Elegant DP Solution with Clear Mathematical & Constructive Proof
Intuition\nThe problem requires splitting an array into subarrays to maximize the total cost, where the cost of a subarray is defined in an alternating addition
# Intuition\nThe problem requires splitting an array into subarrays to maximize the total cost, where the cost of a subarray is defined in an alternating addition-subtraction manner. This can be approached using dynamic programming by considering only subarrays of length 1 or 2, simplifying the calculation of costs.\n\nthat was my notepad notes, it might be helpful for someone.\n```\ncost(l, r) = a[l] - a[l + 1] .... a[r] * -1 ^ ((r % l) % 2)\n\ncost(l, r) = (a[l] - a[l + 1]) + (a[l + 2] - a[l + 3]) .....\n\ncost(l, r) = cost(l, l + 1) + cost(l + 2, l + 3) + .... \n\nso to construct cost(l, r), we can only assume \nthat we will only take subarrays of length 1 or 2.\n\nstate : dp[i] represents maximum sum to make from prefix[0...i]\n\ndp[i] = max(dp[i - 1] + a[i], dp[i - 2] + cost(i - 1, i))\n\n\n```\n\n# Approach\n1. **Cost Calculation**: The cost of a subarray `nums[l..r]` is given by:\n   $ \\text{cost}(l, r) = nums[l] - nums[l + 1] + nums[l + 2] - nums[l + 3] + \\ldots + ((-1)^{r-l} \\cdot nums[r]) $\n   Simplifying this, the cost of a subarray can be broken down into:\n   $ \\text{cost}(l, r) = \\text{cost}(l, l + 1) + \\text{cost}(l + 2, l + 3) + \\ldots $\n\n2. **Dynamic Programming State**: Define `dp[i]` as the maximum sum of costs for subarrays covering the prefix `nums[0..i]`.\n\n3. **Recurrence Relation**:\n   - `dp[i] = dp[i - 1] + nums[i]` if the element `nums[i]` starts a new subarray.\n   - `dp[i] = max(dp[i], dp[i - 2] + (nums[i - 1] - nums[i]))` if the subarray ending at `i` has length 2.\n   This recurrence is derived from the observation that to construct the cost, we can only assume subarrays of length 1 or 2.\n\n4. **Initialization**: Insert a dummy element at the beginning of `nums` for easier handling of the base cases. Initialize the `dp` array with size `n + 1`.\n\n5. **Iteration**: Loop through the array to fill the `dp` table based on the recurrence relations.\n\n6. **Result**: The value `dp[n]` will contain the maximum total cost of subarrays.\n\n# Complexity\n- **Time complexity**: \\(O(n)\\), as we iterate through the array once.\n- **Space complexity**: \\(O(n)\\), for storing the `dp` array.\n\n# Code\n```cpp\nclass Solution {\npublic:\n    long long maximumTotalCost(vector<int> &nums) {\n        int n = nums.size();\n        vector<long long> dp(n + 1);\n        nums.insert(nums.begin() + 0, 0);\n\n        for (int i = 1; i <= n; i++) {\n            dp[i] = dp[i - 1] + nums[i];\n            if (i > 1) {\n                dp[i] = max(dp[i], dp[i - 2] + (nums[i - 1] - nums[i]));\n            }\n        }\n\n        return dp[n];\n    }\n};\n```\n\nIn this solution, the maximum total cost is calculated by dynamically deciding whether to start a new subarray or to extend the current one based on the costs defined in the problem. This approach effectively maximizes the total cost by considering subarrays of length 1 or 2, as derived from your idea.
['Array', 'Dynamic Programming', 'C++']