[Array] Minimum Recolors to Get K Consecutive Black Blocks - Easy

To solve the problem of finding the least number of recolors to get K black blocks in a row in an array, we look at how the blocks are arranged. We can think of this as a binary array. In this array, ‘1’ means a black block and ‘0’ means a white block. Our goal is to find out how many white blocks we need to change to black to have K ’1’s in a row. We can use a sliding window technique to do this well. This way, we can check parts of the array without doing extra work.

In this article, we will look at different ways to solve the problem of minimum recolors for K consecutive black blocks. First, we will explain the problem statement. Then, we will show an optimal sliding window method in Java, Python, and C++. After that, we will give a brute force solution for all three programming languages. We will also check the time and space complexity for each method. Finally, we will answer some common questions about this problem.

  • Array Minimum Recolors to Obtain K Consecutive Black Blocks - Easy
  • Understanding the Problem Statement for Minimum Recolors
  • Optimal Sliding Window Approach for Minimum Recolors in Java
  • Sliding Window Implementation in Python for Minimum Recolors
  • C++ Solution Using Sliding Window Technique for Minimum Recolors
  • Brute Force Approach for Minimum Recolors in Java
  • Brute Force Approach for Minimum Recolors in Python
  • Brute Force Approach for Minimum Recolors in C++
  • Time and Space Complexity Analysis of Solutions
  • Frequently Asked Questions

For more context and similar problems, you can check out resources like Array Two Sum and Array Maximum Subarray.

Understanding the Problem Statement for Minimum Recolors

We want to find the least number of recolors needed to get K black blocks in a row. We have a string of blocks. Each block is either black (‘B’) or white (‘W’). Our goal is to figure out how many white blocks we need to change to black to have at least K black blocks together.

Problem Definition:

  • We get a string called blocks. It has characters ‘B’ and ‘W’.
  • We have an integer K. This tells us how many black blocks we want in a row.
  • We need to find the smallest number of ’W’s to change to ’B’s in any part of the string that is K long.

Example:

  • Input: blocks = "WBBWWBB", K = 3
  • Output: 1
    • Explanation: If we change one ‘W’ in “WBB” to ‘B’, we get “BBB”.

Constraints:

  • The length of blocks is not too long. So, we can use the sliding window method to solve it fast.

We can solve this problem well using different methods. One good way is the sliding window technique. This method helps us keep track of how many white blocks are in the current window of size K. This way, we can make the recoloring faster.

By understanding the problem clearly, we can write good algorithms in different programming languages.

Optimal Sliding Window Approach for Minimum Recolors in Java

We can find the minimum number of recolors needed to get K consecutive black blocks using the sliding window method. This way, we can solve the problem quickly with a time of O(N). Here, N is the length of the array. This method works well for big input sizes.

Algorithm Steps:

  1. First, we set a variable to count the white blocks (whiteCount) in the current window of size K.
  2. Next, we use the sliding window to go through the array:
    • For the first K blocks, we count how many white blocks there are.
    • Then we slide the window one block at a time:
      • We take away the block that is leaving the window.
      • We add the block that is coming into the window.
  3. We keep track of the smallest number of white blocks we find while going through the array.

Java Implementation:

public class MinimumRecolors {
    public int minimumRecolors(String blocks, int K) {
        int n = blocks.length();
        int whiteCount = 0;

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

        int minRecolors = whiteCount;

        // Slide the window across the blocks
        for (int i = K; i < n; i++) {
            // Remove the block going out of window
            if (blocks.charAt(i - K) == 'W') {
                whiteCount--;
            }
            // Add the new block coming into the window
            if (blocks.charAt(i) == 'W') {
                whiteCount++;
            }
            // Update the minimum recolors needed
            minRecolors = Math.min(minRecolors, whiteCount);
        }

        return minRecolors;
    }

    public static void main(String[] args) {
        MinimumRecolors solution = new MinimumRecolors();
        String blocks = "WBBWWBB";
        int K = 3;
        System.out.println("Minimum recolors needed: " + solution.minimumRecolors(blocks, K)); // Output: 2
    }
}

In this code: - The minimumRecolors method finds the minimum number of recolors needed to have K consecutive black blocks (‘B’). - The main method shows how to use this function with an example input.

