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3206. Alternating Groups I

EasyArraySliding Window
Leetcode Link

Problem Description

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.

Intuition

To solve this problem, imagine the circle of tiles as a linear array repeated twice to account for the circular nature. This helps to simply check for alternating groups involving the first and last elements.

We define a group of size 3 as an alternating group if the middle tile is different from its neighbors. By iterating through the extended array, we maintain a count of consecutive pairs that have different colors. Whenever we encounter a pair of tiles with the same color, we reset the count. Any time this count reaches or exceeds 3 after considering the first n tiles and extending beyond them, it indicates we've found an alternating group.

The key is recognizing the circular structure of the problem, which is elegantly handled by doubling the array to streamline the logic of checking neighbors when wrapping around the end.

Learn more about Sliding Window patterns.

Solution Approach

The solution utilizes a single-pass approach with the help of an unfolded circular array. Here's a detailed walkthrough:

  1. Setting Up Variables:

    • k is set to 3, as we're interested in groups of three tiles.
    • n is the length of the colors array, representing the original circle of tiles.
    • ans is initialized to count the number of detected alternating groups.
    • cnt is used to track the current alternating sequence's length.

  2. Unfolding the Circle:

    • By iterating over 2n (doubling the array length), we effectively simulate the circular nature of the tiles. This allows us to seamlessly check the transitions from end to start, considering continuity without manually implementing wrap-around logic.

  3. Iteration:

    • For each index i from 0 to 2n-1, the current tile is colors[i % n]. This modulo operation ensures we wrap around the array naturally.
    • If i > 0 and the current tile is the same color as the previous tile, we reset cnt to 1 since no alternating sequence can continue.
    • Otherwise, increment cnt because the alternating condition is maintained.

  4. Counting Alternating Groups:

    • For any index i that is greater than or equal to n and where cnt is at least k, an alternating group has been confirmed. Increment ans in such cases.

  5. Final Result:

    • Return ans as the total number of detected alternating groups.

This approach effectively and efficiently identifies alternating groups while respecting the circular structure by simulating a linear traversal.

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Example Walkthrough

Let's walk through an example using the array colors = [0, 1, 0, 1, 0] to illustrate the solution approach.

  1. Understanding the Input:

    • The array colors represents a circle of tiles with the sequence of colors as Red, Blue, Red, Blue, Red.

  2. Setting Up Variables:

    • k = 3, representing the length of the alternating group we're inspecting.
    • n = 5, the length of the colors array.
    • Initialize ans = 0 to track the number of alternating groups.
    • Start with cnt = 1, a counter to track consecutive alternating patterns.

  3. Simulating the Circle:

    • We effectively consider the array as colors = [0, 1, 0, 1, 0, 0, 1, 0, 1, 0] by iterating over 2n elements.

  4. Iterating Through the Extended Array:

    • Index 1: current = colors[1 % 5] = 1 (Blue). Different from colors[0] = 0 (Red), so increment cnt to 2.
    • Index 2: current = colors[2 % 5] = 0 (Red). Different from colors[1] = 1 (Blue), increment cnt to 3.
    • Index 3: current = colors[3 % 5] = 1 (Blue). Different from colors[2] = 0 (Red), increment cnt to 4.
    • Index 4: current = colors[4 % 5] = 0 (Red). Different from colors[3] = 1 (Blue), increment cnt to 5.

  5. Counting Alternating Groups:

    • Since we have cnt >= 3 at i = 3 and i = 4, and the index i is greater than n, increment ans each time.

  6. Final Output:

    • By completion, we have found ans = 3 alternating groups in the circle ([0, 1, 0], [1, 0, 1], and repeated [0, 1, 0] when overlapping).

Thus, this walkthrough demonstrates that the solution efficiently counts alternating groups while respecting the circular nature of the tile arrangement.

