Problem Description

You have a string blocks consisting of characters 'W' (white) and 'B' (black), representing the colors of blocks arranged in a line. Given an integer k, your goal is to find at least one occurrence of k consecutive black blocks.

You can perform operations to recolor white blocks to black (but not black to white). The task is to find the minimum number of operations needed to create at least one sequence of k consecutive black blocks.

For example, if blocks = "WBBWWBBWBW" and k = 7, you would need to recolor some white blocks to create 7 consecutive black blocks. The algorithm uses a sliding window approach to find the window of size k that already contains the most black blocks (or equivalently, the fewest white blocks), as this would require the minimum number of recoloring operations.

The solution counts the number of white blocks ('W') in each window of size k as it slides through the string. The window with the fewest white blocks represents the optimal position to create k consecutive black blocks, and the count of white blocks in that window is the answer.

Intuition

To create k consecutive black blocks, we need to select a contiguous segment of length k from the string and ensure all blocks in that segment are black. Since we can only recolor white blocks to black, the number of operations needed for any segment equals the number of white blocks in that segment.

The key insight is that we want to minimize the number of operations, which means we should find the segment of length k that already has the maximum number of black blocks (or equivalently, the minimum number of white blocks). This way, we'll need to recolor the fewest blocks possible.

Instead of checking every possible segment of length k independently, we can use a sliding window technique. We start by counting the white blocks in the first window of size k. Then, as we slide the window one position to the right, we only need to:

• Add 1 to our count if the new block entering the window is white
• Subtract 1 from our count if the block leaving the window is white

This allows us to efficiently track the number of white blocks in each window with O(1) updates rather than recounting the entire window each time. By keeping track of the minimum count across all windows, we find the optimal position that requires the fewest recoloring operations.

The answer is simply the minimum number of white blocks found in any window of size k, as these are the blocks we need to recolor to achieve k consecutive black blocks.

Learn more about Sliding Window patterns.

Solution Approach

The solution implements a sliding window approach to efficiently find the minimum number of white blocks in any window of size k.

Initial Window Setup:

ans = cnt = blocks[:k].count('W')

We start by counting the number of white blocks in the first window (indices 0 to k-1). This count is stored in both cnt (current window count) and ans (minimum count so far).

Sliding the Window:

for i in range(k, len(blocks)):
    cnt += blocks[i] == 'W'
    cnt -= blocks[i - k] == 'W'
    ans = min(ans, cnt)

The loop starts from index k and goes to the end of the string. For each iteration:

• blocks[i] is the new block entering the window from the right
• blocks[i - k] is the block leaving the window from the left

The window update is done efficiently:

• cnt += blocks[i] == 'W': If the new block is white, increment the count by 1 (the expression blocks[i] == 'W' evaluates to True (1) or False (0))
• cnt -= blocks[i - k] == 'W': If the leaving block is white, decrement the count by 1

After updating the current window's white block count, we update ans to maintain the minimum count seen so far using ans = min(ans, cnt).

Time and Space Complexity:

• Time Complexity: O(n) where n is the length of the string, as we visit each character once
• Space Complexity: O(1) as we only use a constant amount of extra space for variables ans and cnt

The final answer is the minimum number of white blocks found in any window of size k, which represents the minimum number of recoloring operations needed.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Example: blocks = "WBWBBBW" and k = 3

We need to find a window of size 3 that requires the minimum number of recoloring operations to create 3 consecutive black blocks.

Step 1: Initialize the first window (indices 0-2)

• First window: "WBW"
• Count white blocks: 2 (positions 0 and 2)
• Set ans = 2 and cnt = 2

Step 2: Slide window to position 1-3

• Window shifts right by 1: "BWB"
• Block leaving (index 0): 'W' → decrement cnt by 1
• Block entering (index 3): 'B' → no change (not white)
• New cnt = 2 - 1 + 0 = 1
• Update ans = min(2, 1) = 1

Step 3: Slide window to position 2-4

• Window: "WBB"
• Block leaving (index 1): 'B' → no change
• Block entering (index 4): 'B' → no change
• New cnt = 1 - 0 + 0 = 1
• Update ans = min(1, 1) = 1

