Problem Description

You have a string road that represents a road condition where:

• Each "x" character represents a pothole
• Each "." character represents smooth road

You're given a budget to repair potholes. The repair cost works as follows:

• You can repair n consecutive potholes in one operation
• The cost for repairing n consecutive potholes is n + 1

For example:

• Repairing 1 pothole costs 2
• Repairing 3 consecutive potholes costs 4
• Repairing 5 consecutive potholes costs 6

Your goal is to find the maximum number of potholes you can repair without exceeding the given budget.

The key insight is that you need to choose which groups of consecutive potholes to repair to maximize the total number of potholes fixed while staying within budget. You don't have to repair all potholes in a group - you can choose to repair them partially or skip them entirely based on your budget constraints.

Intuition

Let's think about the cost efficiency of repairing potholes. When we repair n consecutive potholes, we pay n + 1, which means the cost per pothole is (n + 1) / n.

For different lengths:

• 1 pothole: cost = 2, cost per pothole = 2/1 = 2.00
• 2 potholes: cost = 3, cost per pothole = 3/2 = 1.50
• 3 potholes: cost = 4, cost per pothole = 4/3 = 1.33
• 4 potholes: cost = 5, cost per pothole = 5/4 = 1.25
• 5 potholes: cost = 6, cost per pothole = 6/5 = 1.20

Notice that the cost per pothole decreases as we repair longer consecutive segments. This means repairing longer stretches of potholes is more cost-efficient.

Given this observation, our strategy should be to prioritize repairing the longest consecutive pothole segments first, as they give us the best "bang for our buck."

However, there's a twist: if we can't afford to repair a complete long segment (say length 5), we might consider breaking it down. For instance, if we have a segment of 5 potholes but can only afford to repair 3, we could treat it as a segment of 3 potholes (which we repair) and a segment of 2 potholes (which we leave for later consideration).

This leads us to the approach: first count all consecutive pothole segments by their lengths, then greedily repair from the longest segments to the shortest, breaking down segments when our budget doesn't allow full repairs. This ensures we maximize the number of potholes repaired within our budget constraint.

Learn more about Greedy and Sorting patterns.

Solution Approach

The implementation follows our greedy strategy of prioritizing longer pothole segments:

Step 1: Count consecutive pothole segments

First, we append a "." to the road string to ensure we properly count the last segment if it ends with potholes. We create an array cnt where cnt[k] stores the count of segments that have exactly k consecutive potholes.

We iterate through the road string, counting consecutive "x" characters. When we encounter a ".", we record the count in our cnt array. For example, if the road is "xx.xxx.", we'd have:

• One segment of length 2
• One segment of length 3
• So cnt[2] = 1 and cnt[3] = 1

Step 2: Greedily repair from longest to shortest

We iterate from the longest possible segment length n-1 down to 1. For each length k:

1. Calculate how many complete segments of length k we can afford: t = min(budget // (k + 1), cnt[k])

   • budget // (k + 1) tells us how many segments we can afford
   • We take the minimum with cnt[k] since we can't repair more segments than we have

2. Update our answer: ans += t * k

   • We repaired t segments, each containing k potholes

3. Update the budget: budget -= t * (k + 1)

   • Each segment of length k costs k + 1

4. Handle leftover segments: cnt[k - 1] += cnt[k] - t

   • If we couldn't repair all segments of length k, the remaining cnt[k] - t segments are "broken down" into segments of length k-1
   • This clever trick allows us to consider partial repairs of longer segments

The algorithm continues until we've either exhausted our budget or considered all segment lengths. This greedy approach ensures we maximize the number of potholes repaired by always choosing the most cost-efficient repairs first.

Example Walkthrough

Let's walk through a concrete example with road = "xx.xxx.x" and budget = 7.

