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  • [Swift]LeetCode1007. 行相等的最少多米诺旋转 | Minimum Domino Rotations For Equal Row

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    In a row of dominoes, A[i] and B[i] represent the top and bottom halves of the i-th domino.  (A domino is a tile with two numbers from 1 to 6 - one on each half of the tile.)

    We may rotate the i-th domino, so that A[i] and B[i] swap values.

    Return the minimum number of rotations so that all the values in A are the same, or all the values in B are the same.

    If it cannot be done, return -1.

    Example 1:

    Input: A = [2,1,2,4,2,2], B = [5,2,6,2,3,2]
Output: 2
Explanation: 
The first figure represents the dominoes as given by A and B: before we do any rotations.
If we rotate the second and fourth dominoes, we can make every value in the top row equal to 2, as indicated by the second figure.

    Example 2:

    Input: A = [3,5,1,2,3], B = [3,6,3,3,4]
Output: -1
Explanation: 
In this case, it is not possible to rotate the dominoes to make one row of values equal.

    Note:

    1. 1 <= A[i], B[i] <= 6
    2. 2 <= A.length == B.length <= 20000

    在一排多米诺骨牌中，A[i] 和 B[i] 分别代表第 i 个多米诺骨牌的上半部分和下半部分。（一个多米诺是两个从 1 到 6 的数字同列平铺形成的 —— 该平铺的每一半上都有一个数字。）

    我们可以旋转第 i 张多米诺，使得 A[i] 和 B[i] 的值交换。

    返回能使 A 中所有值或者 B 中所有值都相同的最小旋转次数。

    如果无法做到，返回 -1.

    示例 1：

    输入：A = [2,1,2,4,2,2], B = [5,2,6,2,3,2]
输出：2
解释：
图一表示：在我们旋转之前， A 和 B 给出的多米诺牌。
如果我们旋转第二个和第四个多米诺骨牌，我们可以使上面一行中的每个值都等于 2，如图二所示。

    示例 2：

    输入：A = [3,5,1,2,3], B = [3,6,3,3,4]
输出：-1
解释：
在这种情况下，不可能旋转多米诺牌使一行的值相等。

    提示：

    1. 1 <= A[i], B[i] <= 6
    2. 2 <= A.length == B.length <= 20000
    Runtime: 512 ms
    Memory Usage: 19.2 MB
     1 class Solution {
 2     func minDominoRotations(_ A: [Int], _ B: [Int]) -> Int {
 3         var n:Int = A.count
 4         var best:Int = Int.max
 5         for same in 1...6
 6         {
 7             var a_cost:Int = 0
 8             var b_cost:Int = 0
 9             
10             for i in 0..<n
11             {
12                 if A[i] != same && B[i] != same
13                 {
14                     a_cost = Int.max
15                     b_cost = Int.max
16                     break;
17                 }
18                 if A[i] != same
19                 {
20                     a_cost += 1
21                 }
22                 if B[i] != same
23                 {
24                     b_cost += 1
25                 }
26             }
27             best = min(best, min(a_cost, b_cost))
28         }
29         return best < Int.max / 2 ? best : -1        
30     }
31 }
    512ms 
     1 class Solution {
 2   func minDominoRotations(_ A: [Int], _ B: [Int]) -> Int {
 3     var candidate = A.count + 1
 4     var candidateNum = 0
 5     for num in 1 ... 6 {
 6       var count = 0
 7       var i = 0
 8       var aCount = 0
 9       var bCount = 0
10       while i < A.count {
11         if A[i] != num && B[i] != num {
12           break
13         } 
14         
15         if A[i] != num {
16           aCount += 1
17         } else if B[i] != num {
18           bCount += 1
19         }
20         i += 1
21       }
22       
23       if i == A.count {
24         if aCount < candidate {
25           candidate = aCount
26           candidateNum = num
27         }
28         if bCount < candidate {
29           candidate = bCount
30           candidateNum = num
31         }
32         
33       }
34     }
35     if candidateNum == 0 {
36       return -1
37     } 
38     return candidate
39   }
40 }

