Alternating Groups II

Medium

Samsara

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Topics:

ArraysTwo PointersSliding Windows

There is a circle of red and blue tiles. You are given an array of integers colors and an integer k. The color of tile i is represented by colors[i]:

colors[i] == 0 means that tile i is red.
colors[i] == 1 means that tile i is blue.

An alternating group is every k contiguous tiles in the circle with alternating colors (each tile in the group except the first and last one has a different color from its left and right tiles).

Return the number of alternating groups.

Note that since colors represents a circle, the first and the last tiles are considered to be next to each other.

Example 1:

Input: colors = [0,1,0,1,0], k = 3

Output: 3

Explanation:

Alternating groups:

Example 2:

Input: colors = [0,1,0,0,1,0,1], k = 6

Output: 2

Explanation:

Alternating groups:

Example 3:

Input: colors = [1,1,0,1], k = 4

Output: 0

Explanation:

Constraints:

3 <= colors.length <= 10⁵
0 <= colors[i] <= 1
3 <= k <= colors.length

Try coding it in LeetCode

Alternating Groups II

Medium

Samsara

2 views

Topics:

ArraysTwo PointersSliding Windows

There is a circle of red and blue tiles. You are given an array of integers colors and an integer k. The color of tile i is represented by colors[i]:

colors[i] == 0 means that tile i is red.
colors[i] == 1 means that tile i is blue.

An alternating group is every k contiguous tiles in the circle with alternating colors (each tile in the group except the first and last one has a different color from its left and right tiles).

Return the number of alternating groups.

Note that since colors represents a circle, the first and the last tiles are considered to be next to each other.

Example 1:

Input: colors = [0,1,0,1,0], k = 3

Output: 3

Explanation:

Alternating groups:

Example 2:

Input: colors = [0,1,0,0,1,0,1], k = 6

Output: 2

Explanation:

Alternating groups:

Example 3:

Input: colors = [1,1,0,1], k = 4

Output: 0

Explanation:

Constraints:

3 <= colors.length <= 10⁵
0 <= colors[i] <= 1
3 <= k <= colors.length

Try coding it in LeetCode

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 maximum length of the input binary string `s`?
Can the input string `s` be empty or null?
If no valid partition exists (i.e., even a single substring violates the adjacent condition), what value should I return?
By 'adjacent substrings', do you mean substrings that are directly next to each other in the final partitioned string, or substrings that were originally adjacent in the initial string `s`?
Can you provide a few example inputs and their expected outputs to illustrate the desired behavior, especially edge cases or scenarios where multiple partitions are possible?

Brute Force Solution

Approach

The brute force method involves examining every single possible arrangement of groups within the data. We explore all starting points and lengths for the groups to see if the alternating pattern condition is met. Since every possibility is checked, this guarantees we will find the longest sequence if it exists.

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

Consider every starting position within the data as the beginning of a potential alternating group sequence.
For each starting position, consider every possible length of a potential first group.
Then, for each first group, consider every possible length of a potential second group that immediately follows it.
Continue this process of adding alternating groups and adjusting their lengths as long as the alternating pattern is maintained and you are still within the bounds of the data.
Keep track of the longest alternating group sequence found so far.
After exploring all possible starting positions and group lengths, report the longest alternating group sequence discovered.

Code Implementation

def find_longest_alternating_groups(data):
    longest_sequence_length = 0
    data_length = len(data)

    for start_index in range(data_length):
        for first_group_length in range(1, data_length - start_index + 1):
            current_sequence_length = first_group_length
            previous_group_value = data[start_index]

            # Initialize the next group start position
            next_group_start = start_index + first_group_length
            
            while next_group_start < data_length:
                for next_group_length in range(1, data_length - next_group_start + 1):
                    next_group_value = data[next_group_start]

                    # Check if the new group alternates
                    if next_group_value != previous_group_value:
                        current_sequence_length += next_group_length
                        previous_group_value = next_group_value
                        next_group_start += next_group_length

