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 $1$. However, we can only flip k consecutive bits at a time. So, for a binary array $[1, 1, 0, 0]$ and k = $2$, we can flip the last two bits to get $[1, 1, 1, 1]$. This means we only need a single k flip to turn the entire array into all $1$s.

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

Constraints:

• $1 \leq$ nums.length $\leq 10^3$

• $1 \leq$ k $\leq$ nums.length

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 $-1$ 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 $0$, we need to flip it because we want all bits to be $1$.

  • If the current bit nums[i] is $1$, 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 $0$, it has already been flipped to $1$, so no more flips are needed.

  • If the current bit nums[i] is $1$, it has been flipped to $0$, so we need to flip it again to turn it back to $1$.

• 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 $0$ and has been flipped an even number of times, the algorithm flips the current subarray starting at that position. If the bit is $1$ and has been flipped an odd number of times, it flips again to turn it into $1$. If at any point, flipping k consecutive bits is not possible because there aren’t enough bits left, it returns $-1$. The total number of flips made is tracked, ensuring the appropriate adjustment for each affected bit.