Watch 10 video solutions for Minimum Number of K Consecutive Bit Flips, a hard level problem involving Array, Bit Manipulation, Queue. This walkthrough by Coder Army has 24,266 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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], k = 2 Output: -1 Explanation: 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 Explanation: 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 <= 1051 <= k <= nums.lengthProblem Overview: You are given a binary array nums and an integer k. In one operation you can choose any subarray of length k and flip every bit (0 → 1, 1 → 0). The task is to compute the minimum number of such flips required so the entire array becomes all 1s, or return -1 if it is impossible.
Approach 1: Greedy with Queue (O(n) time, O(k) space)
This method processes the array from left to right and greedily decides whether to start a flip at each index. If the current effective bit is 0, a flip must start here; delaying it would leave that position permanently incorrect. A queue tracks the indices where flips begin so you know when a flip window expires. The parity of active flips determines whether the current bit is effectively flipped. Each step performs constant work: check queue front, update flip parity, and optionally enqueue a new flip. This approach is intuitive and explicitly models the sliding window of active flips using a queue from the queue family of techniques.
Approach 2: Sliding Window In-Place (O(n) time, O(1) extra space)
This optimized greedy technique avoids an explicit queue. Instead, the array itself stores markers indicating where flips start. As you iterate, maintain a variable representing how many flips are currently affecting the position (tracked via parity). When a flip starts, mark the index and toggle the active flip state. When the window moves past k positions, remove the effect of the flip that started there. The logic behaves like a sliding window where each position is influenced by flips started in the previous k indices. The key idea is that only the parity of flips matters, so you never need to actually modify every bit.
The in-place variant is effectively a greedy window combined with prefix-style tracking of flip effects, similar to ideas used in prefix sum difference arrays. It avoids extra data structures while still ensuring each element is processed exactly once.
Recommended for interviews: The greedy sliding-window strategy with O(n) time is the expected solution. Starting with the queue-based version clearly communicates the logic of active flips. Transitioning to the in-place optimization demonstrates deeper understanding of window effects and space optimization. Interviewers typically look for the greedy insight: if the current bit evaluates to 0, you must flip starting there immediately.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Flipping | O(n*k) | O(1) | Conceptual baseline; useful to understand the flipping mechanics but too slow for large inputs. |
| Greedy with Queue | O(n) | O(k) | Clear implementation that explicitly tracks active flip windows using a queue. |
| Sliding Window In-Place | O(n) | O(1) | Optimal approach when memory usage matters; avoids extra structures while maintaining flip parity. |