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Alternating Groups I

Easy

Asked by:

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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. 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.

Every 3 contiguous tiles in the circle with alternating colors (the middle tile has a different color from its left and right tiles) is called an alternating group.

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 = [1,1,1]

Output: 0

Explanation:

Example 2:

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

Output: 3

Explanation:

Alternating groups:

Constraints:

3 <= colors.length <= 100
0 <= colors[i] <= 1

Try coding it in LeetCode

Alternating Groups I

Easy

Asked by:

15 views

Topics:

ArraysTwo PointersSliding Windows

There is a circle of red and blue tiles. You are given an array of integers colors. 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.

Every 3 contiguous tiles in the circle with alternating colors (the middle tile has a different color from its left and right tiles) is called an alternating group.

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 = [1,1,1]

Output: 0

Explanation:

Example 2:

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

Output: 3

Explanation:

Alternating groups:

Constraints:

3 <= colors.length <= 100
0 <= colors[i] <= 1

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 binary string `s`?
Can the input string `s` be empty or null?
If there are multiple substrings with the same maximum number of alternating groups, should I return the number of groups from any of them?
Is a single character considered an alternating group (e.g., "0" or "1")?
If there is no substring with alternating groups (e.g., "0000" or "1111"), what should I return?

Brute Force Solution

Approach

The brute force approach to this problem involves checking every single possible way to split the given input into groups. We'll try all group combinations, and see if any meet the alternating condition.

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

Start by assuming the first group contains only the first element.
Then, assume the first group contains the first two elements, then the first three, and so on, until it contains all the elements.
For each of these possibilities for the first group, consider all possible groupings for the remaining elements.
Repeat the above step until all elements have been assigned to a group.
For each complete grouping, check if the group sizes alternate between even and odd numbers. If they do, store that grouping as a potential solution.
After considering all possible groupings, select the best solution according to whatever criteria the question asks for (e.g., shortest sequence, lexicographically smallest, etc.).

Code Implementation

def alternating_groups_brute_force(input_list):
    all_groupings = []

    def generate_groupings(current_index, current_grouping):
        # Base case: all elements are assigned to a group
        if current_index == len(input_list):
            all_groupings.append(current_grouping)
            return

        # Iterate through all possible sizes for the next group
        for group_size in range(1, len(input_list) - current_index + 1):
            new_group = input_list[current_index:current_index + group_size]

            # Recursively generate groupings for the remaining elements
            generate_groupings(
                current_index + group_size,
                current_grouping + [new_group]
            )

    generate_groupings(0, [])

    alternating_groupings = []

    # Check all possible groupings for the alternating condition
    for grouping in all_groupings:
        is_alternating = True
        for i in range(len(grouping) - 1):
            if len(grouping[i]) % 2 == len(grouping[i+1]) % 2:
                is_alternating = False
                break

        if is_alternating:
            alternating_groupings.append(grouping)

    if not alternating_groupings:
        return []

    # Return the first alternating grouping found
    return alternating_groupings[0]

Big(O) Analysis

Time Complexity

O(2^n) – The described brute force approach explores all possible groupings of the input array of size n. Essentially, for each element, we have two choices: either include it in the current group or start a new group. This branching decision for each of the n elements leads to an exponential number of possible groupings, specifically 2^n. Checking each grouping to see if it meets the alternating condition takes O(n) time in the worst case (to iterate through the groups), but the dominant factor is generating the groupings, making the overall time complexity O(2^n).

Space Complexity

O(N) – The brute force approach described generates all possible groupings by recursively exploring different group sizes. This recursion will create a call stack. In the worst-case scenario, where each group contains only one element, the depth of the recursion will be N, where N is the number of elements in the input. Furthermore, we need to store the current grouping being explored, potentially using a list or similar data structure of size N. Therefore, the space complexity is driven by the recursion depth and the size of the current grouping, both proportional to N.

Optimal Solution

Approach

The key idea is to build the longest possible groups of alternating elements as we go. We achieve this by keeping track of how many groups are still available and prioritizing those that allow us to extend the current alternating sequence.

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

Check if it is possible to form an alternating sequence with remaining groups of elements.
Prioritize using larger groups of elements if using them doesn't ruin our chance to complete the sequence.
If there are only a few elements of one group remaining, use them to finish the sequence.
Repeat until no groups of elements are left.

Code Implementation

def alternating_groups(group_lengths):
    alternating_sequence_length = 0
    current_parity = 0 # 0 or 1 to track alternation

    while True:
        found_next = False
        best_group_index = -1
        best_group_length = -1

        for index, group_length in enumerate(group_lengths):
            if group_length > 0:
                # Check the feasibility of adding group to current sequence
                if index % 2 == current_parity:
                    if best_group_index == -1 or group_length > best_group_length:
                        best_group_index = index
                        best_group_length = group_length
                        found_next = True

        if not found_next:
            break

        # Add the selected group length to the sequence
        group_lengths[best_group_index] -= 1
        alternating_sequence_length += 1

        # Alternate the parity for the next group
        current_parity = 1 - current_parity

    return alternating_sequence_length

Big(O) Analysis

Time Complexity

O(n) – The provided strategy iterates through the groups of elements, attempting to build the longest alternating sequence. Each group is visited and processed at most once in the construction of the alternating sequence. Since we are only looping through the elements once, in effect, the time complexity is directly proportional to the number of groups, n. Thus, the overall time complexity is O(n).

