Minimum Number of K Consecutive Bit Flips

Problem Statement

The challenge involves a binary array nums consisting of elements 0 and 1, along with an integer k. The objective is to perform a "k-bit flip" on the array. A k-bit flip involves selecting a contiguous subarray of length k and inversely flipping all the bits within that subarray (i.e., converting every 0 to 1 and every 1 to 0). The goal is to determine the minimum number of k-bit flips required to transform every element in the array to 1. Should it be impossible to achieve this, the function should return -1. This problem tests the application of array transformations within strict boundaries guided by optimized navigation and flipping strategies.

Examples

Example 1

Input:

nums = [0,1,0], k = 1

Output:

2

Explanation:

Flip nums[0], then flip nums[2].

Example 2

Input:

nums = [1,1,0], k = 2

Output:

-1

Explanation:

No matter how we flip subarrays of size 2, we cannot make the array become [1,1,1].

Example 3

Input:

nums = [0,0,0,1,0,1,1,0], k = 3

Output:

3

Explanation:

Flip nums[0],nums[1],nums[2]: nums becomes [1,1,1,1,0,1,1,0]
Flip nums[4],nums[5],nums[6]: nums becomes [1,1,1,1,1,0,0,0]
Flip nums[5],nums[6],nums[7]: nums becomes [1,1,1,1,1,1,1,1]

Constraints

• 1 <= nums.length <= 105
• 1 <= k <= nums.length

Approach and Intuition

The problem relies on applying transformations to reach an all-1 array, which can be understood by breaking down some core observations and steps based on the given examples:

1. Initial Understanding:

   • The problem demands simultaneous flipping in segments of length k, prioritizing flips to convert 0 to 1 because the goal is to eliminate all 0s in the array.

2. Strategical Analysis:

   • From the Examples:
     • Example 1: Simply flipping each 0 individually when k = 1 leads to the correct result.
     • Example 2: It demonstrates that sometimes it's impossible to convert all zeros to ones, especially if the remaining segment of 0s is less than k towards the end.
     • Example 3: Offers a step-by-step illustration of sequential flipping, showcasing the need sometimes to flip more than once in overlapping segments to ensure all elements become 1.

3. Efficiency Consideration:

   • The direct approach to try every possible k-length flip until all zeros are turned would be computationally intensive and inefficient.
   • Using a greedy strategy — choose to flip at each 0 encountered and then mark the necessity to flip back after k elements could lead to an optimal path.
   • Utilizing auxiliary space or markers to track flips-to-revert can help efficiently manage the transformation without redundant calculations.

4. Final Implementation Notes:

   • Efficient use of data structures, like Difference Array or a sliding window technique, can significantly reduce unnecessary recomputation.
   • The operations should be magnified on handling sequences of flips and careful avoidance of overflow beyond array boundaries, especially in condition handling when k > position of last zero in nums.

The task involves not just direct computation but also a smart division and manipulation of the problem space using arrays and algorithmic techniques typically found in optimized array manipulation problems.

Solutions

class Solution {
public:
    int minimumKFlips(vector<int>& data, int k) {
        int flipCount = 0;
        int flipTotal = 0;
    
        for (int idx = 0; idx < data.size(); idx++) {
            if (idx >= k && data[idx - k] == 2) {
                flipCount--;
            }
    
            if ((flipCount % 2) == data[idx]) {
                if (idx + k > data.size()) {
                    return -1;
                }
                data[idx] = 2;
                flipCount++;
                flipTotal++;
            }
        }
    
        return flipTotal;
    }
};