Maximum Number of Potholes That Can Be Fixed

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Dynamic ProgrammingArrays

You are given a road represented as a binary string road of length n. A '1' represents a section of the road that is in good condition, and a '0' represents a pothole.

You are also given an integer k, which is the maximum number of potholes you can fix.

Your task is to find the maximum number of consecutive sections of the road that can be in good condition after fixing at most k potholes.

Example 1:

Input: road = "1101001110", k = 2
Output: 8
Explanation: We can fix the potholes at indices 2 and 4 to get "1111101110". The longest consecutive sequence of good sections is of length 5. But by fixing the pothole at index 9 as well, we get "1111101111". The longest consecutive sequence of good sections is now of length 8, which is also the maximum possible length.

Example 2:

Input: road = "10101", k = 1
Output: 3
Explanation: We can fix the pothole at index 1 to get "11101". The longest consecutive sequence of good sections is of length 3.

Example 3:

Input: road = "00000", k = 3
Output: 3
Explanation: We can fix the potholes at indices 0, 1, and 2 to get "11100". The longest consecutive sequence of good sections is of length 3.

Constraints:

1 <= n <= 10⁵
road[i] is either '0' or '1'.
0 <= k <= n

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the dimensions of the city grid, and what is the maximum value for the number of potholes in a single cell? Are there any constraints on the size of the grid or the number of potholes?
Can the number of potholes in a cell be zero or negative?
If the number of asphalt buckets is greater than the total number of potholes, what should I return?
Is the input array guaranteed to be valid (e.g., not null or empty)?
Is the number of asphalt buckets guaranteed to be non-negative?

Brute Force Solution

Approach

The brute force approach to fixing the most potholes means we're going to try every single possible combination of which potholes to fix. We'll fix some, then fix others, then fix even more, and so on. After trying every possible combination, we'll simply see which combination allows us to fix the maximum number of potholes within our limits.

Here's how the algorithm would work step-by-step:

First, consider the case where we fix absolutely no potholes. This is one possibility.
Next, consider fixing only the first pothole. That's another possibility.
Then, consider fixing only the second pothole. And then only the third, and so on until you've considered fixing only one pothole at a time.
Now, start looking at pairs of potholes. Consider fixing the first and second, then the first and third, then the first and fourth, and so on. Then consider fixing the second and third, the second and fourth, and so on.
Continue this process, increasing the number of potholes we fix in each combination. Try fixing three potholes at a time, then four, and so on.
For each combination of potholes we consider fixing, check if fixing that particular set of potholes is within the limit we have (such as a budget or a time constraint).
Keep track of the maximum number of potholes fixed among all the valid combinations we've tried.
After trying absolutely every possible combination of potholes and checking if it's within our limit, we'll have found the combination that fixes the most potholes.

Code Implementation

def max_potholes_fixed_brute_force(pothole_costs, budget):
    number_of_potholes = len(pothole_costs)
    max_potholes_fixed = 0

    for i in range(1 << number_of_potholes):
        current_cost = 0
        potholes_fixed_count = 0

        # Iterate through each pothole and check if it's in the current combination
        for j in range(number_of_potholes):
            if (i >> j) & 1:
                current_cost += pothole_costs[j]
                potholes_fixed_count += 1

        # See if cost is within budget
        if current_cost <= budget:

            # Update max potholes fixed if this combination is better
            if potholes_fixed_count > max_potholes_fixed:
                max_potholes_fixed = potholes_fixed_count

    return max_potholes_fixed

Big(O) Analysis

Time Complexity

O(2^n) – The described brute force approach explores every possible combination of potholes to fix. For 'n' potholes, each pothole can either be fixed or not fixed, leading to 2 possibilities for each pothole. Therefore, we have 2 * 2 * ... * 2 (n times) = 2^n possible combinations to consider. Since we iterate through each of these combinations to check feasibility, the time complexity is O(2^n).

Space Complexity

O(1) – The provided brute force approach iterates through all possible combinations of potholes to fix. While it conceptually explores many combinations, it only needs to store a few variables to track the current combination being considered and the maximum number of potholes fixed so far. No auxiliary data structures like arrays or hash maps that scale with the number of potholes (N) are used, and the recursion depth is implicitly managed by iteration, not explicit function calls. Therefore, the space complexity is constant, independent of the input size N.

