2379. Minimum Recolors to Get K Consecutive Black Blocks - Detailed Explanation
Problem Statement
You are given a string called blocks where each character is either 'B' (representing a black block) or 'W' (representing a white block). You are also given an integer k. The goal is to determine the minimum number of recoloring operations needed so that there is at least one substring (or contiguous segment) of k consecutive black blocks. In each operation, you can change a white block ('W') to a black block ('B').
Examples
Example 1
- Input: blocks = "WBBBW", k = 3
- Output: 0
- Explanation:
The substring "BBB" already exists in the blocks, so no recoloring operations are needed.
Example 2
- Input: blocks = "WBWBBBW", k = 2
- Output: 1
- Explanation:
One optimal solution is to recolor the block at index 1 (0-indexed) from 'W' to 'B' so that the substring "BB" is formed (for example, between indices 1 and 2 or by making another consecutive pair). There might be other valid ways, but only one operation is required.
Hints
-
Sliding Window Approach:
Since we are interested in consecutive blocks, try using a sliding window of size k. Evaluate how many white blocks are in the window because these are the blocks that need to be recolored. -
Optimization:
Instead of recalculating the number of white blocks from scratch for each window position, update the count as you slide the window from left to right.
Detailed Explanation
To solve the problem, we need to find a segment of k contiguous blocks that requires the fewest recolorings (i.e., has the minimum number of white blocks). This is because every white block in that segment must be flipped to black in order for it to become all black.
The approach can be summarized in the following steps:
-
Initialize a window of length k:
Count the number of white blocks ('W') in the first k blocks of the string. This count is the number of recolors needed for that window to become all black. -
Slide the window across the string:
For each subsequent window (move one block to the right each time), update the count of white blocks:- If the block that is leaving the window is white, decrease your count by 1.
- If the block that is entering the window is white, increase your count by 1.
Keep track of the minimum count observed across all windows.
-
Return the result:
The minimum count you have recorded is the minimum number of recoloring operations needed to get a window of k consecutive black blocks.
This sliding window method efficiently finds the optimal segment by considering overlapping subarrays instead of evaluating all possible segments separately.
Approaches
Approach 1: Sliding Window
This is the most optimal approach:
-
Time Complexity: O(n) where n is the number of blocks. You only pass through the string once to adjust the window.
-
Space Complexity: O(1) as you only use a few variables to maintain the count.
Steps for Sliding Window:
-
Compute the number of white blocks in the initial window (from index 0 to k-1).
-
Initialize a variable to store the minimum number of whites found so far.
-
Slide the window one position at a time until the end of the string.
-
In each slide, update the count based on the block leaving and the block entering.
-
The answer is the minimum value observed.
Approach 2: Brute Force
A brute force approach might consider every possible substring of length k, count the white blocks in each, and return the minimum count found.
- Time Complexity: O(n * k) which is less efficient, especially for large strings.
- Space Complexity: O(1).
This approach is straightforward but not optimal for large input sizes, which is why the sliding window approach is preferred.
Code Solutions
Python Implementation
Java Implementation
Step-by-Step Walkthrough and Visual Example
Consider Example 2 with blocks = "WBWBBBW" and k = 2:
-
Initial Window (Indices 0 to 1):
- Substring: "WB"
- White count = 1 (only index 0 is 'W').
- Set min_recolors = 1.
-
Slide Window to Indices 1 to 2:
- Outgoing block: index 0 ('W') → white_count decreases by 1: becomes 0.
- Incoming block: index 2 ('W') → white_count increases by 1: becomes 1.
- min_recolors remains 1.
-
Slide Window to Indices 2 to 3:
- Outgoing block: index 1 ('B') → white_count remains 1.
- Incoming block: index 3 ('B') → white_count remains 1.
- min_recolors remains 1.
-
Slide Window to Indices 3 to 4:
- Outgoing block: index 2 ('W') → white_count decreases to 0.
- Incoming block: index 4 ('B') → white_count remains 0.
- min_recolors is updated to 0 as we have found a window with 0 white blocks.
-
Slide Window to Indices 4 to 5:
- Outgoing block: index 3 ('B') → white_count remains 0.
- Incoming block: index 5 ('B') → white_count remains 0.
- min_recolors remains 0.
-
Slide Window to Indices 5 to 6:
- Outgoing block: index 4 ('B') → white_count remains 0.
- Incoming block: index 6 ('W') → white_count increases to 1.
- min_recolors remains 0.
After checking all possible windows, the minimum number of recolors needed is 0.
Common Mistakes
-
Recalculating Window Count from Scratch:
Recomputing the count of white blocks for each window from the beginning leads to inefficient solutions. Update the count incrementally as the window slides. -
Off-by-One Errors:
Make sure the window size exactly matches k and that the indices are correctly managed when moving the window. -
Incorrect Comparison Logic:
Failing to update or compare the minimum recolor count correctly can lead to incorrect results.
Edge Cases
-
All Blocks Already Black:
If the string consists entirely of 'B's, no recoloring is needed, and the answer is 0. -
k Equals 1:
In this case, you only need one black block, so if there is at least one 'B' in the string the answer is 0. Otherwise, if all are 'W', the answer is 1. -
k Greater Than the Number of Blocks:
This case is invalid under the given constraints because k is always less than or equal to the length of blocks.
Alternative Variations
- Instead of counting white blocks, a problem could require you to maximize the number of consecutive black blocks given a limited number of recolors. That variation might use similar sliding window techniques with additional logic.
Related Problems
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