C++ with explanation | O(N*MAX) Time | O(MAX) space
\n\nLet\'s walk through the equation derivation in simple terms:\n\n### Starting Point:\n- You have two arrays, arr1 and arr2, and a given array nums.\n- For ea
\n\nLet\'s walk through the equation derivation in simple terms:\n\n### Starting Point:\n- You have two arrays, `arr1` and `arr2`, and a given array `nums`.\n- For each index `i`, the sum of `arr1[i]` and `arr2[i]` should equal `nums[i]`.\n\n### Step 1: Express `arr2[i]`:\n- Since `arr1[i] + arr2[i] = nums[i]`, we can express `arr2[i]` as:\n  - `arr2[i] = nums[i] - arr1[i]`\n\n### Step 2: Consider Conditions on the Next Elements:\n1. **For `arr1[i+1]`:** \n   - We want `arr1[i+1]` to be at least as large as `arr1[i]` because `arr1` needs to be non-decreasing.\n\n2. **For `arr2[i+1]`:** \n   - We want `arr2[i+1]` to be less than or equal to `arr2[i]` because `arr2` needs to be non-increasing.\n   - Using the expression for `arr2[i]`, we know that:\n     - `arr2[i+1]` should be less than or equal to `nums[i] - arr1[i]`.\n\n### Step 3: Derive the Minimum Value for `arr1[i+1]`:\n1. From the relationship between `arr1` and `arr2` for the next index `i+1`, we know:\n   - `arr1[i+1] = nums[i+1] - arr2[i+1]`\n   \n2. Since `arr2[i+1]` must be less than or equal to `nums[i] - arr1[i]`, substitute this into the equation for `arr1[i+1]`:\n   - `arr1[i+1]` should be greater than or equal to `nums[i+1] - (nums[i] - arr1[i])`.\n\n### Step 4: Combine Conditions:\n- Now, `arr1[i+1]` must satisfy two conditions:\n  1. `arr1[i+1]` must be at least as large as `arr1[i]`.\n  2. `arr1[i+1]` must be at least as large as `nums[i+1] - (nums[i] - arr1[i])`.\n\n- Combining these, the minimum value that `arr1[i+1]` must take is:\n  - `arr1[i+1] >= max(arr1[i], nums[i+1] - (nums[i] - arr1[i]))`\n\n### Step 5: Counting Valid Arrangements:\n- To determine the number of valid ways to fill the arrays starting from `i`, sum up the valid options for `arr1[i+1]` by considering all values greater than or equal to this minimum value.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n    O(N*MAX)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n    O(MAX)\n\n# Code\n```\nclass Solution {\npublic:\n    #define mod 1000000007\n    \n    int countOfPairs(vector<int>& a) {\n        int n = a.size();\n        vector<int> nxtdp(1002, 0), dp(1002, 0);\n\n        for (int i = n - 1; i >= 0; i--) {\n            swap(nxtdp, dp);\n            fill(dp.begin(), dp.end(), 0);  \n\n            for (int up = a[i]; up >= 0; up--) {\n                if (i == n - 1) {\n                    dp[up] = 1;\n                } else {\n                    int down = a[i] - up;\n                    int up_next = max(up, a[i + 1] - down);\n                    \n                    dp[up] = nxtdp[up_next];\n                }\n\n                dp[up] = (dp[up] + dp[up + 1]) % mod;\n            }\n        }\n\n        return dp[0];\n    }\n};\n```
['C++']