What is the time complexity of Minimum Recolors to Get K Consecutive Black Blocks?

The optimal solution has a time complexity of O(n*k).

What data structures are used to solve Minimum Recolors to Get K Consecutive Black Blocks?

This problem uses String, Sliding Window. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Minimum Recolors to Get K Consecutive Black Blocks?

Insight 1: A sliding window approach is ideal for efficiently finding the minimum number of recolorings within a window of size `k`. This avoids redundant calculations. Insight 2: The problem can be solved by iterating through all possible starting positions of a `k`-length window and counting the number of white blocks within each window. The minimum count represents the minimum number of recolorings needed.

Which companies ask LeetCode 2379. Minimum Recolors to Get K Consecutive Black Blocks in interviews?

This problem is frequently asked at HP.

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Minimum Recolors to Get K Consecutive Black Blocks - LeetCode 2379 Solution

Minimum Recolors to Get K Consecutive Black Blocks - Complete Solution Guide

Minimum Recolors to Get K Consecutive Black Blocks is LeetCode problem 2379, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem asks to find the minimum number of operations needed to create at least one sequence of `k` consecutive black blocks ('B') in a string of white ('W') and black blocks. Each operation consists of recoloring a white block to black. The input is a string `blocks` representing the colors of the blocks and an integer `k` representing the desired length of the consecutive black block sequence. The output is the minimum number of recolorings needed.

Solution Approach

The Python and Java solutions employ a brute-force approach: they iterate through all possible `k`-length windows in the string. For each window, they count the number of white blocks and update the minimum number of recolorings found so far. The C++ and C solutions use a sliding window optimization. They first count the white blocks in the initial `k`-length window. Then, as they slide the window one position to the right, they subtract the number of white blocks leaving the window and add the number of white blocks entering the window. This efficiently updates the white block count without recomputing it from scratch for each window. This optimizes the time complexity from O(n*k) to O(n).

Step-by-Step Algorithm

Step 1: Initialize `min_recolors` to a large value (infinity in Python/Java, `k` in C++/C). This will store the minimum number of recolorings found.
Step 2: Iterate through all possible starting positions of a `k`-length window (from index 0 to `n-k`).
Step 3: For each window, count the number of white blocks ('W') within that window.
Step 4: Update `min_recolors` with the minimum between the current `min_recolors` and the number of white blocks in the current window.
Step 5: After iterating through all windows, return `min_recolors`.

Key Insights

Insight 1: A sliding window approach is ideal for efficiently finding the minimum number of recolorings within a window of size `k`. This avoids redundant calculations.
Insight 2: The problem can be solved by iterating through all possible starting positions of a `k`-length window and counting the number of white blocks within each window. The minimum count represents the minimum number of recolorings needed.
Insight 3: The provided C++ and C solutions optimize by using a sliding window to reduce redundant computations. Instead of recounting the number of white blocks in every window, they incrementally update the count as the window slides.

Complexity Analysis

Time Complexity: O(n*k)

Space Complexity: O(1)

Topics

This problem involves: String, Sliding Window.

Companies

Asked at: HP.

Minimum Recolors to Get K Consecutive Black Blocks | LeetCode 2379

Easy

String Sliding Window

Time: O(n*k)Space: O(1)

Solution Code

Available in:

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

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

        return minRecolors;
    }
}

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Problem Info

Description

Problem 2379: Minimum Recolors to Get K Consecutive Black Blocks - Detailed explanation and solution approaches.

Topics

String Sliding Window

Companies

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2379

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Initialize `min_recolors` to a large value (infinity in Python/Java, `k` in C++/C). This will store the minimum number of recolorings found.
2
Step 2: Iterate through all possible starting positions of a `k`-length window (from index 0 to `n-k`).
3
Step 3: For each window, count the number of white blocks ('W') within that window.
4
Step 4: Update `min_recolors` with the minimum between the current `min_recolors` and the number of white blocks in the current window.
5
Step 5: After iterating through all windows, return `min_recolors`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*k)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The space complexity is constant because the amount of extra memory used does not depend on the size of the input string. The time complexity for the optimized solutions is linear because it only iterates through the string once. The unoptimized solutions have a time complexity of O(n*k) because they iterate through each k-sized window.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Prefix Sum

Use a prefix sum array to store the cumulative number of white blocks up to each index. Then calculate the number of white blocks within any window in O(1) time.

Time: O(n)Space: O(n) to store the prefix sum array

Trade-offs: This approach uses more space than the sliding window but provides the same time complexity. It might be preferred if there were many queries for different values of k on the same blocks string.

✨

Key Insights

1Insight 1: A sliding window approach is ideal for efficiently finding the minimum number of recolorings within a window of size `k`. This avoids redundant calculations.
2Insight 2: The problem can be solved by iterating through all possible starting positions of a `k`-length window and counting the number of white blocks within each window. The minimum count represents the minimum number of recolorings needed.
3Insight 3: The provided C++ and C solutions optimize by using a sliding window to reduce redundant computations. Instead of recounting the number of white blocks in every window, they incrementally update the count as the window slides.

📚

Key Terms Explained

For Students

Sliding Window

A moving "window" over data to process subsets efficiently.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: If `k` is 1, the algorithm correctly handles this. It will simply count the number of 'W's in the string.
Edge case 2: If the string already contains `k` consecutive black blocks, the algorithm correctly returns 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle edge cases, such as when k is larger than the length of the string or when there are no white blocks. This can lead to incorrect results or runtime errors.

✗Common mistake 2: Incorrectly implementing the sliding window technique. A common error is not correctly updating the count of white blocks as the window moves, leading to an inaccurate minimum.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing resource allocation. Imagine blocks represent machines in a factory and 'B' represents a working machine and 'W' represents a faulty machine. This algorithm could find the minimum number of machines to fix to get a contiguous sequence of working machines (k consecutive black blocks).

🏦Real-world scenario 2: DNA sequencing. Finding the minimum mutations needed to achieve a specific DNA sequence pattern (represented by consecutive black blocks).

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 10, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

What data structures complement sliding window?

Hash maps track character frequencies within the window. Deques enable O(1) access to both window boundaries. Heaps help find min/max within the window. Choose based on what property you need to maintain as the window slides.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Recolors to Get K Consecutive Black Blocks"?

For this easy-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether sliding window techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?