hamsters at indices 0 and 2
• Bucket at index 3 feeds hamsters at indices 2 and 4
• Note: Hamster at index 2 has buckets on both sides, but only needs one to be fed

This demonstrates the efficiency of the greedy approach - by preferring right placement, we allowed the bucket at index 3 to serve double duty, feeding both the hamster at index 2 (from its left) and the hamster at index 4 (from its right).

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the input string street. The algorithm uses a single while loop that iterates through each character of the string exactly once. In each iteration, it performs constant-time operations (checking characters, incrementing counters). Even though sometimes i is incremented by 2 (when i += 2), each character is still visited at most once, maintaining linear time complexity.

Space Complexity: O(1). The algorithm only uses a fixed amount of extra space regardless of the input size. It maintains three variables: ans (bucket counter), i (index pointer), and n (string length). No additional data structures that scale with input size are created, resulting in constant space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Not Skipping Positions After Right Placement

Pitfall: When placing a bucket to the right of a hamster, forgetting to skip the next position leads to unnecessary bucket placement.

Problem Example:

# Incorrect implementation
if street[i] == 'H':
    if i + 1 < n and street[i + 1] == '.':
        ans += 1
        # Missing: i += 2, only incrementing by 1 at loop end
    i += 1

For input "H.H.H", this would place 3 buckets instead of the optimal 2.

Solution: Always skip 2 positions (i += 2) after placing a bucket to the right, as the next position is either the bucket itself or already covered by it.

2. Checking Left Position Without Verifying Bucket Placement

Pitfall: When checking the left position, not considering whether a bucket was already placed there by a previous hamster.

Why This Works Anyway: The algorithm's structure naturally avoids this issue because:

• We only check left when right placement fails
• If a bucket was placed at position i-1 for a previous hamster, that hamster must have been at position i-2
• Since we're now at position i, we've already moved past that scenario

Example: For ".H.H":

• At index 1 (first H): Can't place right (another H), places left at index 0
• At index 3 (second H): Can place right at index 4 (if it exists) or left at index 2
• The positions don't overlap due to the traversal order

3. Incorrect Boundary Checks

Pitfall: Using wrong conditions for boundary checks:

# Incorrect - using <= instead of <
if i + 1 <= n and street[i + 1] == '.':  # This causes index out of bounds

# Incorrect - checking i > 0 incorrectly
elif i - 1 >= 0 and street[i - 1] == '.':  # Verbose but correct
elif i and street[i - 1] == '.':  # Cleaner, leverages Python's truthiness

Solution: Use i + 1 < n for right boundary and i > 0 (or just i in Python) for left boundary.

4. Not Handling Edge Cases

Pitfall: Missing edge cases like:

• Empty string: "" → Should return 0
• Single hamster with no adjacent spots: "H" → Should return -1
• All dots: "..." → Should return 0

Solution: The algorithm naturally handles these through its logic flow, but it's important to verify during testing.