Problem Description

You are given a string road, consisting only of characters "x" and ".", where each "x" denotes a pothole and each "." denotes a smooth road, along with an integer budget.

In one repair operation, you can repair n consecutive potholes for a price of n + 1.

Return the maximum number of potholes that can be fixed such that the sum of the costs of all repairs doesn't exceed the given budget.

Intuition

To solve this problem optimally, we need to utilize a counting and greedy approach. The idea is to count the number of consecutive potholes of each length and attempt to repair larger groups first since they offer a better ratio of potholes repaired to cost incurred.

1. First, convert each segment of consecutive potholes into a count and store these counts in an array called cnt.
2. cnt[k] indicates the number of segments of consecutive potholes of length k.
3. Start with the longest sequences of potholes (k) and see how many of these you can afford to repair based on the budget.
4. This is done by calculating t, which is the minimum between the number of segments cnt[k] and how many your budget allows (budget // (k + 1)). Fix as many as t segments of length k.
5. For each segment repaired, reduce the budget accordingly and increase the count of repaired potholes (ans).
6. If there are more segments than you can afford to fix, consider them as shorter segments for future budget calculations by updating cnt[k-1].
7. Repeat the process for smaller segments until the budget is exhausted or all segments are considered.

By following the above logic, you maximize the number of potholes repaired within the given budget by focusing on cost-effective repairs.

Learn more about Greedy and Sorting patterns.

Solution Approach

The solution implements a counting and greedy technique to fix potholes optimally. Here's a step-by-step walkthrough of the approach:

1. Augment the Road: Append a "." to the road string to ensure any trailing pothole segments are processed.

2. Initialize Variables:

   • Create an array cnt of size n where n is the length of the augmented road. This will store the counts of consecutive pothole segments.
   • Use a variable k to keep track of the current length of a consecutive pothole segment.
   • Set ans to 0 to track the total number of potholes repaired.

3. Count Consecutive Potholes:

   • Traverse the road character by character using a loop.
   • Increment k for every pothole "x" encountered.
   • When a smooth road "." is encountered after a sequence of potholes, update cnt[k] by the number of such sequences and reset k to 0.

4. Repair Potholes Greedily:

   • Iterate over possible lengths of pothole sequences from n-1 to 1.
   • For each length k, determine the number of segments t that can be completely fixed given the current budget.
     • This is calculated by t = min(budget // (k + 1), cnt[k]).
   • Increment ans by t * k because for each segment of length k, k potholes are repaired.
   • Decrease the budget by t * (k + 1), accounting for the segments repaired.
   • If cnt[k] > t, add the leftover potholes to the next smaller group of potholes by updating cnt[k-1].

5. End Condition:

   • Stop the process if the budget is exhausted.

The algorithm prioritizes longer segments first due to their cost efficiency, ensuring the maximum number of potholes are repaired within the given budget.

Example Walkthrough

Let's walk through a simple example to illustrate the approach. Consider the following input:

• String road: "..xx..xxx...x"
• Budget: 6

Step 1: Augment the Road

To ensure all potholes are considered, append a "." to the road. It becomes "..xx..xxx...x.".

Step 2: Initialize Variables

Create an array, cnt, initialized to zeros with size equal to the length of the augmented road. Initialize k = 0 to count consecutive potholes and ans = 0 to sum repaired potholes.

Step 3: Count Consecutive Potholes

Iterate over the string:

• For index 2: encounter "x", increment k to 1.
• For index 3: encounter another "x", increment k to 2.
• For index 4: encounter "." (end of segment), update cnt[2] to 1, reset k to 0.
• For index 6: encounter "x", increment k to 1.
• For index 7: another "x", increment k to 2.
• For index 8: another "x", increment k to 3.
• For index 9: encounter "." (end of segment), update cnt[3] to 1, reset k to 0.
• For index 12: encounter "x", increment k to 1.
• For index 13: encounter "." (end of segment), update cnt[1] to 1, reset k to 0.

cnt now looks like [0, 1, 1, 1, ..., 0].

Step 4: Repair Potholes Greedily

Starting from length 3 downwards:

• For k = 3: Can fix t = min(budget // (3+1), cnt[3]) = min(6 // 4, 1) = 1. Repair this segment:

  • Update ans += 3 * 1 = 3.
  • Adjust budget -= 4 * 1 = 2.
  • If cnt[3] > t, add excess to cnt[2]. In this case, it's not needed.

• For k = 2: Can fix t = min(budget // (2+1), cnt[2]) = min(2 // 3, 1) = 0. Budget doesn't allow repair of length-2 segments.

• For k = 1: Can fix t = min(budget // (1+1), cnt[1]) = min(2 // 2, 1) = 1. Repair this segment:

  • Update ans += 1 * 1 = 4.
  • Adjust budget -= 2 * 1 = 0.

Step 5: End Condition

The budget is exhausted, so stop the process.

Result

The maximum number of potholes repaired is 4.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n), where n is the length of the string road. This is because the code iterates through the string road a few times: once to count the sequence of potholes and once more to calculate the maximum number of potholes that can be filled within the given budget.

The space complexity is also O(n), as the auxiliary array cnt used in the code has a size proportional to n. This array is utilized to keep track of the lengths of contiguous pothole segments.

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