C++ Solution with Detailed description
Intuition\nThe problem requires to find the minimum number of reverse operations needed to convert the given sequence of integers b to a sequence a of length n,
# Intuition\nThe problem requires to find the minimum number of reverse operations needed to convert the given sequence of integers b to a sequence a of length n, such that a subsequence of length k starting from the index p in the sequence a is strictly increasing. A reverse operation consists of reversing a subsequence in a.\n\n# Approach\nThe given problem can be solved using BFS. Initially, create a set s containing all the elements in the sequence b. Then, create a queue q and a map vis to keep track of visited elements. Initialize the map with the starting element p as visited. Create a vector d of length n initialized to -1 which will be used to store the minimum number of reverse operations needed for each index in the sequence a. Create two sets vs[0] and vs[1] and initialize them with indices of even and odd positions respectively in the sequence a. Remove the indices of elements in the sequence b from their corresponding sets.\n\nNow, while the queue q is not empty, perform the following operations:\n\nRemove the front element from the queue and set the minimum number of reverse operations needed for that index in d to the current distance dis.\nCompute the indices mn and mx such that the subsequence of a starting from index mn and ending at index mx contains k elements including the element at index p.\nFor each index j in the set vs[p % 2] (i.e. the set of indices with the same parity as p), such that j is between mn and mx and (mx + ((j-p)/2)) < n, do the following:\nIf j is not visited, mark it as visited and push it to the queue. Also, add it to a vector v.\nFor each element in the vector v, remove the index from the set vs[j%2].\nIncrement the distance dis.\nFinally, return the vector d which contains the minimum number of reverse operations needed for each index in the sequence a.\n\n# Complexity\n\n- Time complexity:\nThe time complexity of the above solution is O(n log n) where n is the length of the sequence a. The for loop in the beginning has a time complexity of O(n log n) because it uses set data structure which takes log n time for insertion of each element. The while loop has a time complexity of O(n) because each index in the sequence a is visited only once. The inner for loop has a time complexity of O(log n) because the set data structure is used again. Therefore, the overall time complexity is O(n log n).\n\n- Space complexity:\nThe space complexity of the above solution is O(n) where n is the length of the sequence a. The set and vector data structures used in the solution have a space complexity of O(n). The queue and map data structures used in the solution have a space complexity of O(w) where w is the maximum width of the tree during BFS. In the worst case, the maximum width of the tree is n/2 (when k is 2 and p is 0 or 1) which gives a space complexity of O(n). Therefore, the overall space complexity is O(n).\n\n# Code\n```\nclass Solution {\npublic:\n    vector<int> minReverseOperations(int n, int p, vector<int>& b, int k) {\n        // Initialize an array d to store the minimum number of reverse operations needed to sort the array.\n        vector<int> d(n,-1);\n        // Initialize a set s to store the values in the array b.\n        set<int> s;\n        for(int i : b) s.insert(i);\n        // Initialize a queue q to perform BFS.\n        queue<int> q;\n        // Initialize a map vis to store the visited nodes.\n        map<int,bool> vis;\n        // Mark the starting node as visited.\n        vis[p] = true;\n        // Push the starting node into the queue.\n        q.push(p);\n        // Initialize a variable dis to store the current distance.\n        int dis = 0;\n        // Initialize a vector of sets vs to store the indices of odd and even numbers.\n        vector<set<int>> vs(2);\n        for(int i = 0; i < n; i++) {\n            vs[i % 2].insert(i);\n        }\n        // Remove the elements from the set that appear in the array b.\n        for(int j : b) {\n            vs[j % 2].erase(j);\n        }\n        // Remove the starting node from the set.\n        vs[p % 2].erase(p);\n        // Perform BFS.\n        while(!q.empty()) {\n            int sz = q.size();\n            while(sz--) {\n                // Pop the front node from the queue.\n                int a = q.front();\n                q.pop();\n                // Set the distance of the current node.\n                d[a] = dis;\n                // Calculate the range of the current node.\n                int mn = max(0, a - k + 1);\n                int y = a - mn;\n                int j = mn + (k - 1 - y);\n                int p = j;\n                // Initialize a vector v to store the children of the current node.\n                vector<int> v;\n                int mx = mn + k - 1;\n                // Traverse the set to find the children of the current node.\n                for(auto k1 = vs[j % 2].lower_bound(j); k1 != vs[j % 2].end() && *k1 <= (a + k - 1) && (mx + ((*k1 - p) / 2) < n); k1++) {\n                    // If the child node is not visited, mark it as visited and push it into the queue and the vector v.\n                    if(vis[*k1] == false) {\n                        vis[*k1] = true;\n                        q.push(*k1);\n                        v.push_back(*k1);\n                    }\n                }\n                // Remove the children from the set.\n                for(int i : v) {\n                    vs[i % 2].erase(i);\n                }\n            }\n            // Increment the distance.\n            dis++;\n        }\n        // Return the array d.\n        return d;\n    }\n};\n\n```
['C++']