Dynamic Programming: Count Alternating Subarrays Explained
Intuition\nThe problem asks to count the number of alternating subarrays in the given array. An alternating subarray is defined as a contiguous subarray where t
# Intuition\nThe problem asks to count the number of alternating subarrays in the given array. An alternating subarray is defined as a contiguous subarray where the elements alternate between being even and odd.\n\n# Approach\nTo solve this problem, we can use dynamic programming. We\'ll maintain a dynamic programming array `dp`, where `dp[i]` represents the length of the alternating subarray ending at index `i`. We initialize `dp[0]` to 1, as the subarray consisting of only the first element is always alternating.\n\nThen, we iterate through the array starting from index 1. At each index `i`, if the current element is different from the previous element, it means the current element can be added to the alternating subarray ending at index `i-1`, thus increasing its length by 1. Otherwise, if the current element is the same as the previous element, the alternating subarray ends at index `i-1`, and we start a new alternating subarray at index `i`.\n\nFinally, we return the sum of all elements in the `dp` array, which represents the total count of alternating subarrays in the given array.\n\n# Complexity\n- Time complexity: O(n), where n is the length of the input array `nums`.\n- Space complexity: O(n), as we use an additional array of size `n` to store the dynamic programming values.\n\n# Code\n```python\nfrom typing import List\n\nclass Solution:\n    def countAlternatingSubarrays(self, nums: List[int]) -> int:\n        n = len(nums)\n        dp = [0] * n\n        dp[0] = 1\n        for i in range(1, n):\n            if nums[i] != nums[i - 1]:\n                dp[i] = dp[i - 1] + 1\n            else:\n                dp[i] = 1\n        return sum(dp)\n
['Python3']