What is the brute force solution for Minimum Recolors to Get K Consecutive Black Blocks?

The brute force approach for Minimum Recolors to Get K Consecutive Black Blocks is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible substring of length k in the given string. For each substring, we count how many white blocks need to be recolored to achieve k consecutive black blocks. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Recolors to Get K Consecutive Black Blocks?

Minimum Recolors to Get K Consecutive Black Blocks uses the following concepts: String, Sliding Window. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Minimum Recolors to Get K Consecutive Black Blocks?

Always explain your thought process clearly during the interview. Consider edge cases before finalizing your solution. Practice explaining your code as you write it to simulate interview conditions.

What are common mistakes when solving Minimum Recolors to Get K Consecutive Black Blocks?

Not considering edge cases where k is equal to n. Failing to update the count correctly while sliding the window.

#2379

Minimum Recolors to Get K Consecutive Black Blocks

Easy

StringSliding WindowSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a sliding window technique to efficiently count the number of white blocks in a window of size k. This avoids the need to recount for overlapping parts of the string.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable to count the number of white blocks in the first window of size k.
2Step 2: Slide the window across the string, updating the count by removing the block that goes out of the window and adding the new block that comes into the window.
3Step 3: Keep track of the minimum recolors needed as you slide the window.

solution.py14 lines

1def minimumRecolors(blocks, k):
2    min_recolors = 0
3    n = len(blocks)
4    for i in range(k):
5        if blocks[i] == 'W':
6            min_recolors += 1
7    current_recolors = min_recolors
8    for i in range(k, n):
9        if blocks[i] == 'W':
10            current_recolors += 1
11        if blocks[i - k] == 'W':
12            current_recolors -= 1
13        min_recolors = min(min_recolors, current_recolors)
14    return min_recolors

ℹ

Complexity note: This complexity is achieved because we only traverse the string once, maintaining a count of white blocks in the current window without needing nested loops.

1Using a sliding window reduces the number of operations needed to count the recolors.
2Maintaining a running count of changes avoids redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.