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Minimum Operations to Make Binary Array Elements Equal to One I

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Amazon

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ArraysGreedy AlgorithmsBit Manipulation

You are given a binary array nums. You can perform the following operation any number of times: Choose any 3 consecutive elements from the array and flip all of them (0 becomes 1, and 1 becomes 0).

Return the minimum number of operations required to make all elements in nums equal to 1. If it is impossible, return -1.

Example 1:

Input: nums = [0,1,1,1,0,0] Output: 3 Explanation: We can do the following operations:

Choose indices 0, 1, and 2. nums becomes [1,0,0,1,0,0].
Choose indices 1, 2, and 3. nums becomes [1,1,1,0,0,0].
Choose indices 3, 4, and 5. nums becomes [1,1,1,1,1,1].

Example 2:

Input: nums = [0,1,1,1] Output: -1 Explanation: It is impossible to make all elements equal to 1.

Constraints:

3 <= nums.length <= 10^5
0 <= nums[i] <= 1

Minimum Operations to Make Binary Array Elements Equal to One I

Medium

Amazon

2 views

Topics:

ArraysGreedy AlgorithmsBit Manipulation

You are given a binary array nums. You can perform the following operation any number of times: Choose any 3 consecutive elements from the array and flip all of them (0 becomes 1, and 1 becomes 0).

Return the minimum number of operations required to make all elements in nums equal to 1. If it is impossible, return -1.

Example 1:

Input: nums = [0,1,1,1,0,0] Output: 3 Explanation: We can do the following operations:

Choose indices 0, 1, and 2. nums becomes [1,0,0,1,0,0].
Choose indices 1, 2, and 3. nums becomes [1,1,1,0,0,0].
Choose indices 3, 4, and 5. nums becomes [1,1,1,1,1,1].

Example 2:

Input: nums = [0,1,1,1] Output: -1 Explanation: It is impossible to make all elements equal to 1.

Constraints:

3 <= nums.length <= 10^5
0 <= nums[i] <= 1

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 is the range of values allowed in the binary array; specifically, can the array contain values other than 0 and 1?
If the input array is empty or null, what should I return?
Is the goal to make all elements equal to one using only adjacent swaps, or can other types of operations be considered?
If it is impossible to make all elements equal to one, what should I return (e.g., -1, Integer.MAX_VALUE, or an exception)?
Can you provide an example of an input array and its corresponding minimum operations to ensure I understand the problem correctly?

Brute Force Solution

Approach

The most straightforward way to solve this is to try every possible combination of operations. We will systematically explore each option and keep track of how many operations are needed for each combination until we find the one that requires the fewest operations.

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

Consider each pair of adjacent positions in the input.
For each pair, explore both options: either perform an operation on the pair to make both positions equal to one, or leave the pair as they are.
Continue this process for every possible combination of operations across all the adjacent pairs.
For each complete combination, count the total number of operations performed.
Compare the total number of operations needed for each combination.
Choose the combination that requires the smallest number of operations as the solution.

Code Implementation

def min_operations_to_equal_one(numbers):
    minimum_operations = float('inf')

    def solve(current_index, current_numbers, operations_count):
        nonlocal minimum_operations

        # If we've processed all pairs, check if all elements are 1
        if current_index >= len(numbers) - 1:
            if all(number == 1 for number in current_numbers):
                minimum_operations = min(minimum_operations, operations_count)
            return

        # Explore the option of NOT performing an operation on the current pair
        solve(current_index + 1, current_numbers[:], operations_count)

        # Explore the option of performing an operation on the current pair
        # This operation makes adjacent elements 1
        temp_numbers = current_numbers[:]
        temp_numbers[current_index] = 1
        temp_numbers[current_index + 1] = 1

        solve(current_index + 1, temp_numbers, operations_count + 1)

    solve(0, numbers[:], 0)

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

Big(O) Analysis

Time Complexity

O(2^n) – The proposed solution explores every possible combination of operations on adjacent pairs. For an array of size n, there are approximately n-1 adjacent pairs. For each pair, we have two choices: either perform an operation or don't. This leads to 2 * 2 * ... * 2 (n-1) times, or 2^(n-1) possible combinations. Therefore, the time complexity is proportional to the number of combinations explored, making it O(2^n) since the -1 is insignificant in Big O notation.

Space Complexity

O(2^N) – The plain English explanation describes exploring every possible combination of operations on adjacent pairs. In the worst case, for an array of size N, there can be N-1 adjacent pairs. Each pair has two options (operate or don't), which leads to 2^(N-1) different combinations being explored. Each of these combinations implicitly involves tracking which operations were performed, potentially with a stack or similar structure. Therefore, the space complexity is O(2^(N-1)), which simplifies to O(2^N).

Optimal Solution

Approach

The key is to focus on sequences of zeros, because ones don't need any changes. We simply need to count the length of the longest consecutive sequence of zeros in the array.

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

Go through the array and keep track of how many zeros you see in a row.
Whenever you encounter a one, reset your zero count back to zero.
As you go through the array, remember the longest streak of zeros you've found so far.
The length of that longest streak of zeros is the minimum number of changes you need to make.