Optimal Solution

Approach

The best way to fix the most potholes is to focus on repairing sections of the road one at a time, prioritizing sections with the most potholes. We want to efficiently use our repair material without wasting any on sections that don't need it as much.

Here's how the algorithm would work step-by-step:

First, consider the entire road as a series of smaller sections. Each section has a specific number of potholes.
Our goal is to figure out which sections we should repair to fix the maximum number of potholes possible, given the limited amount of repair material we have.
Start by looking at each section individually and determine how much repair material is needed to fix all the potholes in that section.
Next, sort the sections based on how many potholes we can fix per unit of repair material used. This helps us prioritize the sections where we get the most 'bang for our buck'.
Begin repairing sections in order of this priority. As we repair a section, subtract the amount of repair material used from our total available material.
Continue repairing sections until we run out of repair material.
The total number of potholes fixed across all the sections we were able to repair is the maximum number of potholes we can fix.

Code Implementation

def max_potholes_fixed(pothole_depths, repair_material):
    fixed_potholes_count = 0

    for section_index in range(len(pothole_depths)):
        section_potholes = sorted(pothole_depths[section_index])
        section_cost = sum(section_potholes)

        # If enough material exists, fix all potholes in section.
        if section_cost <= repair_material:
            fixed_potholes_count += len(section_potholes)
            repair_material -= section_cost

        else:
            # If not enough material, fix smallest potholes first.
            potholes_fixed_in_section = 0
            material_used = 0

            for depth in section_potholes:
                if material_used + depth <= repair_material:
                    material_used += depth
                    potholes_fixed_in_section += 1
                else:
                    break

            # Update the count of potholes fixed and remaining material.
            fixed_potholes_count += potholes_fixed_in_section
            repair_material -= material_used

        # No material left, so exit the loop early.
        if repair_material == 0:
            break

    return fixed_potholes_count

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm first iterates through n sections to calculate the repair material needed for each section. Then, the sections are sorted based on potholes per unit of repair material using an efficient sorting algorithm like merge sort or quicksort, which takes O(n log n) time. Finally, the algorithm iterates through the sorted sections, repairing them until the repair material is exhausted, which takes at most O(n) time. Since sorting dominates the time complexity, the overall time complexity is O(n log n).

Space Complexity

O(N) – The algorithm sorts the sections based on potholes per unit of repair material, implying the creation of an auxiliary data structure (likely a list or array) to store and sort the sections. This auxiliary data structure stores information for each of the N sections, where N is the number of sections. Therefore, the space complexity is directly proportional to the number of sections, resulting in O(N) auxiliary space.

Edge Cases

Case	How to Handle
grid is null or empty	Return 0 since there are no potholes to fix when the grid is null or empty.
buckets is negative	Treat negative buckets as 0, as you can't use negative asphalt.
grid contains negative pothole values	Treat negative pothole values as 0, as potholes cannot be negative.
buckets is zero	Return 0 since no potholes can be fixed when no asphalt buckets are available.
grid contains very large pothole values exceeding integer limits	Use long or appropriate data type to store pothole values and sums to prevent integer overflow.
buckets is much larger than the total number of potholes	Return the total number of potholes in the grid since we can fix all of them.
All cells in the grid have zero potholes	Return 0 since there are no potholes to fix even if buckets are available.
Maximum grid size causing memory issues	Optimize the algorithm to use less memory (e.g., using an in-place modification or a more memory-efficient data structure than sorting the entire grid).

grid is null or empty

How to Handle:

Return 0 since there are no potholes to fix when the grid is null or empty.

buckets is negative

How to Handle:

Treat negative buckets as 0, as you can't use negative asphalt.

grid contains negative pothole values

How to Handle:

Treat negative pothole values as 0, as potholes cannot be negative.

buckets is zero

How to Handle:

Return 0 since no potholes can be fixed when no asphalt buckets are available.

grid contains very large pothole values exceeding integer limits

How to Handle:

Use long or appropriate data type to store pothole values and sums to prevent integer overflow.

buckets is much larger than the total number of potholes

How to Handle:

Return the total number of potholes in the grid since we can fix all of them.

All cells in the grid have zero potholes

How to Handle:

Return 0 since there are no potholes to fix even if buckets are available.

Maximum grid size causing memory issues

How to Handle:

Optimize the algorithm to use less memory (e.g., using an in-place modification or a more memory-efficient data structure than sorting the entire grid).