Minimum Recolors to Get K Consecutive Black Blocks

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ArraysStringsSliding Windows

You are given a 0-indexed string blocks of length n, where blocks[i] is either 'W' or 'B', representing the color of the i<sup>th</sup> block. The characters 'W' and 'B' denote the colors white and black, respectively.

You are also given an integer k, which is the desired number of consecutive black blocks.

In one operation, you can recolor a white block such that it becomes a black block.

Return the minimum number of operations needed such that there is at least one occurrence of k consecutive black blocks.

For example:

If blocks = "WBBWWBBWBW" and k = 7, the output should be 3. One way to achieve 7 consecutive black blocks is to recolor the 0th, 3rd, and 4th blocks so that blocks = "BBBBBBBWBW". It can be shown that there is no way to achieve 7 consecutive black blocks in less than 3 operations. Therefore, we return 3.
If blocks = "WBWBBBW" and k = 2, the output should be 0. No changes need to be made, since 2 consecutive black blocks already exist. Therefore, we return 0.

Can you provide an algorithm to solve this problem efficiently?

Minimum Recolors to Get K Consecutive Black Blocks

Easy

Google

1 view

Topics:

ArraysStringsSliding Windows

You are given a 0-indexed string blocks of length n, where blocks[i] is either 'W' or 'B', representing the color of the i<sup>th</sup> block. The characters 'W' and 'B' denote the colors white and black, respectively.

You are also given an integer k, which is the desired number of consecutive black blocks.

In one operation, you can recolor a white block such that it becomes a black block.

Return the minimum number of operations needed such that there is at least one occurrence of k consecutive black blocks.

For example:

If blocks = "WBBWWBBWBW" and k = 7, the output should be 3. One way to achieve 7 consecutive black blocks is to recolor the 0th, 3rd, and 4th blocks so that blocks = "BBBBBBBWBW". It can be shown that there is no way to achieve 7 consecutive black blocks in less than 3 operations. Therefore, we return 3.
If blocks = "WBWBBBW" and k = 2, the output should be 0. No changes need to be made, since 2 consecutive black blocks already exist. Therefore, we return 0.

Can you provide an algorithm to solve this problem efficiently?

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 constraints on the size of the input string `blocks` and the value of `k`?
Can the input string `blocks` be empty, or can `k` be zero or greater than the length of `blocks`?
Is the input string `blocks` guaranteed to contain only 'W' and 'B' characters?
If there are multiple possible recolorings that result in the minimum number of recolors, is any one acceptable, or is there a specific one I should return?
Can I assume that both the input string and the integer k are valid and within acceptable ranges, or should I include input validation?

Brute Force Solution

Approach

The brute force approach to this problem is to consider every possible sequence of blocks of a certain length. For each of these sequences, we count how many blocks need to be recolored to make them all black, and then we find the sequence that requires the fewest recolors overall.

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

First, look at the very first sequence of blocks, where the number of blocks in the sequence is the required length.
Count how many blocks are white in that sequence. This tells you how many blocks you would have to recolor to make them all black.
Next, move one block to the right and look at the next sequence of blocks of the required length.
Again, count how many blocks are white in this new sequence.
Keep moving one block to the right, and repeat the counting process for each sequence you find.
As you go, keep track of the smallest number of recolors you've seen so far.
Once you've considered every possible sequence of the required length, the smallest recolor count you saved is your answer.

Code Implementation

def min_recolors_brute_force(blocks, k):
    minimum_recolors = float('inf')

    # Iterate through all possible sub-arrays of length k
    for i in range(len(blocks) - k + 1):
        current_recolors = 0

        # Count the number of 'W' blocks in the current sub-array
        for j in range(i, i + k):
            if blocks[j] == 'W':
                current_recolors += 1

        # Update the minimum recolors if necessary
        minimum_recolors = min(minimum_recolors, current_recolors)

    return minimum_recolors

Big(O) Analysis

Time Complexity

O(n*k) – The problem involves iterating through the blocks array of size n. For each iteration, we consider a window of size k (k consecutive blocks). Within each window, we count the number of white blocks. Since we iterate through n-k+1 windows and perform k operations (counting white blocks) for each window, the overall time complexity is O((n-k+1)*k). Simplifying, this approximates to O(n*k) assuming k is less than or equal to n. In the worst-case scenario where k is close to n, it can be seen as O(n*n) where n is the number of blocks and k is the number of required consecutive black blocks.

