Minimum Number of K Consecutive Bit Flips

Hard

Amazon

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Topics:

ArraysGreedy AlgorithmsSliding Windows

You are given a binary array nums and an integer k.

A k-bit flip is choosing a subarray of length k from nums and simultaneously changing every 0 in the subarray to 1, and every 1 in the subarray to 0.

Return the minimum number of k-bit flips required so that there is no 0 in the array. If it is not possible, return -1.

A subarray is a contiguous part of an array.

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

Minimum Number of K Consecutive Bit Flips

Hard

Amazon

3 views

Topics:

ArraysGreedy AlgorithmsSliding Windows

You are given a binary array nums and an integer k.

A k-bit flip is choosing a subarray of length k from nums and simultaneously changing every 0 in the subarray to 1, and every 1 in the subarray to 0.

Return the minimum number of k-bit flips required so that there is no 0 in the array. If it is not possible, return -1.

A subarray is a contiguous part of an array.

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

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible values of elements in the array, and are they guaranteed to be 0 or 1?
What are the constraints on the value of K? Can K be larger than the size of the input array?
If it's impossible to flip the array to all zeros, what should the function return?
Is the input array guaranteed to be non-null and non-empty?
Should I modify the input array directly, or am I allowed to create a copy?

Brute Force Solution

Approach

The problem requires us to flip sections of a binary sequence and find the fewest flips needed to make all digits ones. The brute force approach essentially tries every possible combination of flips of a given length to see which one results in all ones with the minimum number of flips.

Here's how the algorithm would work step-by-step:

Go through the sequence from the beginning.
At each position, consider flipping the next k digits and also consider *not* flipping them.
For each choice (flip or don't flip), create a copy of the sequence and perform the flip if you chose to do so.
After making a flip, update the sequence appropriately (changing the bits of the flipped section).
Continue doing this recursively until the end of the sequence is reached.
Keep track of the total number of flips performed for each of these possible paths.
After exploring all possibilities, find the path that resulted in all ones and had the fewest flips.

Code Implementation

def min_k_bit_flips_brute_force(binary_sequence, k_value):
    sequence_length = len(binary_sequence)
    minimum_flips = float('inf')

    def flip_subarray(current_sequence, start_index, k_size):
        for i in range(start_index, start_index + k_size):
            current_sequence[i] = 1 - current_sequence[i]
        return current_sequence

    def solve(current_sequence, current_index, flip_count):
        nonlocal minimum_flips

        # Base case: If we've reached the end of the sequence
        if current_index == sequence_length:
            if all(bit == 1 for bit in current_sequence):
                minimum_flips = min(minimum_flips, flip_count)
            return

        # Recursive step: Consider not flipping the current subarray
        solve(current_sequence[:], current_index + 1, flip_count)

        # Only consider a flip if there is enough space
        if current_index + k_value <= sequence_length:
            
            # Create a copy to avoid modifying the original during recursion
            next_sequence = current_sequence[:]

            # Flip the next k bits in a sequence copy
            flipped_sequence = flip_subarray(next_sequence, current_index, k_value)

            #Explore the next possibility after having performed a flip
            solve(flipped_sequence, current_index + 1, flip_count + 1)

    solve(binary_sequence, 0, 0)

    if minimum_flips == float('inf'):
        return -1
    else:
        return minimum_flips

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm explores every possible combination of flips for each position in the array of size n. At each position, there are two choices: to flip or not to flip the next k digits. This branching creates a binary tree of possibilities. The depth of the tree is n, as we iterate through each element of the array. Therefore, the total number of paths explored grows exponentially with n, leading to a time complexity of O(2^n).

Space Complexity

O(2^N * N) – The brute force approach explores every possible combination of flips and non-flips for each position in the sequence. This leads to a recursion tree with a depth of N (the length of the binary sequence). At each node, a copy of the sequence is created to simulate the flip or no-flip decision. Thus the space for copies of the sequence grows to N at each level of the recursion tree. Because it is a binary tree we have 2^N possible copies each of size N. Therefore the space complexity is O(2^N * N).

Optimal Solution

Approach

The key is to only flip a group of bits when absolutely necessary, and to avoid redundant flips. We can do this by keeping track of previous flips using an auxiliary structure to simulate flipping without actually modifying the original data. This reduces unnecessary work, leading to a more efficient solution.

Here's how the algorithm would work step-by-step:

Go through the bits from left to right one at a time.
Keep track of whether a bit has been flipped an odd or even number of times up to the current position. This tells you if the current bit is effectively flipped.
If a bit is a 0 and needs to be a 1 (or is a 1 and needs to be a 0 because of prior flips), you must flip the next group of bits.
If you flip a group of bits, increment your flip counter. Also, remember to record the fact that you have flipped this group.
If you reach a point where you need to flip a group of bits, but there aren't enough bits left, it's impossible to solve the problem.
The total number of flips needed is your answer.

Code Implementation

def min_k_bit_flips(binary_array, k_consecutive):
    number_of_flips = 0
    size_of_array = len(binary_array)
    flipped = [0] * size_of_array

    for i in range(size_of_array):
        if i > 0:
            flipped[i] += flipped[i-1]

        # Check if the current bit needs flipping.
        if (binary_array[i] + flipped[i]) % 2 == 0:
            # Need to flip this bit to a 1

            # If there are not enough bits to flip, return -1.
            if i + k_consecutive > size_of_array:
                return -1

            number_of_flips += 1
            if i < size_of_array - 1:
                # Propagate the flip
                flipped[i] += 1
                if i + k_consecutive < size_of_array:
                  flipped[i + k_consecutive] -= 1

    return number_of_flips

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array A of size n exactly once. During each iteration, a constant amount of work is performed: checking if a flip is needed based on previous flips and potentially initiating a new flip. The simulation of flips is done using an auxiliary data structure (e.g., a queue or array) which, while adding complexity to the implementation, contributes only a constant amount of time for each element considered in A. Therefore, the time complexity is directly proportional to the size of the input array.

Space Complexity

O(N) – The algorithm uses an auxiliary structure to keep track of previous flips. Specifically, the explanation mentions recording the fact that a group of bits has been flipped. This implies storing information for each bit of the input array to denote if it's been flipped, leading to an array or similar data structure of size N, where N is the number of bits in the input array. Therefore, the space complexity is proportional to the input size. This means the space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 if the array is null or empty, as no flips are needed.
K is larger than the size of the array	Return -1 if K is larger than the array size, as a K-sized flip is impossible.
Array contains all 0s and K is greater than 1	Return -1 because if there are no 1s, flipping K bits will not change the fact that all elements will be zero after all possible flips.
Array contains all 1s and K is 1	Return 0 because no flips are required since all bits are already 1.
K is equal to 1 and the array contains 0s	The minimum number of flips is the number of zeros in the array; iterate through the array and count zeros.
Large array size causing potential integer overflow in flip count	Ensure the flip count variable is large enough (e.g., long) to accommodate potentially large numbers of flips.
No solution exists (impossible to make all bits 1)	Return -1 when it's not possible to make all elements 1 after all possible operations.
Array with alternating 0s and 1s and a large K approaching the array length	This requires careful consideration of the final sub-array and may lead to no solution if the tail end is not consistently 1s after flips.