3101. Count Alternating Subarrays

Problem Description

You are given a binary array nums. We define a subarray as alternating if no two adjacent elements in the subarray have the same value. The task is to return the number of alternating subarrays in nums.

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Intuition

To solve this problem, we need to count the number of subarrays where no two consecutive elements are the same. An effective approach is to "enumerate" these subarrays while iterating through the array.

Here's the thought process:

Initial Setup: Start by iterating over nums from the second element, as comparing adjacent elements is fundamental to this task.
Counting Mechanism: Use a variable s to keep track of how many valid alternating subarrays end at the current position. Initially, s is set to 1, representing the single-element subarray starting with the first element.
Comparison Logic:
- As you move to each new element, compare it with its predecessor.
- If the current element differs from the preceding element (nums[i] != nums[i-1]), extend the alternating pattern, thus incrementing s by 1.
- If they are the same (nums[i] == nums[i-1]), reset s to 1 because the alternating pattern breaks.
Summation: For each position, add s to the total count of alternating subarrays because this s represents all possible alternating subarrays ending at that position.
Result Compilation: Repeat this process until the entire array is traversed, accumulating the total number of valid alternating subarrays. This ensures that every possibility is counted accurately.

This approach efficiently counts all valid subarrays in a single pass through the array, balancing simplicity and performance.

Learn more about Math patterns.

Solution Approach

The provided solution employs an Enumeration strategy that carefully counts all alternating subarrays within the given binary array nums. Here's a detailed breakdown of the approach:

Initialization:
- Define two variables: ans to store the total count of alternating subarrays and s to track the number of valid alternating subarrays that end at each element. Both are initialized to 1 because the first element itself is an alternating subarray of length one.
Iterate with pairwise:
- Use a function pairwise(nums) which generates pairs of consecutive elements in nums. This is vital for straightforward comparisons between adjacent elements.
Iterate Over Each Pair:
- For each adjacent pair (a, b) in nums, check if a is not equal to b:
  - If a != b, then they form a valid alternating sequence, so increment s by 1.
  - If a == b, reset s to 1 since the alternation breaks with identical adjacent elements.
Accumulate Results:
- After updating s, add its value to ans. This includes counting all valid subarrays ending at the current position into the total result.
Return the Result:
- Once the loop completes, the variable ans holds the total count of alternating subarrays in nums, which is then returned.

The pattern used here is a variation of a sliding window, where the window is dynamically adjusted based on the condition of elements being alternating. The time complexity is O(n) since the algorithm traverses the array only once. The overall process efficiently tallies subarrays using arithmetic and logic operations.

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Example Walkthrough

Let's consider a small example to illustrate the solution approach. Suppose nums = [1, 0, 1, 0, 1].

Here's how the algorithm proceeds:

Initialization:
- Start with ans = 1 and s = 1. This accounts for the first element being an alternating subarray by itself.
Iteration with Pairwise Comparison:
- First Pair (1, 0):
  - Since 1 != 0, they form a valid alternating subarray.
  - Increment s to 2 (s now reflects two alternating subarrays: [1] and [1, 0]).
  - Add s to ans: ans = 3.
- Second Pair (0, 1):
  - 0 != 1, continuing the alternation.
  - Increment s to 3 (s reflects alternating subarrays: [0], [1, 0, 1], and [0, 1]).
  - Add s to ans: ans = 6.
- Third Pair (1, 0):
  - 1 != 0, maintaining the alternation.
  - Increment s to 4 (s includes [1, 0], [0, 1, 0], [1, 0], and [1, 0, 1, 0]).
  - Add s to ans: ans = 10.
- Fourth Pair (0, 1):
  - 0 != 1, continuing the alternation.
  - Increment s to 5 (s now includes [0, 1], [1, 0, 1], [0, 1, 0, 1], [1, 0, 1, 0, 1], and [0, 1, 0]).
  - Add s to ans: ans = 15.
Result:
- The final value of ans is 15, representing the total number of alternating subarrays in nums.