This sliding window method works well and is easy to understand. We can manage bigger datasets easily. If we want to learn more about array problems, we can check out Array Maximum Subarray or Array Best Time to Buy and Sell Stock.

Sliding Window Implementation in Python for Minimum Recolors

We can use the sliding window method to find out how many white blocks we need to change to black blocks. We want to get K black blocks in a row from an array that shows colors as B (black) and W (white).

Problem Statement

Given a string that shows blocks of colors, we need to find the least number of white blocks (W) we must change to black (B) so that we have at least K black blocks together.

Python Implementation

Here is how we can use the sliding window approach in Python:

def minimumRecolors(blocks: str, k: int) -> int:
    n = len(blocks)
    min_recolors = float('inf')
    current_white_count = 0
    
    # Initial count of white blocks in the first window of size k
    for i in range(k):
        if blocks[i] == 'W':
            current_white_count += 1
            
    min_recolors = current_white_count
    
    # Slide the window across the blocks
    for i in range(k, n):
        if blocks[i] == 'W':
            current_white_count += 1
        if blocks[i - k] == 'W':
            current_white_count -= 1
        
        min_recolors = min(min_recolors, current_white_count)
    
    return min_recolors

Explanation of the Code

  • Initialization:
    • n is the length of the blocks string.
    • We set min_recolors to infinity. This helps us find the minimum.
    • current_white_count counts how many white blocks are in the current window.
  • First Window Calculation:
    • The first loop counts white blocks in the first K blocks.
  • Sliding the Window:
    • The second loop moves the window one block at a time.
    • If a new block comes in the window (right side), we check if it is white and add to the count.
    • We also check if the block that goes out of the window (left side) is white and take it away from the count.
  • Update Minimum:
    • After we adjust the count for the current window, we update the minimum recolors needed.

This method finds the minimum number of recolors in O(n) time. It is good for large inputs.

For more about similar array problems, we can see Array Two Sum or Array Maximum Subarray.

C++ Solution Using Sliding Window Technique for Minimum Recolors

We want to find the least number of recolors needed to get K black blocks in a row from an array of blocks. The blocks are either ‘B’ for black or ‘W’ for white. We can use the sliding window technique in C++. This method helps us go through the array while counting the white blocks in a window of size K.

C++ Code Implementation

#include <iostream>
#include <vector>
#include <string>
using namespace std;

int minimumRecolors(string blocks, int k) {
    int minRecolors = k; // Max recolors if all are white
    int whiteCount = 0;  // Count of white blocks in current window

    // Set up the first window
    for (int i = 0; i < k; i++) {
        if (blocks[i] == 'W') {
            whiteCount++;
        }
    }
    minRecolors = min(minRecolors, whiteCount);

    // Slide the window across the blocks
    for (int i = k; i < blocks.size(); i++) {
        // Remove the leftmost block from the previous window
        if (blocks[i - k] == 'W') {
            whiteCount--;
        }
        // Add the new block to the current window
        if (blocks[i] == 'W') {
            whiteCount++;
        }
        minRecolors = min(minRecolors, whiteCount);
    }

    return minRecolors;
}

int main() {
    string blocks = "WBWBBBW";
    int k = 3;
    cout << "Minimum recolors needed: " << minimumRecolors(blocks, k) << endl;
    return 0;
}

Explanation of the Code

  • Function Definition: The function minimumRecolors takes a string blocks and an integer k. These represent the blocks and the number of consecutive blocks we need.
  • Initial Window Setup: We check the first k blocks to count how many are white.
  • Sliding Mechanism: When the window slides, we take out the left block from the count if it is white. We add the new right block into the count if it is white.
  • Minimum Calculation: As we slide, we keep track of the least number of recolors needed.

Example Usage

If we have the string blocks = "WBWBBBW" and k = 3, the output will be:

Minimum recolors needed: 1

This shows that we need to change at least one block to get three black blocks together.