Space Complexity

O(1) – The provided plain English explanation does not describe the use of any auxiliary data structures such as lists, hash maps, or recursion. The steps focus on prioritizing and using existing groups of elements, not on creating new ones. Thus, the algorithm appears to operate in place, using only a constant amount of extra space for variables tracking the current state. Therefore, the space complexity is O(1).

Edge Cases

Case	How to Handle
Empty string input	Return 0, as there are no groups in an empty string.
Null string input	Return 0, handling null string gracefully by treating it as an empty string.
String with a single character (e.g., '0' or '1')	Return 1, as a single character forms a single group.
String with all identical characters (e.g., '0000' or '1111')	Return 1, as the entire string represents a single group.
String with two alternating groups (e.g., '0011' or '1100')	Return 2, correctly identifying the two alternating groups.
String with many alternating groups (e.g., '010101')	The algorithm must correctly count the groups, which should scale linearly to prevent Time Limit Exceeded.
String with mixed alternating and non-alternating groups (e.g., '0010110')	The solution must correctly identify the longest substring that alternates, requiring careful substring analysis.
Extremely long input string	Ensure the solution uses efficient substring analysis and doesn't have excessive memory usage to prevent Out of Memory errors.

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 binary string `s`?
Can the input string `s` be empty or null?
If there are multiple substrings with the same maximum number of alternating groups, should I return the number of groups from any of them?
Is a single character considered an alternating group (e.g., "0" or "1")?
If there is no substring with alternating groups (e.g., "0000" or "1111"), what should I return?

Brute Force Solution

Approach

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

Start by assuming the first group contains only the first element.
Then, assume the first group contains the first two elements, then the first three, and so on, until it contains all the elements.
For each of these possibilities for the first group, consider all possible groupings for the remaining elements.
Repeat the above step until all elements have been assigned to a group.
For each complete grouping, check if the group sizes alternate between even and odd numbers. If they do, store that grouping as a potential solution.
After considering all possible groupings, select the best solution according to whatever criteria the question asks for (e.g., shortest sequence, lexicographically smallest, etc.).

Code Implementation

def alternating_groups_brute_force(input_list):
    all_groupings = []

    def generate_groupings(current_index, current_grouping):
        # Base case: all elements are assigned to a group
        if current_index == len(input_list):
            all_groupings.append(current_grouping)
            return

        # Iterate through all possible sizes for the next group
        for group_size in range(1, len(input_list) - current_index + 1):
            new_group = input_list[current_index:current_index + group_size]

            # Recursively generate groupings for the remaining elements
            generate_groupings(
                current_index + group_size,
                current_grouping + [new_group]
            )

    generate_groupings(0, [])

    alternating_groupings = []

    # Check all possible groupings for the alternating condition
    for grouping in all_groupings:
        is_alternating = True
        for i in range(len(grouping) - 1):
            if len(grouping[i]) % 2 == len(grouping[i+1]) % 2:
                is_alternating = False
                break

        if is_alternating:
            alternating_groupings.append(grouping)

    if not alternating_groupings:
        return []

    # Return the first alternating grouping found
    return alternating_groupings[0]

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Check if it is possible to form an alternating sequence with remaining groups of elements.
Prioritize using larger groups of elements if using them doesn't ruin our chance to complete the sequence.
If there are only a few elements of one group remaining, use them to finish the sequence.
Repeat until no groups of elements are left.

Code Implementation

def alternating_groups(group_lengths):
    alternating_sequence_length = 0
    current_parity = 0 # 0 or 1 to track alternation

    while True:
        found_next = False
        best_group_index = -1
        best_group_length = -1

        for index, group_length in enumerate(group_lengths):
            if group_length > 0:
                # Check the feasibility of adding group to current sequence
                if index % 2 == current_parity:
                    if best_group_index == -1 or group_length > best_group_length:
                        best_group_index = index
                        best_group_length = group_length
                        found_next = True

        if not found_next:
            break

        # Add the selected group length to the sequence
        group_lengths[best_group_index] -= 1
        alternating_sequence_length += 1

        # Alternate the parity for the next group
        current_parity = 1 - current_parity

    return alternating_sequence_length

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty string input	Return 0, as there are no groups in an empty string.
Null string input	Return 0, handling null string gracefully by treating it as an empty string.
String with a single character (e.g., '0' or '1')	Return 1, as a single character forms a single group.
String with all identical characters (e.g., '0000' or '1111')	Return 1, as the entire string represents a single group.
String with two alternating groups (e.g., '0011' or '1100')	Return 2, correctly identifying the two alternating groups.
String with many alternating groups (e.g., '010101')	The algorithm must correctly count the groups, which should scale linearly to prevent Time Limit Exceeded.
String with mixed alternating and non-alternating groups (e.g., '0010110')	The solution must correctly identify the longest substring that alternates, requiring careful substring analysis.
Extremely long input string	Ensure the solution uses efficient substring analysis and doesn't have excessive memory usage to prevent Out of Memory errors.