Minimum Reverse Operations to Move a Value in an Array
Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem is asking us to find the minimum number of reverse operations needed to mov
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem is asking us to find the minimum number of reverse operations needed to move a value (represented as 1) to each position in an array, given certain positions are banned. The reverse operation can only be performed on a subarray of size k and the value should never go to any of the positions in banned.\n\nOur first thought is to use a breadth-first search (BFS) approach. We can start from the initial position of the value and try to move it to each position in the array using reverse operations. We need to keep track of the minimum number of reverse operations needed to move the value to each position.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWe use a breadth-first search (BFS) approach to solve this problem. We start from the initial position of the value and try to move it to each position in the array using reverse operations. We keep track of the minimum number of reverse operations needed to move the value to each position in a list out.\n\nFor each visited position, there are potentially O(k) target positions that can be reached through reverse operations. To avoid the multiplicative cost of iterating over all these potential positions, we update the nextNode2s array. This array initially points forward by 2, but it is updated dynamically to point beyond all the target positions considered for each visited position. This optimization helps improve the efficiency of the algorithm and avoids unnecessary computations.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of this approach is O(n), where n is the length of the array. This is because we visit each position in the array once.\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity of this approach is also O(n). This is because we need to store the minimum number of reverse operations needed to move the value to each position in a list, and we also need to store the next node for each position in an array.\n# Code\n```\nfrom collections import deque\nclass Solution:\n    def minReverseOperations(self, n, p, banned, k):\n        out = [-1]*n\n        for node in banned:\n            out[node] = -2\n        nodes = [p]\n        depth = 0\n        out[p] = depth\n        step = k - 1\n        nextNode2s = [i+2 for i in range(n)]\n        while nodes:\n            depth += 1\n            newNodes = []\n            for node1 in nodes:\n                loReverseStart = max(node1 - step, 0)\n                hiReverseStart = min(node1, n - k)\n                loNode2 = 2 * loReverseStart + k - 1 - node1\n                hiNode2 = 2 * hiReverseStart + k - 1 - node1\n                postHiNode2 = hiNode2 + 2\n                node2 = loNode2\n                while node2 <= hiNode2:\n                    nextNode2 = nextNode2s[node2]\n                    nextNode2s[node2] = postHiNode2\n                    if node2 >= 0 and node2 < n and out[node2] == -1:\n                        newNodes.append(node2)\n                        out[node2] = depth\n                    node2 = nextNode2\n            nodes = newNodes\n        for i in range(n):\n            if out[i] == -2:\n                out[i] = -1\n        return out\n\n```
['Python3']