Problem Description

Given a binary array A (consisting only of 0s and 1s), you are allowed to perform the following operation any number of times: choose a starting index i and flip the next K consecutive bits (i.e., change 0 to 1 and 1 to 0 for those K bits). Your goal is to make the entire array contain only 1s using the minimum number of such flips. If it is impossible, return -1.

• You may not flip bits outside the array (i.e., i + K must not exceed the array length).
• Each flip must be exactly K consecutive bits.
• Return the minimum number of flips needed, or -1 if it's impossible.

Thought Process

At first glance, a brute-force approach might seem viable: for every 0 in the array, flip the next K bits and count the number of flips. However, this is inefficient and can lead to repeated work.

The key insight is to only flip when you encounter a 0, since flipping a 1 is unnecessary. But flipping changes the state of the next K elements, so we need to keep track of the cumulative effect of flips as we move through the array.

We need a way to efficiently know, at every position, whether the current bit is flipped an odd or even number of times, so we can determine if it is effectively a 0 or 1 at that point.

Solution Approach

We use a greedy sliding window approach with a queue or an auxiliary array to track flips:

1. Track flip effect: Use a variable (e.g., flip) to keep track of the current parity (even or odd) of flips affecting the current position.
2. Iterate through the array: For each position i:
   • If the current bit (with flip parity applied) is 0, we must flip starting here.
   • Increment the flip count.
   • Mark that a flip starts at i (e.g., with an auxiliary array or queue).
   • When moving past position i-K+1, remove the effect of the flip that started K steps ago.
3. Edge case: If a flip would go beyond the array bounds, return -1.

This approach ensures each bit is only considered once, and flips are efficiently tracked.

Example Walkthrough

Let's walk through an example with A = [0,1,0] and K = 1:

1. At index 0: The bit is 0, so we flip starting at 0. The array becomes [1,1,0]. Flip count = 1.
2. At index 1: The bit is 1, no flip needed.
3. At index 2: The bit is 0, so we flip starting at 2. The array becomes [1,1,1]. Flip count = 2.
4. All bits are now 1. Return 2.

For a harder example, A = [1,1,0] and K = 2:

1. At index 0: The bit is 1, no flip needed.
2. At index 1: The bit is 1, no flip needed.
3. At index 2: The bit is 0, but flipping starting at 2 would require flipping bits 2 and 3, but the array ends at index 2. So, it's impossible. Return -1.

Time and Space Complexity

• Brute-force approach: For each 0, flip the next K bits, potentially leading to O(N*K) time complexity and repeated work.
• Optimized approach: The sliding window/greedy method only iterates through the array once, with each flip tracked efficiently. Thus, the time complexity is O(N), and space complexity is O(N) (for the auxiliary array).

Summary

The minimum number of K consecutive bit flips problem is best solved using a greedy, sliding window approach that tracks the cumulative effect of flips. By only flipping when necessary and efficiently tracking the effect of each flip, we achieve an optimal O(N) solution. This method is both elegant and efficient, avoiding the pitfalls of brute-force and making use of clever state management.

Code Implementation

class Solution:
    def minKBitFlips(self, A, K):
        n = len(A)
        flip = 0
        res = 0
        isFlipped = [0] * n
        for i in range(n):
            if i >= K:
                flip ^= isFlipped[i-K]
            if A[i] ^ flip == 0:
                if i + K > n:
                    return -1
                isFlipped[i] = 1
                flip ^= 1
                res += 1
        return res
      

class Solution {
public:
    int minKBitFlips(vector<int>& A, int K) {
        int n = A.size(), flip = 0, res = 0;
        vector<int> isFlipped(n, 0);
        for (int i = 0; i < n; ++i) {
            if (i >= K) flip ^= isFlipped[i-K];
            if ((A[i] ^ flip) == 0) {
                if (i + K > n) return -1;
                isFlipped[i] = 1;
                flip ^= 1;
                ++res;
            }
        }
        return res;
    }
};
      

class Solution {
    public int minKBitFlips(int[] A, int K) {
        int n = A.length, flip = 0, res = 0;
        int[] isFlipped = new int[n];
        for (int i = 0; i < n; ++i) {
            if (i >= K) flip ^= isFlipped[i - K];
            if ((A[i] ^ flip) == 0) {
                if (i + K > n) return -1;
                isFlipped[i] = 1;
                flip ^= 1;
                res++;
            }
        }
        return res;
    }
}
      

var minKBitFlips = function(A, K) {
    let n = A.length, flip = 0, res = 0;
    let isFlipped = new Array(n).fill(0);
    for (let i = 0; i < n; ++i) {
        if (i >= K) flip ^= isFlipped[i - K];
        if ((A[i] ^ flip) === 0) {
            if (i + K > n) return -1;
            isFlipped[i] = 1;
            flip ^= 1;
            res++;
        }
    }
    return res;
};