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Alternating Groups I

Number: 3463

Difficulty: Easy

Paid? No

Companies: Samsara

Problem Description

Given a circular array "colors" of tiles—where 0 represents a red tile and 1 represents a blue tile—every group of 3 contiguous tiles is considered an "alternating group" if the middle tile's color is different from both its left and right neighbors. Since the tiles are arranged in a circle, the first and last tiles are neighbors. The task is to count the number of such alternating groups.

Key Insights

  • The circular nature means that the neighbor of the first tile includes the last tile and vice versa.
  • For each tile (considered the middle of a triple), check if it is not equal to its left neighbor and not equal to its right neighbor.
  • A simple one-pass iteration over the array (using modulus arithmetic) is sufficient to evaluate every group of three.
  • Since only a constant number (3) of elements are compared for each index, the operation is efficient.

Space and Time Complexity

Time Complexity: O(n), where n is the number of tiles. Space Complexity: O(1), as we use only a fixed number of extra variables.

Solution

The idea is to iterate through the array considering each tile as the center of a group of three by using the circular indexing technique. For each index i, determine its left neighbor at (i - 1 + n) % n and right neighbor at (i + 1) % n. Check if the color at the center is different from both of these neighbors. If so, increase the count of alternating groups. This approach only requires a single pass over the array, ensuring an optimal solution with O(n) time complexity.

Code Solutions

# Python solution with comments

def alternatingGroups(colors):
    # Initialize the count of alternating groups
    count = 0
    n = len(colors)
    
    # Traverse each tile of the circular array
    for i in range(n):
        # Compute the indices of the left and right neighbors using modulus arithmetic
        left = (i - 1 + n) % n
        right = (i + 1) % n
        
        # If the middle tile differs from both the left and right neighbors, it's an alternating group
        if colors[i] != colors[left] and colors[i] != colors[right]:
            count += 1
    return count

# Example usage
example_colors = [0, 1, 0, 0, 1]
print(alternatingGroups(example_colors))  # Expected output: 3
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