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.lengthThe key idea for #995 Minimum Number of K Consecutive Bit Flips is to simulate flipping bits while minimizing redundant operations. A naive approach would flip every k segment directly, but that leads to repeated work and an inefficient O(n*k) solution.
A better strategy uses a greedy approach with a sliding window. Traverse the array from left to right and decide whether a flip must start at the current index. Instead of physically flipping each bit, track the number of active flips affecting the current position. This can be done using a queue or a prefix/difference technique that records when a flip window starts and ends.
If the effective bit (after considering previous flips) is 0, you must start a new flip of length k. If there are fewer than k elements remaining, the transformation is impossible. This approach ensures each index is processed once, giving an efficient O(n) time complexity with minimal extra space.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Greedy + Sliding Window (Queue) | O(n) | O(k) |
| Greedy + Prefix/Difference Tracking | O(n) | O(1) |
NeetCodeIO
This approach uses a greedy strategy with the help of a queue to track the index of flips. The idea is to traverse through the array and only flip when necessary, using a queue to efficiently track start points of flips that affect the current position. When flipping, add the index to a queue. Dequeue any index that is outside the current consideration (i.e., more than 'k' steps away).
The time complexity is O(n), where n is the length of nums, because we make a single pass through the array, and the space complexity is O(n) due to the auxiliary array used to track flips.
1#include <vector>
2#include <queue>
3#include <iostream>
4using namespace std;
5
6int minKBitFlips(vector<int>& nums, int k) {
7 int flips = 0, flip_count = 0;
8 vector<int> flip(nums.size(), 0);
9 for (int i = 0; i < nums.size(); i++) {
10 if (i >= k) flip_count ^= flip[i - k];
11 if (nums[i] == flip_count % 2) {
12 if (i + k > nums.size()) return -1;
13 flips++;
14 flip_count ^= 1;
flip[i] = 1;
}
}
return flips;
}
int main() {
vector<int> nums = {0, 0, 0, 1, 0, 1, 1, 0};
int k = 3;
cout << minKBitFlips(nums, k) << endl; // Output: 3
return 0;
}In C++, the strategy is similar to C, utilizing a vector to track flips. The iterator checks the effect of previous flips on the current bit using a XOR mechanism to toggle states. A vector is used akin to a flip queue to ensure each flip affects only the appropriate length `k` segment.
This approach utilizes a sliding window technique with a flipping effect flag, which acts as a state tracker for ongoing flip operations. Using an in-place element flipping marker in the array enables this technique to minimize memory usage and avoid using additional data structures to track state.
Time complexity is O(n) because only single iteration and local modification are performed, while the space complexity is O(1), assisting state via in-place array manipulation.
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Yes, this problem or its variants appear in technical interviews at large tech companies. It tests understanding of greedy algorithms, sliding window techniques, and efficient state tracking using queues or prefix logic.
A queue is commonly used to track when a flip window starts and expires. Alternatively, a prefix or difference array technique can simulate flip effects with constant extra space. Both approaches help efficiently manage active flips during traversal.
The optimal approach uses a greedy strategy combined with a sliding window. Instead of flipping bits directly, it tracks the effect of previous flips using a queue or difference array. This allows each index to be processed once, achieving O(n) time complexity.
A greedy decision ensures that whenever a bit becomes 0 after considering previous flips, a flip must start at that index. Delaying the flip would make it impossible to fix that position later because future windows cannot cover it.
This Java solution employs a dynamic-dependent variable `flip` that toggles based on in-place flips denoted by `A[i] += 2`, simulating a local flip state tracker, reducing memory overhead.