Solution Explanation: Minimum Recolors to Get K Consecutive Black Blocks

This problem asks for the minimum number of operations needed to get at least one sequence of k consecutive black blocks ('B') in a string of black and white blocks. An operation consists of changing a white block ('W') to a black block.

The most efficient approach is using a sliding window technique. We maintain a window of size k and count the number of white blocks within that window. The minimum number of white blocks encountered across all possible windows represents the minimum number of recolors needed.

Algorithm:

1. Initialization:

   • Create a window of size k at the beginning of the string blocks.
   • Count the number of white blocks ('W') within this initial window. This count represents the initial number of recolors needed (min_recolors).

2. Sliding the Window:

   • Iterate through the string, moving the window one position to the right in each iteration.
   • For each move:
     • Subtract 1 from the white_block_count if the element leaving the window is 'W'.
     • Add 1 to the white_block_count if the new element entering the window is 'W'.
     • Update min_recolors with the minimum of the current min_recolors and the updated white_block_count.

3. Return the Result:

   • After iterating through the entire string, return min_recolors. This represents the minimum number of recolors required to achieve at least one sequence of k consecutive black blocks.

Time Complexity: O(n), where n is the length of the string blocks. We iterate through the string once using a sliding window.

Space Complexity: O(1), as we only use a few constant extra variables to store the counts and the minimum number of recolors.

Code Examples:

The following code examples demonstrate the solution in several programming languages. They all follow the same algorithmic structure described above.

Python:

def minimumRecolors(blocks: str, k: int) -> int:
    min_recolors = blocks[:k].count('W')  # Initialize with white blocks in the first window
    white_block_count = min_recolors
    for i in range(k, len(blocks)):
        white_block_count -= (blocks[i - k] == 'W')  # Remove block leaving the window
        white_block_count += (blocks[i] == 'W')      # Add block entering the window
        min_recolors = min(min_recolors, white_block_count)  # Update minimum
    return min_recolors
 

Java:

class Solution {
    public int minimumRecolors(String blocks, int k) {
        int minRecolors = 0;
        for (int i = 0; i < k; i++) {
            if (blocks.charAt(i) == 'W') {
                minRecolors++;
            }
        }
        int whiteBlockCount = minRecolors;
        for (int i = k; i < blocks.length(); i++) {
            whiteBlockCount -= (blocks.charAt(i - k) == 'W' ? 1 : 0);
            whiteBlockCount += (blocks.charAt(i) == 'W' ? 1 : 0);
            minRecolors = Math.min(minRecolors, whiteBlockCount);
        }
        return minRecolors;
    }
}

C++:

#include <string>
#include <algorithm>
 
class Solution {
public:
    int minimumRecolors(string blocks, int k) {
        int min_recolors = 0;
        for (int i = 0; i < k; ++i) {
            if (blocks[i] == 'W') min_recolors++;
        }
        int white_count = min_recolors;
        for (int i = k; i < blocks.length(); ++i) {
            white_count -= (blocks[i - k] == 'W');
            white_count += (blocks[i] == 'W');
            min_recolors = min(min_recolors, white_count);
        }
        return min_recolors;
    }
};

These examples are concise and demonstrate the core sliding window logic efficiently. Other languages would follow a similar pattern, adapting the syntax as needed.