Step 4: Slide window to position 3-5

• Window: "BBB"
• Block leaving (index 2): 'W' → decrement cnt by 1
• Block entering (index 5): 'B' → no change
• New cnt = 1 - 1 + 0 = 0
• Update ans = min(1, 0) = 0

Step 5: Slide window to position 4-6

• Window: "BBW"
• Block leaving (index 3): 'B' → no change
• Block entering (index 6): 'W' → increment cnt by 1
• New cnt = 0 - 0 + 1 = 1
• Update ans = min(0, 1) = 0

Result: The minimum number of operations needed is 0, as we found a window "BBB" (positions 3-5) that already contains 3 consecutive black blocks.

This example demonstrates how the sliding window efficiently tracks white blocks in each window of size k, updating the count with O(1) operations as we add/remove elements, ultimately finding the optimal position requiring the fewest recoloring operations.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the string blocks. The algorithm performs an initial count operation on the first k characters which takes O(k) time, then iterates through the remaining n - k characters with constant time operations for each character (checking if a character equals 'W', incrementing/decrementing counter, and updating minimum). Since k ≤ n, the overall time complexity is O(k) + O(n - k) = O(n).

The space complexity is O(1). The algorithm only uses a fixed number of variables (ans, cnt, and i) regardless of the input size. The string slicing operation blocks[:k] creates a temporary substring, but since we're only counting characters and not storing it, and Python's count method operates in constant extra space, the overall space usage remains constant.

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

Common Pitfalls

1. Off-by-One Errors in Window Boundaries

A common mistake is incorrectly calculating which element is leaving the window. When at index i, the element leaving is at index i - k, not i - k + 1 or i - k - 1.

Incorrect Example:

# Wrong: Using i - k + 1
for i in range(k, len(blocks)):
    cnt += blocks[i] == 'W'
    cnt -= blocks[i - k + 1] == 'W'  # Error: wrong index

Solution: Remember that if your window ends at index i, and has size k, it starts at index i - k + 1. Therefore, when moving to include blocks[i], the element being removed is at blocks[i - k].

2. Not Handling Edge Case When k Equals String Length

If k == len(blocks), the initial window is the entire string, and the loop doesn't execute at all. This could lead to issues if not properly initialized.

Problematic Pattern:

# If initialization is done incorrectly
ans = float('inf')  # Wrong initialization
cnt = 0
for i in range(k, len(blocks)):
    # This loop never runs if k == len(blocks)
    ...
return ans  # Returns infinity instead of actual count

Solution: Always initialize ans with the count from the first valid window:

ans = cnt = blocks[:k].count('W')

3. Using Boolean Addition Incorrectly

While blocks[i] == 'W' elegantly returns 1 or 0 for True/False, some might try to "optimize" this incorrectly.

Incorrect Example:

# Wrong: Trying to combine operations
cnt += (blocks[i] == 'W') - (blocks[i - k] == 'W')  
# This works but is less readable and more error-prone

Solution: Keep the operations separate for clarity and debugging:

cnt += blocks[i] == 'W'
cnt -= blocks[i - k] == 'W'

4. Forgetting to Update the Minimum

It's easy to forget updating the answer variable after adjusting the window count.

Incorrect Example:

for i in range(k, len(blocks)):
    cnt += blocks[i] == 'W'
    cnt -= blocks[i - k] == 'W'
    # Forgot: ans = min(ans, cnt)
return ans  # Returns only the initial window's count

Solution: Always update the minimum after each window adjustment:

ans = min(ans, cnt)

5. Misunderstanding the Problem Goal

Some might think we need to find k consecutive blocks that are already black, rather than finding where to create them with minimum operations.

Wrong Interpretation:

# Looking for existing consecutive black blocks
for i in range(len(blocks) - k + 1):
    if 'W' not in blocks[i:i+k]:
        return 0  # Found k consecutive blacks

Solution: The problem asks for the minimum number of operations (white to black recolors) needed to create k consecutive black blocks anywhere in the string, not to find existing ones.