I want to say this is really hard for me in Rust_I'd like to share with you
Intuition\nIn my initial attempts, I overlooked the significance of the parity of k and its implications on the movement of the number 1. This oversight led to 
# Intuition\nIn my initial attempts, I overlooked the significance of the parity of k and its implications on the movement of the number 1. This oversight led to incorrect results in certain test cases. \nAt the other side, the presence of banned positions, which the number 1 cannot occupy, added a layer of complexity. Additionally, the parity of k plays a crucial role, as it dictates how the number 1 can move within the array.\n\n# Approach\nParity Grouping: I started by categorizing all potential positions into two groups based on their parity (even or odd). In my opinion, this is crucial because when k is even, the position of 1 changes its parity with every operation. By doing this grouping upfront, I could preemptively determine which positions were unreachable.\nBreadth-First Search (BFS): Beginning from the initial position, I employed BFS to identify all reachable positions. To optimize the search, I utilized a BTreeSet to store positions within each parity group, allowing for swift identification of positions within the reachable range from the current position.\nDetermining Reachable Range: The get_range function was instrumental in ascertaining the range of positions reachable from the current position. This function, taking into account the value of k and the current position, returns a range indicating where 1 can be moved.\n\n# Complexity\n- Time complexity:\nO(nlogn) - The use of BTreeSet for storing positions means that each insertion and deletion operation has a time complexity of O(nlogn). Given that we perform such operations at most n times, that means the overall time complexity is O(nlogn).\n\n- Space complexity:\nO(n) - Space is required for storing all positions, the answer array, and the BFS queue.\n\n# Code\n```\nuse std::collections::{BTreeSet, VecDeque};\n\nimpl Solution {\n    pub fn min_reverse_operations(n: i32, p: i32, banned: Vec<i32>, k: i32) -> Vec<i32> {\n        let mut even_indices = BTreeSet::new();\n        let mut odd_indices = BTreeSet::new();\n        for i in 0..n {\n            if i % 2 == 0 {\n                even_indices.insert(i);\n            } else {\n                odd_indices.insert(i);\n            }\n        }\n        for &b in &banned {\n            if b % 2 == 0 {\n                even_indices.remove(&b);\n            } else {\n                odd_indices.remove(&b);\n            }\n        }\n\n        let mut ans = vec![-1; n as usize];\n        let mut queue = VecDeque::new();\n        queue.push_back(p);\n        ans[p as usize] = 0;\n\n        if p % 2 == 0 {\n            even_indices.remove(&p);\n        } else {\n            odd_indices.remove(&p);\n        }\n\n        while let Some(current) = queue.pop_front() {\n            let (l, r) = Self::get_range(current, n, k);\n            let current_set = if l % 2 == 0 { &mut even_indices } else { &mut odd_indices };\n            let mut to_remove = Vec::new();\n            for &i in current_set.range(l..=r) {\n                ans[i as usize] = ans[current as usize] + 1;\n                queue.push_back(i);\n                to_remove.push(i);\n            }\n            for i in to_remove {\n                current_set.remove(&i);\n            }\n        }\n\n        ans\n    }\n\n    fn get_range(x: i32, n: i32, k: i32) -> (i32, i32) {\n        let l = std::cmp::max(x - k + 1, 0);\n        let r = std::cmp::min(x, n - k);\n        let l = 2 * l + k - 1 - x;\n        let r = 2 * r + k - 1 - x;\n        (l, r)\n    }\n}\n\n\n\n\n\n\n\n\n```
['Rust']