Using the sliding window technique helps us make the time needed go down to O(n). This is good for larger input sizes. If you want to explore more about similar array problems, check out the article on Array Two Sum.

Brute Force Approach for Minimum Recolors in Java

In the brute force method to find the minimum recolors needed to get K black blocks in a row from a string of blocks made of ‘B’ (black) and ‘W’ (white), we will check every possible segment of length K. For each segment, we will count the white blocks and remember the smallest count we find.

Java Implementation

Here is how we can do the brute force method in Java:

public class MinimumRecolors {
    public int minimumRecolors(String blocks, int K) {
        int minRecolors = Integer.MAX_VALUE;
        int n = blocks.length();

        // Check each starting point for segments of length K
        for (int i = 0; i <= n - K; i++) {
            int recolors = 0;

            // Count the white blocks in the current segment
            for (int j = i; j < i + K; j++) {
                if (blocks.charAt(j) == 'W') {
                    recolors++;
                }
            }

            // Update the minimum recolors needed
            minRecolors = Math.min(minRecolors, recolors);
        }

        return minRecolors;
    }

    public static void main(String[] args) {
        MinimumRecolors solution = new MinimumRecolors();
        String blocks = "WBBWWBB";
        int K = 4;
        int result = solution.minimumRecolors(blocks, K);
        System.out.println("Minimum recolors needed: " + result);
    }
}

Explanation of the Code

  • The outer loop goes through each starting point i for the segments of length K.
  • The inner loop counts the number of ‘W’ blocks in the current segment starting at i.
  • We keep a variable minRecolors to track the lowest number of recolors needed for any segment we check.
  • At the end, the method gives back the minimum recolors we need.

Complexity Analysis

  • Time Complexity: O(N * K), where N is the length of the blocks string. Each segment of length K is checked one by one.
  • Space Complexity: O(1), because we use a fixed amount of space for the variables.

This brute force method is simple but might not be the best for large inputs. A sliding window method could work better for big sizes. For better solutions, look at the section on Optimal Sliding Window Approach for Minimum Recolors in Java.

Brute Force Approach for Minimum Recolors in Python

To find how many recolors we need to get K black blocks next to each other in an array, we can use a brute force method. This way, we look at every possible part of the array that is K long. Then we count how many blocks need to change color in each part.

Implementation

The brute force solution goes through the array. It checks every part that is K long. For each part, it counts the number of white blocks (which we show as 0) and finds the smallest count from all parts.

Here is the Python code for this brute force method:

def minimumRecolors(blocks, k):
    n = len(blocks)
    min_recolors = float('inf')  # Start with a big number

    for i in range(n - k + 1):
        recolors = 0
        # Count the number of white blocks in the current part of length k
        for j in range(i, i + k):
            if blocks[j] == '0':  # 0 means a white block
                recolors += 1
        min_recolors = min(min_recolors, recolors)  # Update minimum recolors

    return min_recolors

# Example usage
blocks = "WBBWWBB"
k = 3
print(minimumRecolors(blocks, k))  # Output: 1

Explanation

  • Input: The function takes a string called blocks. Here ‘B’ means a black block and ‘W’ means a white block. The integer k shows how many black blocks we want together.
  • Loop: The outer loop goes through the blocks array. The inner loop counts white blocks in each part that is k long.
  • Result: The function gives back the smallest number of recolors needed from all parts.

This brute force way is simple but it has a time cost of O(n * k). Here, n is the length of the blocks string. This makes it slower for bigger inputs. If you want faster ways, you can look at the sliding window method in other sections.

For more information on how to work with arrays, check out the article on Array Maximum Subarray.

Brute Force Approach for Minimum Recolors in C++

The brute force method to solve the problem of minimum recolors for K consecutive black blocks checks every possible starting point in the blocks array. We look at each segment of length K and count how many white blocks we need to change to get K black blocks in a row.

Implementation Steps:

  1. We loop through the array from index 0 to n - K.
  2. For each starting index, we count the white blocks in the segment of length K.
  3. We keep track of the smallest number of recolors needed.

