Solution: Minimum Number of K Consecutive Bit Flips

Let's solve the Minimum Number of K Consecutive Bit Flips problem using the Bitwise Manipulation pattern.

Statement

We are given a binary arrayAn array consisting of 0s and 1s only. nums and an integer k. Our task is to find the minimum number of flipsChanging a 0 to a 1 or a 1 to a 0. needed to make all the bits in the array equal to 11. However, we can only flip k consecutive bits at a time. So, for a binary array [1,1,0,0][1, 1, 0, 0] and k = 22, we can flip the last two bits to get [1,1,1,1][1, 1, 1, 1]. This means we only need a single k flip to turn the entire array into all 11s.

If nums cannot be converted to an array with all 11s with the given k, return −1-1.

Constraints:

  • 1≤1 \leq nums.length ≤103\leq 10^3

  • 1≤1 \leq k ≤\leq nums.length

Solution

The main challenge is to decide when and where to flip the k consecutive bits in the array so that we use the minimum number of total k flips. So, at any point during the traversal, we need to track:

  • Whether we have k consecutive bits with us or not.

  • The current state of the k bits, whether they need to be flipped or not.

  • The total number of k flips that have been made.

While we keep track of the above-listed things, we need to consider a few pointers:

  • As we traverse nums, for each position i, we need to check if it’s possible to flip k bits starting from that position. If i + k exceeds the length of the array n, it means there aren’t enough bits left to flip, and we should return −1-1 to indicate that the task is impossible.

  • If a bit has been flipped an even number of times, it remains in its original state.

    • If the current bit nums[i] is 00, we need to flip it because we want all bits to be 11.

    • If the current bit nums[i] is 11, no flip is needed because it’s already in the desired state.

  • If a bit has been flipped an odd number of times, it is now inverted.

    • If the current bit nums[i] is 00, it has already been flipped to 11, so no more flips are needed.

    • If the current bit nums[i] is 11, it has been flipped to 00, so we need to flip it again to turn it back to 11.

  • If we flip a bit at index i, then bits at index i+1 and i+2 are also affected. So, we must ensure that once i is greater than k, i.e., we have moved to the next k consecutive bits, we undo the affect that was done at the index i-k.

We iterate through the array nums and check at each position if the current bit needs to be flipped. We use some tracking mechanism (in this solution, a deque) to track the flipped bits and how flipping a bit at i has affected the bits at i+1 and i+2. For every i that requires a bit flip, we store 1 in the deque at the same index. If the bit is 00 and has been flipped an even number of times, the algorithm flips the current subarray starting at that position. If the bit is 11 and has been flipped an odd number of times, it flips again to turn it into 11. If at any point, flipping k consecutive bits is not possible because there aren’t enough bits left, it returns −1-1. The total number of flips made is tracked, ensuring the appropriate adjustment for each affected bit.

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