#995

Hard

auto_awesomeSliding window with running state updates

LeetCode Problem Workspace

Minimum Number of K Consecutive Bit Flips

Determine the minimum number of k-length consecutive bit flips needed to convert all zeros to ones in a binary array efficiently.

array bit manipulation queue sliding window prefix sum

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

Metric	Value
Time	O(n)
Space	O(1)

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.

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.

terminal

Solution

Solution 1: Difference Array

We notice that the result of reversing several consecutive elements is independent of the order of the reversals. Therefore, we can greedily consider the number of reversals needed at each position.

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class Solution:
    def minKBitFlips(self, nums: List[int], k: int) -> int:
        n = len(nums)
        d = [0] * (n + 1)
        ans = s = 0
        for i, x in enumerate(nums):
            s += d[i]
            if s % 2 == x:
                if i + k > n:
                    return -1
                d[i] += 1
                d[i + k] -= 1
                s += 1
                ans += 1
        return ans

Solution 2: Sliding Window

We can use a variable $\textit{flipped}$ to indicate whether the current position has been flipped. If $\textit{flipped} = 1$, it means the current position has already been flipped; otherwise, it means the current position has not been flipped. For positions that have been flipped, we can set their value to $-1$, allowing us to distinguish which positions have been flipped.

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class Solution:
    def minKBitFlips(self, nums: List[int], k: int) -> int:
        n = len(nums)
        d = [0] * (n + 1)
        ans = s = 0
        for i, x in enumerate(nums):
            s += d[i]
            if s % 2 == x:
                if i + k > n:
                    return -1
                d[i] += 1
                d[i + k] -= 1
                s += 1
                ans += 1
        return ans

Continue Practicing

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