Solution Implementation

1from typing import List
2
3class Solution:
4    def numberOfAlternatingGroups(self, colors: List[int]) -> int:
5        k = 3  # Minimum number of consecutive different colors to form an alternating group
6        n = len(colors)  # Length of the list of colors
7        ans = 0  # Counter for the number of alternating groups
8        cnt = 0  # Counter to track the length of the current alternating sequence
9      
10        # Loop through the colors list twice to handle wraparound at the end
11        for i in range(2 * n):
12            # Check if the current color is the same as the previous one
13            if i > 0 and colors[i % n] == colors[(i - 1) % n]:
14                # Reset the counter if colors are the same
15                cnt = 1
16            else:
17                # Increment the counter if colors are different
18                cnt += 1
19          
20            # Check if we have completed one full loop and the current sequence is at least 'k' long
21            if i >= n and cnt >= k:
22                ans += 1
23      
24        return ans
25
1class Solution {
2    public int numberOfAlternatingGroups(int[] colors) {
3        int k = 3; // Minimum length of alternating group
4        int n = colors.length; // Length of colors array
5        int ans = 0; // Number of alternating groups
6        int count = 0; // Count of current alternating segment length
7
8        // Loop through twice the length of the array to handle cyclic behavior
9        for (int i = 0; i < n * 2; ++i) {
10            int currentColor = colors[i % n]; // Current color
11            int previousColor = colors[(i - 1) % n]; // Previous color
12
13            // Check if the current color equals the previous color
14            if (i > 0 && currentColor == previousColor) {
15                count = 1; // Reset count to 1 if the same color repeats
16            } else {
17                ++count; // Increment count for alternating sequence
18            }
19
20            // Check if we are past the first length of the array and the count is sufficient
21            if (i >= n && count >= k) {
22                ans++; // Increment the number of alternating groups
23            }
24        }
25
26        return ans; // Return the number of alternating groups found
27    }
28}
29
1#include <vector>
2
3class Solution {
4public:
5    int numberOfAlternatingGroups(std::vector<int>& colors) {
6        // Set the minimum length of alternating groups
7        int minLength = 3;
8      
9        // Get the size of the input colors vector
10        int totalColors = colors.size();
11      
12        // Initialize the count of alternating groups
13        int alternatingGroupCount = 0;
14      
15        // Initialize the count of consecutive different colors
16        int currentCount = 0;
17      
18        // Loop through the colors array twice to handle circular groups
19        for (int i = 0; i < totalColors * 2; ++i) {
20            // Use modulo to cycle through the colors
21            if (i > 0 && colors[i % totalColors] == colors[(i - 1) % totalColors]) {
22                // Reset the count if the current color is the same as the previous color
23                currentCount = 1;
24            } else {
25                // Increment the count for consecutive different colors
26                ++currentCount;
27            }
28            // Check if a valid alternating group is found when half the cycle is complete
29            if (i >= totalColors && currentCount >= minLength) {
30                // Increment the number of alternating groups
31                ++alternatingGroupCount;
32            }
33        }
34      
35        // Return the total number of alternating groups
36        return alternatingGroupCount;
37    }
38};
39
1function numberOfAlternatingGroups(colors: number[]): number {
2    // Define the minimum length of the alternating sequence required
3    const k = 3;
4    // Get the total number of colors
5    const n = colors.length;
6    // Initialize the answer (number of groups) and a counter for consecutive different colors
7    let ans = 0;
8    let cnt = 0;
9
10    // Iterate over each color twice (simulate a circular array)
11    for (let i = 0; i < n * 2; ++i) {
12        // If the current and previous colors are the same, reset the counter
13        if (i > 0 && colors[i % n] === colors[(i - 1) % n]) {
14            cnt = 1;
15        } else {
16            // Otherwise, increment the counter
17            ++cnt;
18        }
19        // Once we're past the first round and have an alternating sequence of at least 'k' colors, increment the answer
20        if (i >= n && cnt >= k) {
21            ans++;
22        }
23    }
24    return ans;
25}
26

Time and Space Complexity

The time complexity is O(n) because the loop iterates through the array colors twice (due to n << 1), but this still results in a linear relationship with respect to the length of the list. The constants in time complexity, such as multiplying by 2, do not affect the asymptotic complexity.

The space complexity is O(1) because the solution only utilizes a fixed amount of additional space (for variables like k, n, ans, and cnt), regardless of the size of the input list colors.

Learn more about how to find time and space complexity quickly.

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