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Minimum Recolors to Get K Consecutive Black Blocks

EasyUpdated: Aug 2, 2025
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Problem

You are given a 0-indexed string blocks of length n, where blocks[i] is either 'W' or 'B', representing the color of the ith 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.

Examples

Example 1:

Input: blocks = "WBBWWBBWBW", k = 7
Output: 3
Explanation:
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.

Example 2:

Input: blocks = "WBWBBBW", k = 2
Output: 0
Explanation:
No changes need to be made, since 2 consecutive black blocks already exist.
Therefore, we return 0.

Constraints:

  • n == blocks.length
  • 1 <= n <= 100
  • blocks[i] is either 'W' or 'B'.
  • 1 <= k <= n

Solution

Method 1 - Using Sliding Window

Code

Java
class Solution {
    public int minimumRecolors(String blocks, int k) {
        int n = blocks.length();
        int wCount = 0; // Count of white blocks in the current window
        int ans = Integer.MAX_VALUE;

        // Count 'W' in the first window of size k
        for (int i = 0; i < k; i++) {
            if (blocks.charAt(i) == 'W') {
                wCount++;
            }
        }
        ans = wCount;

        // Slide the window over the string
        for (int i = k; i < n; i++) {
            // Add the new block to the window
            if (blocks.charAt(i) == 'W') {
                wCount++;
            }
            // Remove the first block of the previous window
            if (blocks.charAt(i - k) == 'W') {
                wCount--;
            }
            // Update the minimum operations
            ans = Math.min(ans, wCount);
        }

        return ans;
    }
}
Python
class Solution:
    def minimumRecolors(self, blocks: str, k: int) -> int:
        n = len(blocks)
        w_count: int = 0  # Count of white blocks in the current window
        ans: int = float('inf')

        # Count 'W' in the first window of size k
        for i in range(k):
            if blocks[i] == 'W':
                w_count += 1

        ans = w_count

        # Slide the window over the string
        for i in range(k, n):
            # Add the new block to the window
            if blocks[i] == 'W':
                w_count += 1
            # Remove the first block of the previous window
            if blocks[i - k] == 'W':
                w_count -= 1
            # Update the minimum operations
            ans = min(ans, w_count)

        return ans[LEO](https://leo.ai.pleo.io/chat/leo/a53889f0-556f-4fd9-8742-5b4aaa324b56)

Complexity

  • ⏰ Time complexity: O(n), where n is the length of the string blocks, because we traverse the string once.
  • 🧺 Space complexity: O(1) since we only store variables for the count of white blocks and the minimum operations.

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