What is the time complexity of Minimum Number of K Consecutive Bit Flips?

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

What data structures are used to solve Minimum Number of K Consecutive Bit Flips?

This problem uses Array, Bit Manipulation, Queue, Sliding Window, Prefix Sum. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Number of K Consecutive Bit Flips?

Insight 1: We need to efficiently track the current state of each element after considering all preceding flips. Instead of actually flipping elements multiple times and recalculating, we can maintain a running count of active flips at each index. Insight 2: A sliding window approach allows for efficient tracking of the net effect of flips. As we move through the array, we need to account for flips that have 'expired' (the subarray is no longer overlapping with the current index).

Which companies ask LeetCode 995. Minimum Number of K Consecutive Bit Flips in interviews?

This problem is frequently asked at thoughtspot.

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Minimum Number of K Consecutive Bit Flips - LeetCode 995 Solution

Minimum Number of K Consecutive Bit Flips - Complete Solution Guide

Minimum Number of K Consecutive Bit Flips is LeetCode problem 995, a Hard 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 binary array nums and an integer k . A k-bit flip is choosing a subarray of length k from nums and simultaneously changing every 0 in the subarray to 1 , and every 1 in the subarray to 0 . Return the minimum number of k-bit flips required so that there is no 0 in the array . If it is not possible, return -1 . A subarray is a contiguous part of an array. Example 1: Input: nums = [0,1,0], k = 1 Output: 2 Explanation: Flip nums[0], then flip nums[2]. Example 2: Input: nums = [1,1,0]

Detailed Explanation

The problem requires finding the minimum number of 'k-bit flips' needed to transform a binary array `nums` such that no element is 0. A k-bit flip involves selecting a subarray of length `k` and flipping the bits (0 to 1, 1 to 0) within that subarray. If it's impossible to achieve the desired state, the function should return -1. The input consists of the binary array `nums` and the integer `k`, representing the length of the subarray to be flipped. The output is either the minimum number of flips required or -1 if it's impossible.

Solution Approach

The solution uses a greedy approach combined with a sliding window technique. We iterate through the array, maintaining a `current_flips` variable which tracks the number of active flips affecting the current index. If an element needs to be flipped (i.e., it is effectively 0 after considering preceding flips), we flip the subarray of length `k` starting at the current index. The 'flip' is implemented by incrementing the `current_flips` count and marking the current element as flipped by adding 2 to its value (this doesn't actually change the underlying array values permanently for later calculations). When the sliding window moves past the start of a previously flipped subarray, `current_flips` is decremented. If at any point we need to perform a flip but there aren't `k` elements remaining, it is impossible to solve the problem, so we return -1.

Step-by-Step Algorithm

Step 1: Initialize `flips` to 0 (total number of flips) and `current_flips` to 0 (number of active flips at the current index).
Step 2: Iterate through the array `nums` using index `i`.
Step 3: If `i` is greater than or equal to `k` and `nums[i - k]` is greater than 1, it means a flip started at index `i - k` has ended. Decrement `current_flips` to account for this.
Step 4: Check if the current element effectively needs a flip. This is determined by `(nums[i] + current_flips) % 2 == 0`. If it is true, the current element is effectively 0 and requires a flip.
Step 5: If a flip is needed, check if there are enough elements remaining in the array to perform the flip. If `i + k > n`, return -1 because it's impossible to flip the subarray.
Step 6: If a flip is possible, increment `flips`, increment `current_flips`, and mark that a flip started at index `i` by adding 2 to `nums[i]`.
Step 7: After iterating through the entire array, return the final value of `flips`.

Key Insights

Insight 1: We need to efficiently track the current state of each element after considering all preceding flips. Instead of actually flipping elements multiple times and recalculating, we can maintain a running count of active flips at each index.
Insight 2: A sliding window approach allows for efficient tracking of the net effect of flips. As we move through the array, we need to account for flips that have 'expired' (the subarray is no longer overlapping with the current index).
Insight 3: Using in-place modification (adding 2 to flipped elements) to mark flips can be a space-efficient way to track the flipped status, avoiding extra data structures for that purpose.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Bit Manipulation, Queue, Sliding Window, Prefix Sum.

Companies

Asked at: thoughtspot.

