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Minimum Number of K Consecutive Bit Flips

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 <= 105
  • 1 <= k <= nums.length

Solution Explanation for Minimum Number of K Consecutive Bit Flips

This problem asks for the minimum number of k-bit flips needed to make all bits in the input array nums equal to 1. A k-bit flip inverts the bits within a subarray of length k.

Two efficient solutions are presented below: a difference array approach and a sliding window approach. Both achieve a linear time complexity.

Solution 1: Difference Array

This approach uses a difference array d to track the cumulative effect of flips. d[i] represents the net number of flips that affect the bit at index i. A positive value means an odd number of flips (resulting in a bit flip), and a zero or negative value implies an even number of flips (no net change).

The algorithm iterates through nums. For each bit:

  1. Calculate the current bit value: The current bit's value is determined by summing the difference array up to that point (s = sum(d[:i+1])). If s is even, the bit retains its original value; if s is odd, the bit is flipped.

  2. Check for necessary flips: If the calculated bit value is 0 (meaning it needs flipping), we perform a k-bit flip:

    • Check for out-of-bounds: If i + k exceeds the array length, it's impossible to achieve the goal; return -1.
    • Increment d[i] and decrement d[i + k]: This adds a flip to the range [i, i + k). The decrement at d[i + k] ensures that the flip's effect ends correctly.
    • Increment ans (the total number of flips) and s (to reflect the current bit flip).

  3. Continue: Repeat steps 1 and 2 for all bits.

Time Complexity: O(n), where n is the length of nums. We iterate through the array once.

Space Complexity: O(n) to store the difference array d.

Solution 2: Sliding Window

This method is more concise. It uses a flipped variable to track whether the current position's bit has been flipped an odd or even number of times. It leverages the fact that consecutive k-bit flips on overlapping segments effectively cancel or accumulate.

  1. Iterate through the array: For each bit x at index i:

    • Adjust flipped based on previously flipped segments: If i >= k and the (i - k)th bit was flipped (indicated by it being -1), then flipped is toggled.

    • Check if the bit needs flipping: If x matches the current flipped state (meaning it needs flipping), check for out-of-bounds and perform a flip:

      • Toggle flipped, increment ans, and set the current bit to -1 (to mark it as flipped).

  2. Return: After processing all bits, return ans.

Time Complexity: O(n). We iterate through the array once.

Space Complexity: O(1). We use only a few constant extra variables.

Both solutions provide a correct and efficient way to solve the problem. The sliding window approach might be slightly more intuitive and space-efficient. The difference array method offers a more generalizable approach for similar problems involving cumulative effects over ranges.