BEATS 100% || LEARN INTUITION BUILDING || DETAILED
Intuition\n Describe your first thoughts on how to solve this problem. \nApproaching problems like counting alternating subarrays involves thinking like a detec
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nApproaching problems like counting alternating subarrays involves thinking like a detective. Imagine scanning the array for "clues" - elements differing by 1 alternately. A sliding window helps us focus on potential subarrays as we move along. We track their length and any relevant details (like consecutive differences here). As we gather evidence (valid subarrays), we accumulate the count. Finally, at the scene\'s end (array\'s end), we adjust for any misleading clues (edge cases) to reveal the final count of these alternating subarray patterns.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe provided code effectively calculates the number of subarrays in an array `nums` where elements alternate by differing exactly by 1. Here\'s a detailed explanation:\n\n**Functionality:**\n\n- It counts the number of subarrays within `nums` that satisfy the condition of having elements differing by 1 alternately.\n\n**Breakdown:**\n\n1. **Initialization:**\n   - `n = nums.length`: Stores the length of the input array.\n   - `ans = 0`: Initializes a variable to accumulate the count of valid subarrays.\n   - `alt = 0`: Tracks the length of the current alternating subarray (initialized to 0).\n   - `consecutives = 0`: Tracks the number of consecutive elements with the same difference (initialized to 0).\n   - `if (n == 1)`: Base case - if there\'s only one element, it\'s considered a valid subarray, so return 1.\n\n2. **Iterating Through the Array:**\n   - `int i = 1`: Starts the loop from the second element (`i = 1`) as the first element is already considered in the base case.\n   - `while (i < n)`: The loop iterates as long as `i` is within the array bounds (`i < n`).\n\n3. **Checking for Alternating Difference:**\n   - `alt = 1`: Resets `alt` to 1 for each new potential subarray starting at `i`.\n   - `while (i < n && nums[i] == Math.abs(nums[i-1] - 1))`: This inner loop continues as long as:\n     - `i` is within the array bounds (`i < n`).\n     - The current element (`nums[i]`) differs by exactly 1 from the previous element (`nums[i-1]`).\n   - Inside the inner loop:\n     - `alt++`: Increments `alt` as the current element satisfies the alternating difference condition, extending the potential subarray.\n     - `consecutives++`: Increments `consecutives` to track the number of consecutive elements with the same difference (increasing or decreasing by 1).\n     - `ans += alt`: Adds the current length of the alternating subarray (`alt`) to the total count (`ans`). This considers all possible subarrays ending at each index `i` within the current potential subarray.\n     - `i++`: Increments `i` to move to the next element for checking the next potential subarray.\n\n4. **Handling the End and Returning the Result:**\n   - After the outer loop completes, the code considers elements that might not be part of a longer alternating subarray due to reaching the end of the array.\n   - `return ans + n - consecutives`:\n     - `ans`: The count of valid subarrays accumulated during the loop.\n     - `n`: The total number of elements in the array.\n     - `consecutives`: The number of consecutive elements with the same difference (these might not be part of a longer valid subarray).\n   - Subtracting `consecutives` from `n` effectively removes any potential overcounting of subarrays due to considering consecutive elements at the end that might not be part of a valid alternating subarray.\n\n**Key Points:**\n\n- The code leverages sliding window logic to efficiently identify alternating subarrays.\n- The `alt` variable keeps track of the current subarray\'s length as long as the alternating difference condition is met.\n- The `consecutives` variable helps in adjusting the final count to avoid overcounting subarrays at the end of the array.\n- Overall, the code provides an accurate and efficient solution for counting alternating subarrays with a difference of 1.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n# GET MY LINKEDIN IN COMMENTS. LETS CONQUER DSA TOGETHER AND PLS UPVOTE\nO(n)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(1)\n# Code\n```\nclass Solution {\n    public long countAlternatingSubarrays(int[] nums) {\n        long n=nums.length;\n        long ans=0;\n        long alt=0;\n        long consecutives=0;\n        if(n==1){\n            return 1;\n        }\n        int i=1;\n        while(i<n){\n            alt=1;\n            while(i<n && nums[i]==Math.abs(nums[i-1]-1)){\n                alt++;\n                consecutives++;\n                ans+=alt;\n                i++;\n            }\n            i++;\n        }\n        return ans+n-consecutives;\n    }\n}\n```
['Array', 'Java']