[Swift]LeetCode995. K 連續位的最小翻轉次數 | Minimum Number of K Consecutive Bit Flips

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In an array A containing only 0s and 1s, a K-bit flip consists of choosing a (contiguous) subarray of length K and simultaneously changing every 0 in the subarray to 1, and every 1 in the subarray to 0.

Return the minimum number of K-bit flips required so that there is no 0 in the array.  If it is not possible, return -1.

Example 1:

Input: A = [0,1,0], K = 1 Output: 2 Explanation: Flip A[0], then flip A[2]. 

Example 2:

Input: A = [1,1,0], K = 2 Output: -1 Explanation: No matter how we flip subarrays of size 2, we can't make the array become [1,1,1]. 

Example 3:

Input: A = [0,0,0,1,0,1,1,0], K = 3 Output: 3 Explanation: Flip A[0],A[1],A[2]: A becomes [1,1,1,1,0,1,1,0] Flip A[4],A[5],A[6]: A becomes [1,1,1,1,1,0,0,0] Flip A[5],A[6],A[7]: A becomes [1,1,1,1,1,1,1,1] 

Note:

1. 1 <= A.length <= 30000
2. 1 <= K <= A.length

---

在僅包含 0 和 1 的數組 A 中，一次 K 位翻轉包括選擇一個長度爲 K的（連續）子數組，同時將子數組中的每一個 0 更改成 1，而每一個 1 更改成 0。

返回所需的 K 位翻轉的次數，以便數組沒有值爲 0 的元素。若是不可能，返回 -1。 

示例 1：

輸入：A = [0,1,0], K = 1
輸出：2
解釋：先翻轉 A[0]，而後翻轉 A[2]。

示例 2：

輸入：A = [1,1,0], K = 2
輸出：-1
解釋：不管咱們怎樣翻轉大小爲 2 的子數組，咱們都不能使數組變爲 [1,1,1]。

示例 3：

輸入：A = [0,0,0,1,0,1,1,0], K = 3
輸出：3
解釋：
翻轉 A[0],A[1],A[2]: A變成 [1,1,1,1,0,1,1,0]
翻轉 A[4],A[5],A[6]: A變成 [1,1,1,1,1,0,0,0]
翻轉 A[5],A[6],A[7]: A變成 [1,1,1,1,1,1,1,1] 

提示：

1. 1 <= A.length <= 30000
2. 1 <= K <= A.length

---

Runtime: 632 ms

Memory Usage: 19 MB

 1 class Solution {
 2     func minKBitFlips(_ A: [Int], _ K: Int) -> Int {
 3         var ans:Int = 0
 4         var n:Int = A.count
 5         var f:[Bool] = [Bool](repeating:false,count:n + 1)
 6         var sum:Bool = false
 7         for i in 0..<n
 8         {
 9             //異或
10             sum = (sum == f[i]) ? false : true
11             if sum == (A[i] == 1)
12             {
13                 if i + K > n {return -1}
14                 f[i + K] = !f[i + K]
15                 sum = !sum
16                 ans += 1
17             }
18         }
19         return ans        
20     }
21 }

---

744ms 

 1 class Solution {
 2     func minKBitFlips(_ A: [Int], _ K: Int) -> Int {
 3         var left = 0
 4         var flips = 0
 5         var totalFlips = 0
 6         var a = A
 7         
 8         while left < a.count {
 9             totalFlips = left < K ? totalFlips : totalFlips - a[left - K ]
10         
11             if totalFlips % 2 == 0 && a[left] == 0 {
12                 if left > a.count - K {
13                     return -1
14                 }
15                 flips += 1
16                 totalFlips += 1
17                 a[left] = 1
18             } else if totalFlips % 2 == 1 && a[left] == 1 {
19                 if left > a.count - K {
20                     return -1
21                 }
22                 
23                 flips += 1
24                 totalFlips += 1
25                 a[left] = 1
26             } else {
27                 a[left] = 0
28             }
29             
30             left += 1
31         }
32         
33         return flips
34     }
35 }

---

748ms

 1 class Solution {
 2     func minKBitFlips(_ A: [Int], _ K: Int) -> Int {
 3         var A = A
 4         var count = 0
 5         var i = 0
 6         var flipHint = A.map { _ in false }
 7         var flipped = false
 8         while i < A.count {
 9             if flipHint[i] == true {
10                 flipped.toggle()
11             }
12 
13             if A[i] == (flipped ? 1 : 0) {
14                 flipped.toggle()
15                 if i + K < A.count {
16                     flipHint[i+K] = true
17                 } else if i + K > A.count {
18                     return -1
19                 }
20                 count += 1
21             }
22             i += 1
23         }
24         return count
25     }
26 }

---

900ms

 1 class Solution {
 2     func minKBitFlips(_ A: [Int], _ K: Int) -> Int {
 3         var count = 0, firstIndex = 0, A = A
 4         var lastFlip: [Int] = []
 5 
 6         while firstIndex < A.count {
 7             defer { firstIndex += 1 }
 8 
 9             if let first = lastFlip.first,
10                 firstIndex >= first {
11                 lastFlip.removeFirst()
12             }
13 
14             guard (A[firstIndex] + lastFlip.count) % 2 == 0 else {
15                 continue
16             }
17             if firstIndex + K > A.endIndex {
18                 return -1
19             } else {
20                 lastFlip.append(firstIndex + K)
21                 count += 1
22             }
23         }
24 
25         return count
26     }
27 }