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LeetCode 3101: Count Alternating Subarrays Solution

Master LeetCode problem 3101 (Count Alternating Subarrays), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3101. Count Alternating Subarrays

Medium
ArrayMath

Problem Explanation

Explanation:

To solve this problem, we can iterate through the array and count the number of alternating subarrays at each index. We can use a sliding window approach to keep track of the current alternating subarray, and update the count accordingly. The key idea is to consider the current element and the previous element to determine if they form an alternating subarray.

Algorithmic Idea:

  1. Initialize a variable count to keep track of the total number of alternating subarrays.
  2. Initialize a variable prev to store the value of the previous element.
  3. Initialize a variable currentLength to keep track of the length of the current alternating subarray.
  4. Iterate through the array:
    • If the current element is different from the previous element, increment currentLength by 1.
    • Otherwise, update the count by adding the number of subarrays that can be formed with the current alternating subarray length, and reset currentLength.
    • Update prev to the current element.
  5. After the loop ends, add the remaining subarrays that end at the last index to the count.
  6. Return the total count.

Time Complexity:

The time complexity of this solution is O(n) where n is the length of the input array.

Space Complexity:

The space complexity of this solution is O(1) as we are using a constant amount of extra space.

:

Solution Code

class Solution {
    public int countAlternatingSubarrays(int[] nums) {
        int count = 0;
        int prev = -1;
        int currentLength = 0;
        
        for (int num : nums) {
            if (num != prev) {
                currentLength++;
            } else {
                count += (currentLength * (currentLength + 1)) / 2;
                currentLength = 1;
            }
            prev = num;
        }
        
        count += (currentLength * (currentLength + 1)) / 2;
        
        return count;
    }
}

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