    520ms

     1 class Solution {
 2     func minDominoRotations(_ A: [Int], _ B: [Int]) -> Int {
 3         guard A.count == B.count else { return -1 }
 4         var candidate_a = A[0]
 5         var candidate_b = B[0]
 6         var result_A = 0
 7         var result_B = 0
 8         for i in 1..<A.count {
 9             if candidate_a != -1 {
10                 if A[i] != candidate_a, B[i] != candidate_a {
11                     result_A = Array(A[0..<i]).filter({$0 == candidate_a}).count
12                     result_B = Array(B[0..<i]).filter({$0 == candidate_a}).count
13                     candidate_a = -1
14                 }
15             }
16             if candidate_b != -1 {
17                 if A[i] != candidate_b, B[i] != candidate_b {
18                     
19                     result_A = Array(A[0..<i]).filter({$0 == candidate_b}).count
20                     result_B = Array(B[0..<i]).filter({$0 == candidate_b}).count
21                     candidate_b = -1
22                 }
23             }
24             if candidate_a == -1, candidate_b == -1 {
25                 return -1
26             }
27             if candidate_a == -1, candidate_b != -1{
28                 if A[i] != candidate_b {
29                     result_A += 1
30                 }
31                 if B[i] != candidate_b {
32                     result_B += 1
33                 }
34             } else if candidate_b == -1, candidate_a != -1{
35                 if A[i] != candidate_a {
36                     result_A += 1
37                 }
38                 if B[i] != candidate_a {
39                     result_B += 1
40                 }
41                 
42             } else {
43                 if A[i] != candidate_a{
44                     result_A += 1
45                 }
46                 if B[i] != candidate_b{
47                     result_B += 1
48                 }
49             }
50         }
51         return min(result_A, result_B)
52     }
53 }

    536ms

     1 class Solution {
 2     func minDominoRotations(_ A: [Int], _ B: [Int]) -> Int {
 3         var arA: [Int] = [0,0,0,0,0,0,0]
 4         var arB: [Int] = [0,0,0,0,0,0,0]
 5         var ab = 0
 6         for i in (0..<A.count) {
 7             arA[A[i]] += 1
 8             arB[B[i]] += 1
 9         }
10         var freq = -1
11         for i in (1..<7) {
12             if arA[i] + arB[i] >= A.count {
13                 freq = i
14                 if arA[i] >= arB[i] { ab = 0 } else { ab = 1}
15             }
16         }
17         if freq == -1 { return -1 }
18         print(freq, ab)
19         var res = 0
20         
21         for i in (0..<A.count) {
22             if A[i] == freq && B[i] == freq { continue }
23             if A[i] != freq && B[i] != freq { return -1 }
24             if ab == 0 && A[i] != freq { res += 1 }
25             if ab != 0 && B[i] != freq { res += 1 }
26         }
27         
28         return res
29     }
30 }

    568ms

     1 class Solution {
 2     func minDominoRotations(_ A: [Int], _ B: [Int]) -> Int {
 3         let len = A.count
 4         var mapA = [Int: Int](), mapB = [Int: Int]()
 5         var mapSame = [Int: Int]()
 6         
 7         var diffCount = 0
 8         for i in 0..<len {
 9             if A[i] != B[i] {
10                 mapA[A[i], default: 0] += 1
11                 mapB[B[i], default: 0] += 1
12                 diffCount += 1
13             } else {
14                 mapSame[A[i], default: 0] += 1
15             }
16         }
17 
18         var result = -1
19         for i in 1...6 where mapA[i, default: 0] + mapB[i, default: 0] == diffCount && diffCount + mapSame[i, default: 0] == len {
20             let swapCount = min(mapA[i, default: 0], mapB[i, default: 0])
21             if result == -1 {
22                 result = swapCount
23             } else {
24                 result = min(result, swapCount)
25             }
26         }
27         return result
28     }
29 }