Code Implementation

def minimum_operations(numbers):
    longest_zero_sequence_length = 0
    current_zero_sequence_length = 0

    for number in numbers:
        if number == 0:
            # Count consecutive zeros.
            current_zero_sequence_length += 1

            # Update the longest sequence if needed.
            if current_zero_sequence_length > longest_zero_sequence_length:
                longest_zero_sequence_length = current_zero_sequence_length

        else:
            # Reset count when a one is encountered.
            current_zero_sequence_length = 0

    return longest_zero_sequence_length

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n exactly once to count the longest consecutive sequence of zeros. Within the loop, simple comparisons and assignments are performed, taking constant time. Therefore, the time complexity is directly proportional to the number of elements in the input array, resulting in O(n).

Space Complexity

O(1) – The algorithm uses a constant amount of extra space. It only needs to store two integer variables: one to keep track of the current consecutive zero count and another to store the maximum consecutive zero count encountered so far. These variables occupy a fixed amount of memory, independent of the input array's size N. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as no operations are possible on an empty array.
Array with only one element.	If the single element is 1, return 0; otherwise, return 1 as one operation is needed to change it to 1.
Array with all elements already equal to 1.	Return 0 since no operations are required.
Array with all elements equal to 0.	Return the length of the array, as each element must be flipped to 1.
Array with very large size (performance consideration).	The solution should iterate through the array once, resulting in O(n) time complexity, which is efficient for large arrays.
Array contains a mix of 0s and 1s with 0s clustered together.	The solution should iterate and flip each zero regardless of clustering, ensuring all elements become 1.
Integer overflow if the array is extremely large.	Ensure the chosen data type (e.g., int in Java/C++) can accommodate the array's size; consider long if necessary.
Input array is read-only or immutable.	The solution must create a mutable copy of the array to perform the flipping operations, ensuring the original array remains unchanged.

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 is the range of values allowed in the binary array; specifically, can the array contain values other than 0 and 1?
If the input array is empty or null, what should I return?
Is the goal to make all elements equal to one using only adjacent swaps, or can other types of operations be considered?
If it is impossible to make all elements equal to one, what should I return (e.g., -1, Integer.MAX_VALUE, or an exception)?
Can you provide an example of an input array and its corresponding minimum operations to ensure I understand the problem correctly?

Brute Force Solution

Approach

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

Consider each pair of adjacent positions in the input.
For each pair, explore both options: either perform an operation on the pair to make both positions equal to one, or leave the pair as they are.
Continue this process for every possible combination of operations across all the adjacent pairs.
For each complete combination, count the total number of operations performed.
Compare the total number of operations needed for each combination.
Choose the combination that requires the smallest number of operations as the solution.

Code Implementation

def min_operations_to_equal_one(numbers):
    minimum_operations = float('inf')

    def solve(current_index, current_numbers, operations_count):
        nonlocal minimum_operations

        # If we've processed all pairs, check if all elements are 1
        if current_index >= len(numbers) - 1:
            if all(number == 1 for number in current_numbers):
                minimum_operations = min(minimum_operations, operations_count)
            return

        # Explore the option of NOT performing an operation on the current pair
        solve(current_index + 1, current_numbers[:], operations_count)

        # Explore the option of performing an operation on the current pair
        # This operation makes adjacent elements 1
        temp_numbers = current_numbers[:]
        temp_numbers[current_index] = 1
        temp_numbers[current_index + 1] = 1

        solve(current_index + 1, temp_numbers, operations_count + 1)

    solve(0, numbers[:], 0)

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

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

The key is to focus on sequences of zeros, because ones don't need any changes. We simply need to count the length of the longest consecutive sequence of zeros in the array.

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

Go through the array and keep track of how many zeros you see in a row.
Whenever you encounter a one, reset your zero count back to zero.
As you go through the array, remember the longest streak of zeros you've found so far.
The length of that longest streak of zeros is the minimum number of changes you need to make.

Code Implementation

def minimum_operations(numbers):
    longest_zero_sequence_length = 0
    current_zero_sequence_length = 0

    for number in numbers:
        if number == 0:
            # Count consecutive zeros.
            current_zero_sequence_length += 1

            # Update the longest sequence if needed.
            if current_zero_sequence_length > longest_zero_sequence_length:
                longest_zero_sequence_length = current_zero_sequence_length

        else:
            # Reset count when a one is encountered.
            current_zero_sequence_length = 0

    return longest_zero_sequence_length

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as no operations are possible on an empty array.
Array with only one element.	If the single element is 1, return 0; otherwise, return 1 as one operation is needed to change it to 1.
Array with all elements already equal to 1.	Return 0 since no operations are required.
Array with all elements equal to 0.	Return the length of the array, as each element must be flipped to 1.
Array with very large size (performance consideration).	The solution should iterate through the array once, resulting in O(n) time complexity, which is efficient for large arrays.
Array contains a mix of 0s and 1s with 0s clustered together.	The solution should iterate and flip each zero regardless of clustering, ensuring all elements become 1.
Integer overflow if the array is extremely large.	Ensure the chosen data type (e.g., int in Java/C++) can accommodate the array's size; consider long if necessary.
Input array is read-only or immutable.	The solution must create a mutable copy of the array to perform the flipping operations, ensuring the original array remains unchanged.