Monotonic Pairs Count
Intuition\nThe task is to find the number of valid pairs (arr1, arr2) such that:\n1. arr1 is monotonically non-decreasing.\n2. arr2 is monotonically non-increas
# Intuition\nThe task is to find the number of valid pairs `(arr1, arr2)` such that:\n1. `arr1` is monotonically non-decreasing.\n2. `arr2` is monotonically non-increasing.\n3. For each index `i`, `arr1[i] + arr2[i]` equals `nums[i]`.\n\nThe challenge here is that `nums` values can be as large as 1000, making a direct approach impractical. We use dynamic programming with prefix sums to handle this efficiently.\n\n# Approach\n1. **Initialization:**\n   - Create a DP array `dp` where `dp[j]` represents the number of ways to form a valid `arr1` up to the current index such that `arr1[i] = j`.\n\n2. **Base Case:**\n   - Initialize the DP array for the first element of `nums`. All possible values of `arr1[0]` are valid for the initial DP setup.\n\n3. **DP Transition:**\n   - For each subsequent index `i`, calculate the difference `d` between the current and previous `nums` values. \n   - Compute prefix sums to efficiently calculate the number of valid pairs that end at each possible value of `arr1[i]`.\n   - Update the DP table `dp` for the current index based on the prefix sums.\n\n4. **Compute Result:**\n   - Sum up the values in the DP array corresponding to the last element of `nums` to get the total number of valid pairs.\n\n5. **Modulo Operation:**\n   - Return the result modulo \\(10^9 + 7\\) to handle large numbers.\n\n# Complexity\n- **Time complexity:** `O(n * max_value)`\n  - Where `n` is the length of `nums` and `max_value` is the maximum value in `nums` (in this case, 1000). The time complexity is dominated by the DP transitions and prefix sum calculations.\n\n- **Space complexity:** `O(max_value)`\n  - The space complexity is due to the DP array and prefix sum array, which both have a size proportional to `max_value`.\n\n\n# Code\n```\nclass Solution:\n    def countOfPairs(self, nums: List[int]) -> int:\n        MOD = 10**9 + 7\n        n = len(nums)\n        max_val = max(nums)     \n        dp = [1] * (max_val + 1)        \n        for i in range(1, n):\n            d = max(0, nums[i] - nums[i - 1])\n            prefix_sum = [0] * (max_val + 1)\n            prefix_sum[0] = dp[0]\n            for j in range(1, max_val + 1):\n                prefix_sum[j] = (prefix_sum[j-1] + dp[j]) % MOD\n            dp2 = [(prefix_sum[j - d] if j >= d else 0) for j in range(nums[i] + 1)]\n            dp = dp2 + [0] * (max_val - nums[i] + 1)  \n        return sum(dp[:nums[-1] + 1]) % MOD\n\n```
['Python3']