Space Complexity

O(1) – The provided algorithm iterates through the input blocks and keeps track of the minimum recolors needed. It only uses a few variables to store the current count of white blocks in a window of size k and the minimum number of recolors seen so far. Since the space used by these variables is constant and does not depend on the input size N (the number of blocks), the space complexity is O(1).

Optimal Solution

Approach

The goal is to find the smallest number of changes needed to get a certain number of consecutive black blocks. Instead of checking every possible set of blocks, we use a 'sliding window' technique. We move this window across the blocks, keeping track of how many changes are needed within that window, and then find the minimum.

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

Imagine a 'window' that's the exact size of the number of consecutive black blocks we want.
Place the window at the very beginning of the blocks.
Count how many white blocks are inside the window. This is the number of changes needed to make all blocks inside the window black.
Slide the window one block to the right.
Recalculate the number of white blocks in the new window by subtracting the contribution of the block leaving the window and adding the contribution of the block entering the window.
Keep track of the smallest number of white blocks you've seen in any window position so far. This is the minimum number of changes needed.
Continue sliding the window until it reaches the end of the blocks.
The smallest number of white blocks you tracked represents the fewest changes needed to get the desired consecutive black blocks.

Code Implementation

def minimum_recolors(blocks, consecutive_blocks):
    number_of_blocks = len(blocks)
    minimum_recolors_needed = consecutive_blocks

    # Initial count of white blocks in the first window.
    white_blocks_in_window = 0
    for i in range(consecutive_blocks):
        if blocks[i] == 'W':
            white_blocks_in_window += 1

    minimum_recolors_needed = white_blocks_in_window

    # Slide the window and update the count of white blocks.
    for i in range(1, number_of_blocks - consecutive_blocks + 1):
        # Subtract the contribution of the block leaving the window.
        if blocks[i - 1] == 'W':
            white_blocks_in_window -= 1

        # Add the contribution of the block entering the window.
        if blocks[i + consecutive_blocks - 1] == 'W':
            white_blocks_in_window += 1

        # Update the minimum recolors needed.
        minimum_recolors_needed = min(minimum_recolors_needed, white_blocks_in_window)

    return minimum_recolors_needed

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the blocks once using a sliding window of size k. In each iteration, it performs a constant number of operations: subtracting the contribution of the block leaving the window and adding the contribution of the block entering. Therefore, the time complexity is directly proportional to the number of blocks, n. The algorithm processes each block a fixed number of times, leading to a linear time complexity.

Space Complexity

O(1) – The algorithm maintains a sliding window and tracks the minimum number of white blocks encountered. It uses only a few integer variables to store the window's white block count and the overall minimum. The space used by these variables is constant and does not depend on the input size N (the number of blocks). Thus, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty blocks string	Return 0 if k is also 0, otherwise return -1 (or throw an exception) to indicate invalid input.
k is 0	Return 0, as no recoloring is needed for a zero-length consecutive block.
k is greater than the length of blocks	Return -1 (or throw an exception) as it is impossible to get k consecutive black blocks.
blocks contains only 'B' characters	Return 0, as no recoloring is needed.
blocks contains only 'W' characters	Return k, as all k consecutive blocks will need to be recolored.
blocks contains a very long string with alternating 'B' and 'W' characters and large k	Sliding window approach should efficiently calculate the minimum recolors by iterating through each possible window of size k.
k is equal to the length of blocks	Calculate the number of 'W' characters in the entire string and return that count.
Integer overflow if k is very large (although string size limits this to a degree)	Ensure that the data type used to store the count of recolors is large enough to prevent overflow, such as using long.