BFS and TreeSet based solution (O(n) time, O(n) space) easy understand
Intuition\n Describe your first thoughts on how to solve this problem. \nWe can use a breadth-first search (BFS) approach to solve this problem. Since we are al
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe can use a breadth-first search (BFS) approach to solve this problem. Since we are allowed to perform multiple reverse operations of size k, we will use BFS to find the minimum number of operations needed to reach each index i, starting from the index p. We will maintain a TreeSet of available even and odd positions to facilitate searching for the next index to reverse to.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Initialize the \'arr\' to hold the results for each index, initially filled with -2.\n1. Set the positions specified in the \'banned\' array to -1 and the position p to 0.\n1. Create a Queue \'q\' to perform BFS and add the starting position \'p\'.\n1. Create TreeSets \'odd\' and \'even\' to store the available odd and even positions.\n1. Perform BFS using the queue \'q\' to find the minimum number of reverse operations for each index.\n1. After BFS is complete, set any index i with a value of -2 in the \'arr\' to -1, as it is impossible to reach that position.\n# Complexity\n- Time complexity: O(n) - where n is the length of the input array \'nums\'.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n) - since we store the input elements in the TreeSet, Queue and the result array \'arr\'.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    public int[] minReverseOperations(int n, int p, int[] banned, int k) {\n         // Step 1: Initialize the \'arr\' to hold the results for each index, initially filled with -2\n    int[] arr = new int[n];\n    for (int i = 0; i < n; i++) {\n        arr[i] = -2;\n    }\n\n    // Step 2: Set the positions specified in the \'banned\' array to -1 and the position p to 0\n    for (int num : banned) {\n        arr[num] = -1;\n    }\n    arr[p] = 0;\n\n    // Step 3: Create a Queue \'q\' to perform BFS and add the starting position \'p\'\n    Queue<Integer> q = new LinkedList<>();\n    q.add(p);\n\n    // Step 4: Create TreeSets \'odd\' and \'even\' to store the available odd and even positions\n    TreeSet<Integer> odd = new TreeSet<>();\n    TreeSet<Integer> even = new TreeSet<>();\n    for (int i = 0; i < n; i++) {\n        if (arr[i] == -2) {\n            if (i % 2 == 0) {\n                even.add(i);\n            } else {\n                odd.add(i);\n            }\n        }\n    }\n\n    // Step 5: Perform BFS using the queue \'q\' to find the minimum number of reverse operations for each index\n        \n        while (!q.isEmpty()) {\n            int origin = q.remove();\n            int min = 0;\n            if (origin < k - 1) {\n                min = k - 1 -origin;\n            } else {\n                min = origin - k + 1;\n            }\n            \n            int max = 0;\n            int rightDistance = n - 1 - origin;\n            if (rightDistance < k - 1) {\n                max = k - 1 - rightDistance;\n            } else {\n                max = rightDistance - k + 1;\n            }\n            max = n - 1 - max;\n        \n            TreeSet<Integer> set = max % 2 == 0 ? even : odd;\n            Integer next = set.ceiling(min);\n            while (next != null && next <= max) {\n                arr[next] = arr[origin] + 1;\n                q.add(next);\n                set.remove(next);\n                next = set.ceiling(min);\n            }\n        }\n// Step 6: After BFS is complete, set any index i with a value of -2 in the \'arr\' to -1\n        for (int i = 0; i < n; i++) {\n            if (arr[i] == -2) {\n                arr[i] = -1;\n            }\n        }\n        return arr;\n    }\n}\n```
['Java']