C++ | Dynamic Programming |
Intuition\n Describe your first thoughts on how to solve this problem. Upon reviewing the provided code, it appears to be a solution to a problem that involves 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->Upon reviewing the provided code, it appears to be a solution to a problem that involves counting the number of ways to arrange a given sequence of numbers. The solution seems to be using a combination formula and a recursive approach to calculate the desired result.\n\nThe code initializes a 2D vector called `comb` with dimensions `(n + 1) \xD7 (n + 1)`, where `n` is the size of the input vector `nums`. This vector will be used to store the combination values for later calculations.\n\nThe outer loop iterates from 1 to `n` and the inner loop iterates from 1 to `i`. It populates the `comb` vector using the formula `(comb[i - 1][j - 1] + comb[i - 1][j]) % 1000000007`. This formula calculates the combination value `C(i, j)` by summing the combination values of the previous row and the previous row with an offset of -1.\n\nAfter initializing the `comb` vector, the code defines a lambda function `dfs` that takes a reference to a vector `nums` and returns an integer. This function recursively divides the input vector into two parts, `left` and `right`, based on whether each element is less than or greater than the first element of the vector. It calculates the number of arrangements for each subvector and multiplies them together, considering the combination values from the `comb` vector.\n\nFinally, the main function `numOfWays` calls the `dfs` function on the input vector `nums` and subtracts 1 from the result before returning it.\n\nOverall, the code aims to count the number of ways to arrange the elements of the `nums` vector according to certain conditions, using a combination formula and a recursive approach.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->The given code is implementing a recursive approach to solve a problem related to counting the number of ways to arrange a sequence of numbers. Let\'s break down the approach:\n\n1. The code initializes a 2D vector called `comb` with dimensions `(n + 1) \xD7 (n + 1)`, where `n` is the size of the input vector `nums`. This vector will be used to store precalculated combination values.\n\n2. The `comb` vector is populated using a combination formula `(comb[i - 1][j - 1] + comb[i - 1][j]) % 1000000007`. This formula calculates the combination value `C(i, j)` by summing the combination values of the previous row and the previous row with an offset of -1.\n\n3. The code defines a lambda function called `dfs` that takes a reference to a vector `nums` and returns an integer. This function is responsible for the recursive division of the input vector into two parts, `left` and `right`, based on whether each element is less than or greater than the first element of the vector.\n\n4. Within the `dfs` function, the base case is when the size of the input vector `nums` is less than or equal to 2. In this case, the function returns 1, indicating that there is only one possible arrangement.\n\n5. If the input vector size is greater than 2, the function proceeds to divide the vector into two subvectors, `left` and `right`, by comparing each element to the first element of the vector. Elements smaller than the first element are added to the `left` subvector, while elements greater than or equal to the first element are added to the `right` subvector.\n\n6. The function then calculates the number of arrangements for each subvector by recursively calling itself on the `left` and `right` subvectors. It multiplies these results together.\n\n7. Additionally, the function multiplies the result by the combination value `comb[n - 1][left.size()]`, which is obtained from the precalculated `comb` vector. This combination value represents the number of ways to arrange the elements in the subvectors `left` and `right` relative to each other.\n\n8. The final result is returned as `(int)res`, where `res` is a long long integer, and it is subtracted by 1 outside the `dfs` function before returning from the `numOfWays` function.\n\nThe approach combines the use of combination formulas and recursion to count the number of arrangements based on certain conditions. The recursive division of the vector into subvectors allows for exploring all possible arrangements and calculating the number of ways for each arrangement.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ --> O(n^2)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --> O(n^2)\n\n# Code\n```\nclass Solution {\npublic:\n        int numOfWays(vector<int>& nums) \n        {\n        int n = nums.size();\n        vector<vector<int>> comb(n + 1, vector<int>(n + 1));\n        comb[0][0] = 1;\n        for (int i = 1; i <= n; ++i) \n        {\n            comb[i][0] = 1;\n            for (int j = 1; j <= i; ++j)\n             {\n                comb[i][j] = (comb[i - 1][j - 1] + comb[i - 1][j]) % 1000000007;\n            }\n        }\n        function<int(vector<int>&)> dfs = [&](vector<int>& nums)\n         {\n            int n = nums.size();\n            if (n <= 2) return 1;\n            vector<int> left, right;\n            for (int i = 1; i < n; ++i) \n            {\n                if (nums[i] < nums[0]) left.push_back(nums[i]);\n                else right.push_back(nums[i]);\n            }\n            long long res = comb[n - 1][left.size()];\n            res = res * dfs(left) % 1000000007;\n            res = res * dfs(right) % 1000000007;\n            return (int)res;\n        };\n        return dfs(nums) - 1;\n    }\n};\n```\n![Leetcode.jpeg](https://assets.leetcode.com/users/images/aa253232-46b2-42c2-8335-fe8118d4eda6_1686897124.426299.jpeg)\n
['Array', 'Dynamic Programming', 'Recursion', 'Counting', 'C++']