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.
We'll cover the following
Statement
We are given a nums
and an integer k
. Our task is to find the minimum number of k
consecutive bits at a time. So, for a binary array k
= k
flip to turn the entire array into all
If nums
cannot be converted to an array with all k
, return
Constraints:
nums.length
k
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 positioni
, we need to check if it’s possible to flipk
bits starting from that position. Ifi + k
exceeds the length of the arrayn
, it means there aren’t enough bits left to flip, and we should returnto 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, we need to flip it because we want all bits to be . If the current bit
nums[i]
is, 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, it has already been flipped to , so no more flips are needed. If the current bit
nums[i]
is, it has been flipped to , so we need to flip it again to turn it back to .
If we flip a bit at index
i
, then bits at indexi+1
andi+2
are also affected. So, we must ensure that oncei
is greater thank
, i.e., we have moved to the nextk
consecutive bits, we undo the affect that was done at the indexi-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 k
consecutive bits is not possible because there aren’t enough bits left, it returns
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