Dynamic Programming (DP) for Counting Monotonic Subsequences by coderkb
Intuition:\nThe problem requires counting the number of valid subsequences where each subsequence is monotonic (either strictly increasing or non-decreasing). T
**Intuition:**\nThe problem requires counting the number of valid subsequences where each subsequence is monotonic (either strictly increasing or non-decreasing). The challenge is to efficiently count all such valid subsequences modulo \\(10^9 + 7\\).\n\n**Approach:**\n1. **Dynamic Programming Setup:**\n   - We use a DP table `dp[i][j]` where `i` represents the index in the array `nums`, and `j` represents a possible value that a subsequence can end with.\n   - `dp[i][j]` stores the number of valid subsequences ending at index `i` with the value `j`.\n\n2. **Initialization:**\n   - Initialize the DP table for the first element `nums[0]`. For every value less than or equal to `nums[0]`, set `dp[0][j] = 1`.\n\n3. **DP Transition:**\n   - For each element in `nums`, if the previous element is less than the current one, we update the DP table by summing up the counts from previous valid states.\n   - If the previous element is greater than or equal to the current one, we update the DP table by summing up all previous states less than or equal to the current value.\n\n4. **Final Count:**\n   - The final count of all valid subsequences is obtained by summing up all values in `dp[n-1]`, where `n` is the size of the array.\n\n**Time Complexity:** \n- **Time Complexity:** $$(O(n \\times 1000)$$, where $$n$$ is the size of the input array `nums`. The DP table is of size $$(n \\times 1000)$$.\n\n**Space Complexity:** \n- **Space Complexity:** $$(O(n \\times 1000)$$ due to the DP table used to store the counts of subsequences.\n\n**Code with Comments:**\n\n```cpp\nclass Solution {\n    int mod = 1e9 + 7; // Modulus for large numbers to avoid overflow\npublic:\n    int countOfPairs(vector<int>& nums) {\n        int n = nums.size();\n        vector<vector<int>> dp(n + 1, vector<int>(1001, 0)); // DP table to store subsequence counts\n\n        // Initialize DP table for the first element\n        for(int j = 0; j <= nums[0]; ++j) {\n            dp[0][j] = 1;\n        }\n\n        // DP transition to fill the table\n        for(int i = 1; i < n; ++i) {\n            if(nums[i - 1] < nums[i]) {\n                // Case where nums[i-1] < nums[i]\n                int diff = nums[i] - nums[i - 1];\n                int sum = 0;\n                for(int j = diff; j <= nums[i]; ++j) {\n                    sum = (sum + dp[i - 1][j - diff]) % mod; // Accumulate valid subsequences\n                    dp[i][j] = sum; // Store the count in the DP table\n                }\n            } else {\n                // Case where nums[i-1] >= nums[i]\n                int sum = 0;\n                for(int j = 0; j <= nums[i]; ++j) {\n                    sum = (sum + dp[i - 1][j]) % mod; // Accumulate valid subsequences\n                    dp[i][j] = sum; // Store the count in the DP table\n                }\n            } \n        }\n\n        // Sum up all valid subsequences ending at the last element\n        long ans = 0;\n        for(int j = 0; j < 1001; ++j) {\n            ans = (ans + dp[n - 1][j]) % mod; // Sum all the values in the last row of DP table\n        }\n\n        return ans; // Return the final count\n    }\n};\n```\n\n**Explanation:**\n- **DP Initialization:** The first element is straightforward, allowing subsequences ending in any value up to `nums[0]`.\n- **DP Transition:** For each subsequent element, we consider whether the current subsequence can continue in a monotonic fashion based on the previous element.\n- **Final Count:** After processing all elements, the final count of valid subsequences is derived from the last row of the DP table.\n\nThis solution efficiently counts all valid monotonic subsequences using dynamic programming, ensuring that the result is computed within the given constraints.
['Dynamic Programming', 'C++']