Step 1: Identify consecutive pothole segments

First, we append a "." to get "xx.xxx.x." and scan for consecutive potholes:

• "xx" → segment of length 2
• "xxx" → segment of length 3
• "x" → segment of length 1

Our count array becomes:

• cnt[1] = 1 (one segment of length 1)
• cnt[2] = 1 (one segment of length 2)
• cnt[3] = 1 (one segment of length 3)

Step 2: Process segments from longest to shortest

Round 1: Length 3

• We have cnt[3] = 1 segment of length 3
• Cost to repair one segment: 3 + 1 = 4
• Can we afford it? budget // 4 = 7 // 4 = 1 segment
• We can repair: min(1, 1) = 1 segment
• Potholes repaired: 1 × 3 = 3
• Budget remaining: 7 - 4 = 3
• Running total: 3 potholes repaired

Round 2: Length 2

• We have cnt[2] = 1 segment of length 2
• Cost to repair one segment: 2 + 1 = 3
• Can we afford it? budget // 3 = 3 // 3 = 1 segment
• We can repair: min(1, 1) = 1 segment
• Potholes repaired: 1 × 2 = 2
• Budget remaining: 3 - 3 = 0
• Running total: 3 + 2 = 5 potholes repaired

Round 3: Length 1

• We have cnt[1] = 1 segment of length 1
• Cost to repair one segment: 1 + 1 = 2
• Can we afford it? budget // 2 = 0 // 2 = 0 segments
• Budget exhausted, can't repair any more

Final Result: We repaired 5 potholes total (the segments of length 3 and 2), leaving only the single pothole unrepaired. This demonstrates how the greedy approach prioritizes longer segments for better cost efficiency.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm consists of two main passes:

1. The first loop iterates through the road string once to count consecutive potholes, taking O(n) time where n is the length of the road string.
2. The second loop iterates through the cnt array from index n-1 down to 1. In the worst case, this is O(n) iterations. Although there's a redistribution operation cnt[k - 1] += cnt[k] - t within the loop, each pothole segment is processed at most once and potentially redistributed, but the total work across all iterations remains bounded by O(n).

Therefore, the overall time complexity is O(n) + O(n) = O(n).

Space Complexity: O(n)

The algorithm uses an array cnt of size n to store the count of pothole segments of each length. This array requires O(n) space. The other variables (k, ans, budget, t) use constant space O(1).

Therefore, the overall space complexity is O(n).

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

Common Pitfalls

1. Forgetting to Handle the Last Pothole Sequence

One of the most common mistakes is failing to process the last sequence of potholes when the road string ends with "x" characters. Without proper handling, the algorithm would miss counting these final potholes.

Why it happens: When iterating through the string and counting consecutive potholes, we only record a sequence when we encounter a ".". If the string ends with potholes, there's no final "." to trigger the recording.

Example of the bug:

# Buggy version - misses last sequence
for char in road:
    if char == "x":
        current_sequence_length += 1
    elif current_sequence_length > 0:
        sequence_count[current_sequence_length] += 1
        current_sequence_length = 0
# If road = "xx.xxx", only the first sequence (length 2) gets counted

Solution: Add a sentinel character (like ".") to the end of the road string to ensure all sequences are processed:

road += "."  # Ensures last sequence is counted

2. Incorrect Partial Repair Logic

Another critical pitfall is mishandling the "leftover" sequences when we can't afford to repair all sequences of a given length. The algorithm cleverly breaks down unrepaired sequences into shorter ones for future consideration.

Why it happens: It's tempting to simply skip sequences we can't fully afford, but this misses optimization opportunities where we could repair them as shorter sequences.

Example of the bug:

# Buggy version - doesn't handle partial repairs
for length in range(road_length - 1, 0, -1):
    sequences_to_fix = min(budget // (length + 1), sequence_count[length])
    fixed_potholes += sequences_to_fix * length
    budget -= sequences_to_fix * (length + 1)
    # Missing: sequence_count[length - 1] += sequence_count[length] - sequences_to_fix

Impact: If we have a sequence of 5 potholes but can only afford to repair sequences of length 3, the buggy version would skip it entirely instead of considering it as a sequence of length 4 in the next iteration.

Solution: Always transfer unrepaired sequences to the next shorter length:

sequence_count[length - 1] += sequence_count[length] - sequences_to_fix

3. Off-by-One Error in Cost Calculation

A subtle but impactful mistake is using the wrong cost formula, such as using length instead of length + 1 for the repair cost.

Example of the bug:

# Wrong: using length instead of length + 1
sequences_to_fix = min(budget // length, sequence_count[length])  # Bug!
budget -= sequences_to_fix * length  # Bug!

Solution: Always remember the cost formula is n + 1 for repairing n consecutive potholes:

sequences_to_fix = min(budget // (length + 1), sequence_count[length])
budget -= sequences_to_fix * (length + 1)