                        # Break to restart from the outter loop, length found
                        break
                    else:
                        # If next group does not alternate, sequence is broken
                        next_group_length = data_length
                else:
                    # No valid next group found from this starting point
                    next_group_start = data_length

            # Compare current sequence length
            longest_sequence_length = max(longest_sequence_length, current_sequence_length)

    return longest_sequence_length

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm iterates through every possible starting position in the input array of size n. For each starting position, it explores all possible lengths for the first group. Then, it considers all possible lengths for the subsequent alternating groups. The maximum length of the groups considered will be bounded by n, thus leading to a nested iteration based on possible sub-lengths of groups, leading to cubic time complexity. Therefore the overall time complexity can be represented as O(n^3).

Space Complexity

O(1) – The brute force approach described focuses on iterating through all possible starting positions and group lengths. It appears to only maintain a variable to store the longest alternating group sequence found so far, and does not use any auxiliary data structures that scale with the input size N (where N is the size of the input data). Thus, the space required is constant regardless of the input size. We can assume only a few variables for indexing and length tracking are used, leading to constant space complexity.

Optimal Solution

Approach

The problem asks to find the longest sequence that alternates between two groups. The key idea is to use a simple counting method while iterating through the sequence. We keep track of the length of the current group and update the maximum length whenever we find a valid alternating sequence.

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

Start by assuming the first element belongs to the first group.
Go through each element one by one.
Check if the current element belongs to the same group as the previous element. If it does, the current sequence is broken, so reset the count of current sequence to 1.
If the element belongs to a different group, it extends the alternating sequence, so increment the count.
Keep track of the maximum length found so far during the process.
Return the maximum length after checking all elements.

Code Implementation

def alternating_groups_ii(sequence):
    max_length = 0
    current_length = 0
    if not sequence:
        return 0

    # Assume the first element belongs to group 1
    previous_group = sequence[0] % 2
    current_length = 1
    max_length = 1

    for i in range(1, len(sequence)):
        current_element = sequence[i]
        current_group = current_element % 2

        # Check if the current element belongs to the same group
        if current_group == previous_group:
            current_length = 1

        # Extend the alternating sequence
        else:
            current_length += 1

        previous_group = current_group

        # Update the maximum length
        max_length = max(max_length, current_length)

    return max_length

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input sequence once. For each of the n elements, a constant amount of work is done: a comparison to determine if the group has changed and potentially updating a counter. Therefore, the time complexity is directly proportional to the number of elements in the sequence, resulting in O(n) time complexity.

Space Complexity

O(1) – The algorithm's space complexity is constant because it only uses a few variables to store the current group length and the maximum length encountered so far. These variables require a fixed amount of memory, irrespective of the input sequence's size, denoted as N. The provided plain English description does not mention any auxiliary data structures, temporary lists, or recursive calls. Therefore, the space used does not grow with the input size, resulting in O(1) space complexity.

Edge Cases

Case	How to Handle
Empty string input	Return 0, as there are no substrings to create.
String of length 1	Return 1, as the single character forms a valid substring.
String with all '0's or all '1's	Return 1, as any partition would have adjacent substrings with the same number of 0s and 1s.
String with alternating '0' and '1' (e.g., '010101')	Return the length of the string, as each pair of characters can be a substring.
String with long sequences of '0's and '1's (e.g., '0000011111')	The solution should be able to determine the optimal split points to maximize the number of substrings.
String with '0' and '1' counts that are drastically different (e.g., '0000000001')	The optimal solution might involve making one large substring, or splitting to achieve alternating groups.
Maximum length string (consider memory and time complexity)	Ensure the algorithm's time and space complexity are efficient enough to handle large input strings, possibly O(n) time and O(1) or O(n) space depending on implementation.
String starts or ends with a very long sequence of identical characters	Algorithm should correctly identify the first/last splitting opportunity after the initial long sequence.