Minimum Number of K Consecutive Bit Flips

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;
    }
};

The given C++ code defines a function within a class that calculates the minimum number of consecutive K bit flips required to transform a given binary array so that all elements become 1. If not possible, the function returns -1.

Start by initializing two counters: flipCount and flipTotal. The flipCount tracks the number of flips at the current position in the array, while flipTotal counts the total number of flips performed.

Iterate through each element of the array. If the index is beyond the period k, adjust flipCount by decrementing it if a flipped mark (indicated by the value 2) appears k positions earlier.

Check if the current number of flips (flipCount) is even or odd, and compare this to the current element by checking if they're equal. If they're equal, and if it's feasible to flip from the current element to k elements ahead (i.e., it doesn't go out of bounds of the array):

• Mark this element as flipped (set to 2).
• Increment both the flipCount and flipTotal.

Continue until you've processed all elements of the array.

Finally, return the flipTotal if flipping is successful for all elements; otherwise, return -1 if at some position flipping k consecutive elements isn't feasible. This optimizes the process by flipping only when necessary and checking all constraints to ensure all subsequent bits can be properly flipped.

class Solution {
    
    public int minimumFlips(int[] data, int len) {
        int flipCount = 0; 
        int flipTotal = 0;
    
        for (int index = 0; index < data.length; ++index) {
            if (index >= len && data[index - len] == 2) {
                flipCount--;
            }
    
            if ((flipCount % 2) == data[index]) {
                if (index + len > data.length) {
                    return -1;
                }
                data[index] = 2;
                flipCount++;
                flipTotal++;
            }
        }
    
        return flipTotal;
    }
}

The given Java program is designed to solve the problem of flipping consecutive bits in a binary array, specifically to ensure that there are no 0s left when reading the array from left to right. The solution involves determining the minimum number of flips required, where each flip inverts the value of len consecutive elements in the array.

Key Implementation Details:

• The function minimumFlips, takes two parameters: data (an array of integers consisting of 0s and 1s) and len (the length of consecutive bits to be flipped each time).
• The solution utilizes two main counters:
  • flipCount to track the number of flips made in the current sliding window.
  • flipTotal to keep a total count of all flips.
• A for loop iterates through the array:
  • If the loop index is greater than or equal to len and the element at index - len is 2, it means a flip has been completed for these bits, so flipCount is decremented by one.
  • The condition checks if the current bit should be flipped based on the flipCount (i.e., it checks if the number of flips applied so far results in a 1), and if the current bit matches this result, a flip operation is initiated.
    • If initiating a flip results in going past the end of the array (index + len > data.length), it's not possible to flip the required number of bits, and the function returns -1.
    • Otherwise, the current bit is marked with a placeholder value 2, indicating that a flip operation has affected this bit, and flipCount and flipTotal are updated.
• After iterating through the array, flipTotal, representing the total flips necessary to ensure that all positions of the array are 1 after applying the flip operation according to the given rule, is returned.

In summary, this Java solution efficiently manages consecutive bit flips in an array using a sliding window approach, modifying bits as necessary while keeping track of operations' impacts using counters and conditions to handle edge cases. This approach ensures minimal performance overhead by avoiding redundant operations and directly checking the conditions required to determine if a flip should occur.

class Solution:
    def minimumKBitFlips(self, bits: List[int], window_size: int) -> int:
        flip_effect = 0  # Active flips effect tracker
        count_flips = 0  # Total flips counter
    
        for index in range(len(bits)):
            # Reduce flip effect when the window moves beyond the influence of an earlier flip
            if index >= window_size and bits[index - window_size] == 2:
                flip_effect -= 1
    
            # Determine if the current element needs flipping
            if (flip_effect % 2) == bits[index]:
                if index + window_size > len(bits):  # Check bounds
                    return -1
                bits[index] = 2  # Mark the bit as flipped
                flip_effect += 1
                count_flips += 1
    
        return count_flips

This Python solution focuses on solving the problem of determining the minimum number of consecutive K-bit flips required to transform a binary array such that all the elements become 1's. The approach used in the solution follows a greedy algorithm enhanced with an efficient marking strategy using a flip effect tracker which helps in minimizing unnecessary operations.

Here's an outline of how the solution operates:

• Initialize a flip_effect variable to track the current net effect of all previous flips.
• Use a count_flips variable to record the total number of flips made.
• Iterate over each bit in the array:
  • If moving beyond the current window, adjust the flip_effect by subtracting the impact of the bit that's now out of the flip window.
  • Check if the current bit, when considering the flip_effect, remains unchanged. If unchanged and flipping is feasible within the bounds, mark the bit and adjust counters and effects accordingly.
  • If a flip is required but not feasible due to boundary constraints, return -1 indicating the transformation isn't possible under given conditions.

The final count of flips, held by count_flips, gives the minimum number of flip operations needed, or -1 if the transformation isn't achievable. This solution ensures a direct and efficient processing path, avoiding excess operations and maintaining clarity regarding the flip operations' net effects across the array.