Minimum Number of K Consecutive Bit Flips | LeetCode 995

Hard

Array Bit Manipulation Queue Sliding Window Prefix Sum

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

Solution Code

Available in:

class Solution {
    public int minKBitFlips(int[] nums, int k) {
        int n = nums.length;
        int flips = 0;
        int current_flips = 0;

        for (int i = 0; i < n; i++) {
            if (i >= k && nums[i - k] > 1) {
                current_flips -= 1;
            }
            
            if ((nums[i] + current_flips) % 2 == 0) {
                if (i + k > n) {
                    return -1;
                }
                
                current_flips += 1;
                flips += 1;
                nums[i] += 2;
            }
        }
        
        return flips;
    }
}

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Problem Info

Description

Problem 995: Minimum Number of K Consecutive Bit Flips - Detailed explanation and solution approaches.

Topics

Array Bit Manipulation Queue Sliding Window Prefix Sum

Companies

thoughtspot

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:995

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

7 steps

1
Step 1: Initialize `flips` to 0 (total number of flips) and `current_flips` to 0 (number of active flips at the current index).
2
Step 2: Iterate through the array `nums` using index `i`.
3
Step 3: If `i` is greater than or equal to `k` and `nums[i - k]` is greater than 1, it means a flip started at index `i - k` has ended. Decrement `current_flips` to account for this.
4
Step 4: Check if the current element effectively needs a flip. This is determined by `(nums[i] + current_flips) % 2 == 0`. If it is true, the current element is effectively 0 and requires a flip.
5
Step 5: If a flip is needed, check if there are enough elements remaining in the array to perform the flip. If `i + k > n`, return -1 because it's impossible to flip the subarray.
6
Step 6: If a flip is possible, increment `flips`, increment `current_flips`, and mark that a flip started at index `i` by adding 2 to `nums[i]`.
7
Step 7: After iterating through the entire array, return the final value of `flips`.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because each element in the input array is visited and processed once. The sliding window operations (incrementing and decrementing `current_flips`) take constant time. The space complexity is constant because the algorithm only uses a few variables (`flips`, `current_flips`, `i`, `n`, `k`) to store intermediate results, and the space used by these variables does not depend on the size of the input.

📖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

Using a Queue to track flips

Instead of `current_flips`, a queue can store the starting indices of the flipped subarrays. When the window moves past the start of a flip, the index is removed from the queue, effectively decrementing the `current_flips` effect.

Time: O(n)Space: O(k) in the worst case (when nearly all subarrays are flipped).

Trade-offs: The queue approach can be more intuitive for some but uses O(k) space instead of O(1). It's preferable if space is not a major concern, and readability is favored.

✨

Key Insights

1Insight 1: We need to efficiently track the current state of each element after considering all preceding flips. Instead of actually flipping elements multiple times and recalculating, we can maintain a running count of active flips at each index.
2Insight 2: A sliding window approach allows for efficient tracking of the net effect of flips. As we move through the array, we need to account for flips that have 'expired' (the subarray is no longer overlapping with the current index).
3Insight 3: Using in-place modification (adding 2 to flipped elements) to mark flips can be a space-efficient way to track the flipped status, avoiding extra data structures for that purpose.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Queue

FIFO structure - First In, First Out. Like a line at a store!

Sliding Window

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

Edge Cases

Edge case 1: When it's impossible to make all elements 1, due to the constraint `k` and array length. For example, `nums = [1,1,0], k = 2`. The solution correctly returns -1 when `i + k > n` during a flip attempt.
Edge case 2: When the array is already all 1s (or all effectively 1s considering prior flips). The `if ((nums[i] + current_flips) % 2 == 0)` condition ensures no unnecessary flips occur.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to decrement `current_flips` when the sliding window moves past the start of a flip. This leads to incorrect counts of active flips and potentially unnecessary flips.

✗Common mistake 2: Not handling the case where `i + k > n` when attempting to perform a flip. This leads to incorrect results or errors when trying to access elements outside the array bounds.

✅

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: Data correction in communication systems. When transmitting binary data, errors can occur, resulting in flipped bits. This algorithm (or similar concepts) can be used to determine the minimum number of 'burst' corrections (contiguous bit flips) needed to restore the original data, given a maximum burst size k.

🏦Real-world scenario 2: Optimizing manufacturing processes. Consider a production line where a certain operation needs to be applied to consecutive segments of items. The algorithm can help minimize the number of times the operation (the 'flip') needs to be applied to achieve a desired overall result.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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: September 4, 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

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

@DevExpert92Jul 13

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

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 Number of K Consecutive Bit Flips"?

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

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