This walkthrough demonstrates how the solution efficiently counts all potential alternating subarrays using a single traversal of the array.

Solution Implementation

1from typing import List
2from itertools import pairwise
3
4class Solution:
5    def countAlternatingSubarrays(self, nums: List[int]) -> int:
6        # Initialize the answer and the current subsequence length
7        ans = 1  # start with the first element as a trivial subsequence
8        current_length = 1
9      
10        # Iterate through consecutive pairs of numbers in 'nums'
11        for first, second in pairwise(nums):
12            # If the current and next elements are different,
13            # increase the length of the current alternating subsequence
14            if first != second:
15                current_length += 1
16            else:
17                # If they are the same, reset the subsequence length to 1
18                current_length = 1
19          
20            # Add the current subsequence length to the total count
21            ans += current_length
22      
23        return ans
24

1class Solution {
2    public long countAlternatingSubarrays(int[] nums) {
3        long ans = 1; // Initialize answer to 1, considering the first element
4        long s = 1; // s keeps track of the length of the current alternating subarray
5
6        // Loop through each element starting from the second one
7        for (int i = 1; i < nums.length; ++i) {
8            // If current element is different from the previous one, increment s
9            if (nums[i] != nums[i - 1]) {
10                s += 1; // Increase length of current alternating subarray
11            } else {
12                s = 1; // Reset s because the alternating sequence is broken
13            }
14            ans += s; // Add the length of the current alternating subarray to answer
15        }
16      
17        return ans; // Return the total count of alternating subarrays
18    }
19}
20

1#include <vector>
2using namespace std;
3
4class Solution {
5public:
6    // Function to count the number of alternating subarrays in a given vector
7    long long countAlternatingSubarrays(vector<int>& nums) {
8        long long totalAlternatingSubarrays = 1; // Start count with the first element
9        long long currentAlternatingLength = 1;  // Length of the current alternating subarray
10
11        // Iterate through the vector nums starting from the second element
12        for (int i = 1; i < nums.size(); ++i) {
13            // Check if the current element is different from the previous element
14            if (nums[i] != nums[i - 1]) {
15                currentAlternatingLength += 1; // Increment length if it's alternating
16            } else {
17                currentAlternatingLength = 1;  // Reset length if no alternation
18            }
19            // Accumulate the total count with the current alternating length
20            totalAlternatingSubarrays += currentAlternatingLength;
21        }
22      
23        return totalAlternatingSubarrays; // Return the total count of alternating subarrays
24    }
25};
26

1/**
2 * This function counts the total number of alternating subarrays in a given array of numbers.
3 * An alternating subarray is one where consecutive elements are different from each other.
4 * 
5 * @param nums - An array of numbers where we will find alternating subarrays.
6 * @returns The total count of alternating subarrays.
7 */
8function countAlternatingSubarrays(nums: number[]): number {
9    // Initialize `ans` to count alternating subarrays, and `s` to track the length of the current alternating streak.
10    let ans = 1;
11    let s = 1;
12
13    // Iterate through the array starting from the second element.
14    for (let i = 1; i < nums.length; ++i) {
15        // If the current element is different from the previous one, increase the streak length `s`.
16        // Otherwise, reset `s` to 1 since the alternating sequence breaks.
17        s = nums[i] !== nums[i - 1] ? s + 1 : 1;
18
19        // Add the current length of the alternating sequence `s` to the total count `ans`.
20        ans += s;
21    }
22
23    // Return the total count of alternating subarrays.
24    return ans;
25}
26

Time and Space Complexity

The time complexity of the code is O(n), where n is the length of the array nums. This is because the pairwise function processes each pair (a, b) of consecutive elements in nums exactly once. The loop iterates through the array nums once, leading to linear time complexity.

The space complexity of the code is O(1), as it uses a fixed amount of additional space regardless of the input size. Only a few integer variables (ans and s) are used, and no additional data structures are utilized.

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