How to Solve Minimum Number of K Consecutive Bit Flips

This problem requires calculating the least number of k-length consecutive bit flips to turn every 0 in the array into 1. The key is maintaining a running state of flips using a sliding window approach to avoid reprocessing elements unnecessarily. If some zeros cannot be flipped due to array bounds, the solution should correctly return -1.

Problem Statement

Given a binary array nums and an integer k, you can flip any subarray of length k, converting all 0s to 1s and all 1s to 0s simultaneously. Determine the fewest flips needed to make the entire array consist only of 1s.

Return the minimum number of k-bit flips required. If it is impossible to achieve all 1s using any sequence of k-length flips, return -1.

Examples

Example 1

Input: nums = [0,1,0], k = 1

Output: 2

Flip nums[0], then flip nums[2].

Example 2

Input: nums = [1,1,0], k = 2

Output: -1

No matter how we flip subarrays of size 2, we cannot make the array become [1,1,1].

Example 3

Input: nums = [0,0,0,1,0,1,1,0], k = 3

Output: 3

Flip nums[0],nums[1],nums[2]: nums becomes [1,1,1,1,0,1,1,0] Flip nums[4],nums[5],nums[6]: nums becomes [1,1,1,1,1,0,0,0] Flip nums[5],nums[6],nums[7]: nums becomes [1,1,1,1,1,1,1,1]

Constraints

• 1 <= nums.length <= 105
• 1 <= k <= nums.length

Solution Approach

Sliding Window with Running Flip Count

Iterate through the array and maintain a count of flips affecting the current index. Flip at the current index only if the effective value (after previous flips) is 0. Track the flip ending positions using a queue to know when flips no longer affect future elements.

Use Queue to Track Active Flips

Push the index of each flip into a queue and remove indices whose flip effect has passed. This ensures constant-time lookup for the current flip state, allowing the algorithm to run in O(n) without modifying the array repeatedly.

Early Termination for Impossible Cases

If a flip is required at an index where there aren’t enough remaining elements to form a length k subarray, immediately return -1. This handles edge cases efficiently without iterating unnecessarily.

Complexity Analysis

The algorithm processes each element once and maintains a queue of active flips, resulting in O(n) time complexity. The space is O(1) excluding the queue for indices, since flip tracking uses constant additional space proportional to k.

What Interviewers Usually Probe

• Check if the candidate correctly maintains flip parity without physically flipping the array.
• Watch for proper handling of edge indices where a k-length flip may exceed array bounds.
• Observe if the candidate optimizes from a brute-force O(n*k) approach to the efficient O(n) sliding window solution.

Common Pitfalls or Variants

Common pitfalls

• Forgetting to consider flips already applied when deciding to flip the current element.
• Attempting to flip subarrays without tracking previous flips, leading to incorrect results or O(n*k) complexity.
• Not returning -1 when a flip is required but insufficient elements remain for a k-length subarray.

Follow-up variants

• Allow flips of variable length subarrays, requiring dynamic tracking of active flips.
• Count minimum flips to make all elements 0 instead of 1, which mirrors the pattern but inverts the condition.
• Apply the same k-length flip pattern on circular arrays, introducing wrap-around tracking for flips.

How GhostInterview Helps

• Automatically tracks running flip state, helping you focus on identifying where flips are necessary without manual simulation.
• Highlights edge cases where a k-length flip cannot be applied, reducing debugging time for impossible scenarios.
• Provides step-by-step verification of flip application sequences, ensuring you understand the sliding window with running state updates pattern.

Topic Pages

• Array
• Bit Manipulation
• Queue
• Sliding Window
• Prefix Sum

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FAQ

What is the main strategy for Minimum Number of K Consecutive Bit Flips?

The key strategy is a sliding window with running state updates, using a queue to track active flips and avoid redundant operations.

Why do we need a queue in this problem?

The queue tracks indices where flips end, allowing efficient determination of the effective bit state at each position without modifying the array directly.

What should I return if flipping is impossible?

Return -1 whenever a flip is required at an index but there aren’t enough remaining elements to form a k-length subarray.

Can this approach handle large arrays efficiently?

Yes, by maintaining running flip parity and using a queue, the algorithm runs in O(n) time and O(1) additional space, suitable for arrays up to 10^5 elements.

How do I verify flips for correctness?

Simulate the effect of each flip on the current bit using the running flip count. Only flip when the effective value is 0, and track end indices with a queue.