[Swift]LeetCode753. 破解保险箱 | Cracking the Safe

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There is a box protected by a password. The password is n digits, where each letter can be one of the first k digits 0, 1, ..., k-1.

You can keep inputting the password, the password will automatically be matched against the last n digits entered.

For example, assuming the password is "345", I can open it when I type "012345", but I enter a total of 6 digits.

Please return any string of minimum length that is guaranteed to open the box after the entire string is inputted.

Example 1:

Input: n = 1, k = 2
Output: "01"
Note: "10" will be accepted too. 

Example 2:

Input: n = 2, k = 2
Output: "00110"
Note: "01100", "10011", "11001" will be accepted too. 

Note:

1. n will be in the range [1, 4].
2. k will be in the range [1, 10].
3. k^n will be at most 4096.

---

有一个需要密码才能打开的保险箱。密码是 n 位数, 密码的每一位是 k 位序列 0, 1, ..., k-1 中的一个 。

你可以随意输入密码，保险箱会自动记住最后 n 位输入，如果匹配，则能够打开保险箱。

举个例子，假设密码是 "345"，你可以输入 "012345" 来打开它，只是你输入了 6 个字符.

请返回一个能打开保险箱的最短字符串。 

示例1:

输入: n = 1, k = 2
输出: "01"
说明: "10"也可以打开保险箱。 

示例2:

输入: n = 2, k = 2
输出: "00110"
说明: "01100", "10011", "11001" 也能打开保险箱。 

提示：

1. n 的范围是 [1, 4]。
2. k 的范围是 [1, 10]。
3. k^n 最大可能为 4096。

---

12ms

func deBruijnSequence<Alphabets: RandomAccessCollection>(of alphabets:Alphabets, length: Int) -> [Alphabets.Element] {
    typealias Alphabet = Alphabets.Element
    
    let alphabetCount = alphabets.count
    let cycleCount = repeatElement(alphabetCount, count: length - 1).reduce(1, *)
    let debruijnLength = cycleCount * alphabetCount
    
    var used = Array(repeating: false, count: debruijnLength)
    var result: [Alphabet] = []
    result.reserveCapacity(debruijnLength)
    
    for index in 0..<debruijnLength {
        var current = index
        while !used[current] {
            used[current] = true
            
            let elementIndex = current / cycleCount
            current = (current % cycleCount) * alphabetCount + elementIndex
            result.append(alphabets[alphabets.index(alphabets.startIndex, offsetBy: elementIndex)])
        }
        assert(current == index)
    }
    
    assert(result.count == debruijnLength)
    return result
}

class Solution {
    func crackSafe(_ n: Int, _ k: Int) -> String {
        let result = deBruijnSequence(of: (0..<k).map(String.init), length: n)
        return (result + repeatElement("0", count: n - 1)).joined()
    }
}

---

36ms

class Solution {
    func crackSafe(_ n: Int, _ k: Int) -> String {
        let total = Int(pow(Double(k), Double(n)))
        var current = [Int](repeating: 0, count: n)
        var used = Set<String>()
        used.insert(current.reduce("") { $0 + String($1) })
        dfs(n, k, total, &used, &current)
        return current.reduce("") { $0 + String($1) }
    }
        
    private func dfs(_ n: Int, _ k: Int, _ total: Int, _ used: inout Set<String>, _ current: inout [Int]) -> Bool {
        guard used.count < total else {
            return true
        }
        var prefix = Array(current[(current.count - n + 1)...]).reduce("") { $0 + String($1) }
        for num in 0..<k {
            let currentSegment = prefix + String(num)
            guard !used.contains(currentSegment) else {
                continue
            }
            used.insert(currentSegment)
            current.append(num)
            if dfs(n, k, total, &used, &current) {
                return true
            }
            current.removeLast()
            used.remove(currentSegment)
        }
        return false
    }
}

---

152ms

class Solution {
    func dfs(_ n: Int, _ k: Int, _ len: Int, _ data: inout Set<String>, _ res:inout String) -> Bool {
        if (res.count == len) { return true; }
        let suf = res.suffix(n-1)
        for c in 0..<k {
            if !data.contains(suf + "(c)") {
                data.insert(suf + "(c)")
                res += "(c)"
                if (dfs(n, k, len, &data, &res)) {
                    return true
                }
                data.remove(suf + "(c)")
                res.removeLast()
            }
        }
        return false
    }
    func crackSafe(_ n: Int, _ k: Int) -> String {
        let len = Int(pow(Double(k), Double(n))) + n - 1
        var res = String((0..<n).map{_ in Character("0")})
        var data = Set([res])
        dfs(n, k, len, &data, &res)
        return res
    }
}

---

Runtime: 320 ms

Memory Usage: 22.2 MB

class Solution {
    func crackSafe(_ n: Int, _ k: Int) -> String {
        var res:String = "0".repeatString(n - 1)
        var visited:Set<String> = [res]
        var num:Int = Int(pow(Double(k), Double(n)))
        helper(n, k,num, &visited, &res)
        return res
    }
    
    func helper(_ n:Int,_ k:Int,_ total:Int,_ visited:inout Set<String>,_ res:inout String)
    {
        if visited.count == total
        {
            return
        }
        var pre:String = res.subString(res.count - n + 1, n - 1)
        for i in stride(from:k - 1,through:0,by:-1)
        {
            var cur:String = pre + String(i)
            if visited.contains(cur) {continue}
            visited.insert(cur)
            res += String(i)
            helper(n, k, total, &visited, &res)
        }        
    }
}

extension String {
    //获取重复指定次数的字符串
    func repeatString(_ times: Int ) -> String
    {
        var result = String()
        for i in 0...times {
            result += self
        }
        return result 
    }
    
    // 截取字符串：指定索引和字符数
    // - begin: 开始截取处索引
    // - count: 截取的字符数量
    func subString(_ begin:Int,_ count:Int) -> String {
        let start = self.index(self.startIndex, offsetBy: max(0, begin))
        let end = self.index(self.startIndex, offsetBy:  min(self.count, begin + count))
        return String(self[start..<end]) 
    }    
}