Problem Description

In this problem, we're given a string hamsters, which represents a row of positions that can either contain a hamster, denoted by 'H', or be empty, denoted by '.'. We need to place food buckets at some of these empty positions. There are two conditions to ensure that a hamster can be fed:

• There must be at least one food bucket immediately to the left (i - 1) or immediately to the right (i + 1) of the hamster's position.
• We want to place the minimum number of food buckets necessary to feed all the hamsters.

Our task is to find this minimum number of food buckets or determine that it's impossible to feed all hamsters, in which case we return -1.

Intuition

The solution approach is to iterate through the string and decide the placement of the food buckets optimally to minimize their number while ensuring all hamsters can access at least one bucket. This optimality comes from the following rule: whenever we have a choice to place a food bucket either to the left or the right of a hamster, we prefer to place it to the right. Doing so might enable us to feed the next hamster with the same bucket, hence reducing the total number.

Here's the approach in detail:

• We iterate through each position in the string. If we find a hamster ('H'), we check its neighbouring positions.
• If we have an empty position to the right of the hamster, we always place a bucket there, as it could potentially feed another hamster to the right. We increment the answer by 1 and skip checking the next position by advancing two steps in the iteration.
• If the right side is not available, we check the left side. If it's empty and we haven't already placed a bucket there while placing it for a previous hamster, we place a bucket there and increment the answer by 1.
• If neither side is available for placing a bucket, we return -1 as it's impossible to feed this hamster.

The critical insight is that by considering the right side first, we utilize each bucket to its maximum potential, thereby ensuring the least number of buckets is used.

Learn more about Greedy and Dynamic Programming patterns.

Solution Approach

The solution uses a greedy algorithm, which is a common approach for optimization problems where we aim to make the locally optimal choice at each stage with the hope of finding a global optimum.

Here's a step-by-step walkthrough of the implementation based on the reference solution:

1. Initialize a counter ans to 0, which will hold the minimum number of food buckets required.

2. Start iterating through the string street using variable i that begins from index 0.

3. Check if the current character at index i is a hamster ('H').

   • If there is a hamster at index i and the next index i + 1 is within bounds and empty (contains '.'), place a food bucket at i + 1. This is because placing the bucket to the right could potentially also feed a hamster at index i + 2.

     • After placing the bucket, increase ans by 1 to count the newly placed bucket.
     • Move i two positions forward to i + 2, as we have already dealt with hamster at i and potentially at i + 1 if there is one.

   • If you can't place the bucket to the right because i + 1 is out of bounds or not empty, check if the left of the hamster at i is available (make sure you're not out of bounds and the index i - 1 is empty).

     • If a bucket can be placed at i - 1, increase ans by 1. There's no need to move i forward by two positions in this case since we are only dealing with the current hamster.

   • If neither the left nor the right side of the current hamster can have a bucket placed, it is impossible to feed this hamster, so we return -1.

4. If the current character is not a hamster, we simply move to the next index by incrementing i.

5. Continue this process until we have iterated over the entire string. The ans variable by the end of iteration corresponds to the minimum number of buckets required. We return ans.

No additional data structures are used apart from variables for indexing and counting. This makes the algorithm efficient both in terms of time and space complexity, achieving (O(n)) time complexity, where (n) is the length of the street string, and (O(1)) space complexity since we only use a fixed number of extra variables.

Example Walkthrough

Let's go through an example to illustrate the solution approach.

Assume the input string representing the row of positions is: "H..HH..H.."

Following the solution approach, we'll determine where to place the food buckets:

1. Starting at index 0, we see 'H', the position contains a hamster.

   • i + 1 is within bounds and empty, so we place a bucket at i + 1 (position 1).
   • We increment ans by 1 (now ans is 1).
   • We skip the next position by moving two positions forward to index 2.

2. At index 2, there's no hamster, just a food bucket, so we move to the next index.

3. At index 3, again, there's no hamster; thus, we move to index 4.

4. At index 4, there's a hamster 'H'.

   • i + 1 is within bounds and empty, so we put a bucket at i + 1 (position 5).
   • We increment ans by 1 (now ans is 2).
   • We move two positions forward to index 6.

5. At index 6, there's no hamster; we move to index 7.

6. At index 7, there's another hamster 'H'.

   • i + 1 is within bounds but contains another hamster, so we can't place a bucket there.
   • We look to the left (i - 1), but there's already a food bucket there placed for the previous hamster, so no new bucket is needed here.

7. We move to the next index 8 and find it empty. Move to index 9.

8. At index 9, there's a hamster 'H'.

   • i + 1 is out of bounds, so we look at i - 1, which is within bounds and empty.
   • We place a bucket at i - 1 (position 8).
   • We increment ans by 1 (now ans is 3).

Having gone through the string, we needed a minimum of 3 food buckets to feed all hamsters in the given row of positions. Hence, we return 3.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code involves a single loop through the length of the string street. Within this loop, the operations are constant time checks and assignments. The loop iterates at most n times, where n is the length of street. In the worst case, every character is visited once, and update operations are constant time O(1). Therefore, the time complexity is O(n).

Space Complexity

The space complexity of the code is O(1). This is because the code only uses a fixed number of extra variables (ans, i, and n) that do not depend on the size of the input street. There are no data structures used that grow with the input size. Hence, the space required does not scale with the size of the input.

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