Alternating Groups I — Easy Problem & Solution

Problem statement

Tiles are arranged in a circle, coloured red (0) or blue (1). An alternating group is three contiguous tiles — wrapping around the circle — whose middle tile differs in colour from both of its neighbours.

Return the number of alternating groups.

Example 1

Input: colors = [1,1,1]
Output: 0
Explanation: Every tile matches its neighbours.

Example 2

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

Example 3

Input: colors = [0,1,0,1]
Output: 4
Explanation: A fully alternating circle: every tile is the middle of a group.

Constraints

• 3 <= colors.length <= 100
• colors[i] is either 0 or 1.

How to solve Alternating Groups I

Every tile is the middle of exactly one group, so scan the tiles and count those that differ from both neighbours, indexing modulo n to close the circle.

Approach

1. For each index i, read colors[(i - 1 + n) % n] and colors[(i + 1) % n].
2. Count i when its colour differs from both.

Why it works

Adding n before the modulus is what keeps the left index correct at i == 0, since a plain -1 % n is negative in most languages. Counting by middle tile rather than by window start makes each group counted exactly once.

Complexity

• Time — O(n)
• Space — O(1)

Pitfalls

• A plain (i - 1) % n goes negative at the start of the array.
• With only two colours, "differs from both neighbours" also means the neighbours match each other.
• The circle means there are n groups to consider, not n - 2.

Reference solution

Python

from typing import List

def numberOfAlternatingGroups(colors: List[int]) -> int:
    n = len(colors)
    count = 0
    for i in range(n):
        left = colors[(i - 1) % n]
        right = colors[(i + 1) % n]
        if colors[i] != left and colors[i] != right:
            count += 1
    return count

JavaScript

var numberOfAlternatingGroups = function(colors) {
    var n = colors.length, count = 0;
    for (var i = 0; i < n; i++) {
        var left = colors[(i - 1 + n) % n];
        var right = colors[(i + 1) % n];
        if (colors[i] !== left && colors[i] !== right) count++;
    }
    return count;
};

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