C++ Code Example:

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int minimumRecolors(vector<char>& blocks, int k) {
    int n = blocks.size();
    int minRecolors = INT_MAX;

    for (int i = 0; i <= n - k; i++) {
        int recolors = 0;
        for (int j = 0; j < k; j++) {
            if (blocks[i + j] == 'W') {
                recolors++;
            }
        }
        minRecolors = min(minRecolors, recolors);
    }

    return minRecolors;
}

int main() {
    vector<char> blocks = {'W', 'B', 'B', 'W', 'B', 'W', 'W', 'B'};
    int k = 3;
    cout << "Minimum Recolors: " << minimumRecolors(blocks, k) << endl; // Output: 1
    return 0;
}

Explanation of the Code:

  • We have a function called minimumRecolors. It takes a vector of characters (blocks) and an integer k.
  • We start by setting minRecolors to a big number.
  • For each starting index i, we count the white blocks in the next k blocks.
  • If the current count of recolors is less than what we had before, we update minRecolors.
  • In the end, the function gives back the minimum number of recolors we need.

This brute force method has a time complexity of O(n * k). It is not very efficient for big arrays when we compare with faster methods like the sliding window approach. For other array problems, we can find helpful tips in articles like Array Two Sum and Array Maximum Subarray.

Time and Space Complexity Analysis of Solutions

When we solve the problem of finding the minimum recolors to get K consecutive black blocks, we can look at the time and space complexity of different methods.

Sliding Window Approach

  1. Time Complexity:
    • The sliding window method goes through the array one time. It keeps a window of size K. So, the time complexity is O(N). Here, N is the length of the array.
  2. Space Complexity:
    • The space complexity is O(1) because we only use a few extra variables to track the state of the current window.

Brute Force Approach

  1. Time Complexity:
    • The brute force method checks every possible starting point for K consecutive blocks. This means it checks around N-K+1 windows. So, the time complexity is O(N * K). Here, N is the length of the array and K is the size of the blocks we want to check.
  2. Space Complexity:
    • The space complexity is O(1) because it uses a fixed amount of extra space for temporary variables.

Summary of Complexities

Approach Time Complexity Space Complexity
Sliding Window O(N) O(1)
Brute Force O(N * K) O(1)

We prefer the sliding window approach for its speed, especially with larger arrays. The brute force method can be easier for small inputs.

For more related problems, we can check these articles: Array Two Sum, Array Maximum Subarray, and Array Best Time to Buy and Sell Stock.

Frequently Asked Questions

1. What is the Minimum Recolors to Get K Consecutive Black Blocks problem?

The Minimum Recolors to Get K Consecutive Black Blocks is a problem. You get a string that shows blocks of colors. ‘B’ means black and ‘W’ means white. The goal is to find out how many white blocks we need to repaint. We need at least K black blocks in a row. We can solve this problem well using sliding window methods.

2. How can the Sliding Window Technique optimize the solution for Minimum Recolors?

The Sliding Window Technique helps us make the solution better for the Minimum Recolors to Get K Consecutive Black Blocks problem. It does this by keeping a window that is size K. As we move the window across the string, we can count the white blocks fast. We do not need to check the whole window again. This way, we make the time it takes much less to O(N).

3. What are the time and space complexities for the Sliding Window solution?

The time complexity for the Sliding Window solution is O(N). Here, N is the number of blocks. The space complexity is O(1). This is because we only need a few variables to keep track of the white blocks and the window size.

4. Can a Brute Force approach solve the Minimum Recolors problem?

Yes, a Brute Force approach can solve the Minimum Recolors to Get K Consecutive Black Blocks problem. It checks every possible substring that is K long. Then it counts the white blocks in each substring. But this way is not very good. It has a time complexity of O(N*K), which is slower than the sliding window method.

5. Are there any variations of the Minimum Recolors problem in coding interviews?

Yes, we can find variations of the Minimum Recolors to Get K Consecutive Black Blocks problem in coding interviews. These variations may have different rules about colors. They may also ask for different lengths for the sequence or other rules like maximizing black blocks while minimizing recolors. Knowing about the sliding window and brute force can help a lot in solving these variations well.

For more problems and solutions that are related, check out Array Two Sum and Array Maximum Subarray.