Minimum Number Of K Consecutive Bit Flips - Leetcode Solution

Problem statement:

In this problem, we are given an array of 0s and 1s, and a number ‘K’. Our task is to find the minimum number of bit flips required to convert the entire array into 1s, where a bit flip is defined as inverting a range of K consecutive bits in the array.

Approach:

The solution to this problem requires some observation. We can see that changing a bit twice is equivalent to not changing it at all. So, we should only flip a bit once. If we flip a bit more than once, then it may be equivalent to not flipping it at all or flipping it once, depending on the number of flips.

One more observation is that if we flip a bit at position i, then all the bits that lie in the i+1 to i+K-1 range will be flipped as well. So, we can keep track of a flip count for every bit in the array, which tells us how many times that bit has been flipped till now.

We can then start from the leftmost bit (i.e. index 0) and check if it needs to be flipped. If it needs to be flipped, then we update its flip count and increment the total flip count. We then update the flip counts of all the bits in the i+1 to i+K-1 range. We continue this process till we reach the end of the array.

If we reach the end of the array and there are still some 0s left, then it is not possible to convert the array into all 1s by flipping K consecutive bits, so we return -1.

Algorithm:

1. Initialize the flip count of all the bits to 0 and total flip count to 0.
2. For i = 0 to n-K, where n is the size of the array: a. If the bit at index i needs to be flipped (i.e. the flip count is odd), then update its flip count and increment the total flip count. b. Update the flip counts of all the bits in the i+1 to i+K-1 range.
3. Check if there are any 0s left in the array.
4. If there are no 0s left, return the total flip count.
5. If there are still some 0s left, return -1.

Code implementation:

def minKBitFlips(A, K): n = len(A) flip_count = [0] * n total_flips = 0 for i in range(n-K+1): if (A[i] + flip_count[i]) % 2 == 0: flip_count[i] += 1 total_flips += 1 for j in range(i+1, i+K): flip_count[j] += 1 for i in range(n-K+1, n): if (A[i] + flip_count[i]) % 2 == 0: return -1 return total_flips

Time Complexity:

The time complexity of this algorithm is O(n * K), where n is the size of the array and K is the length of the range of bits that can be flipped.

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