    576ms

     1 class Solution {
 2     func minDominoRotations(_ A: [Int], _ B: [Int]) -> Int {
 3         var countA = [Int](repeating: 0, count: 7)
 4         var countB = [Int](repeating: 0, count: 7)
 5         for i in A.indices {
 6             countA[A[i]] += 1
 7             countB[B[i]] += 1
 8         }
 9 
10         var ans = Int.max
11         for i in 1...6 {
12             if countA[i] + countB[i] >= A.count {
13                 for j in A.indices {
14                     if A[j] != i && B[j] != i  {
15                         break
16                     }
17                     if j == A.count - 1 {
18                         let curmin = min(A.count - countB[i], A.count - countA[i])
19                         ans = min(ans, curmin)
20                     }
21                 }
22             }
23         }
24         return ans == Int.max ? -1 : ans    
25     }
26 }

    620ms

     1 class Solution {
 2     func mask(_ a : Int, _ b : Int ) -> Int {
 3         return (1 << a) | (1 << b)
 4     }
 5     func minDominoRotations(_ A: [Int], _ B: [Int]) -> Int {
 6         guard A.count > 0 && A.count == B.count else {
 7             return -1
 8         }
 9         var totalMask = 255
10         
11         for (a,b) in zip(A, B) {
12             totalMask = totalMask & mask(a, b)
13         }
14         
15         guard totalMask != 0 else {
16             return -1
17         }
18         
19         var target = B[0]
20         if totalMask & (1 << A[0]) != 0 {
21             target = A[0]
22         }
23         
24         var aCount = 0
25         var bCount = 0
26         
27         for (a,b) in zip(A, B) {
28             if a == target {
29                 aCount += 1
30             }
31             if b == target {
32                 bCount += 1
33             }
34         }
35         
36         return A.count - max(aCount, bCount)
37     }
38 }

    676ms

     1 class Solution {
 2   
 3   func minDominoRotations(_ A: [Int], _ B: [Int]) -> Int {
 4     let maxA = A.mostRepeatingElement()
 5     let maxB = B.mostRepeatingElement()
 6     var traversalArray = A
 7     var referenceArray = B
 8     var needle = maxB.element
 9     var swapsExpected = referenceArray.count - maxB.count
10     if maxA.count > maxB.count {
11       traversalArray = B
12       referenceArray = A
13       needle = maxA.element
14       swapsExpected = referenceArray.count - maxA.count
15     }
16     var numberOfSwaps = 0
17     for (index, element) in traversalArray.enumerated() {
18       guard element == needle, referenceArray[index] != needle else { continue }
19       numberOfSwaps += 1
20     }
21     return numberOfSwaps < swapsExpected ? -1 : numberOfSwaps
22   }
23 }
24 
25 extension Array where Element == Int {
26   
27   func mostRepeatingElement() -> (element: Element, count: Int) {
28     var table: [Element : Int] = [:]
29     forEach { element in
30       if let existingElement = table[element] {
31         table[element] = existingElement + 1
32       } else {
33         table[element] = 1
34       }
35     }
36     var maxCount = 0
37     var maxElement = -1
38     table.keys.forEach { key in
39       let count = table[key] ?? 0
40       if count > maxCount {
41         maxCount = count
42         maxElement = key
43       }
44     }
45     return (element: maxElement, count: maxCount)
46   }
47 }

    760ms

     1 class Solution {
 2     func minDominoRotations(_ A: [Int], _ B: [Int]) -> Int {
 3         guard !A.isEmpty && !B.isEmpty else {
 4             return 0
 5         }
 6         
 7         var set = Set([A[0], B[0]])
 8         
 9         for i in 1..<A.count {
10             set.formIntersection([A[i], B[i]])
11         }
12         
13         guard set.count > 0 else { return -1 }
14         
15         var m = Int.max
16         
17         for common in set {
18             let aCount = A.filter { common != $0 }.count
19             let bCount = B.filter { common != $0 }.count
20             m = min(min(aCount, bCount), m)
21         }
22         
23         return m
24     }
25 }

    796ms

     1 class Solution {
 2 func minDominoRotations(_ A: [Int], _ B: [Int]) -> Int {
 3     var common: Set = [1, 2, 3, 4, 5, 6]
 4     for (tile1, tile2) in zip(A, B) {
 5         common = common.intersection(Set([tile1, tile2]))
 6         if common.count == 0 {
 7             return -1
 8         }
 9     }
10 
11     var m: Int = -1
12     for tile in common {
13         m = max(A.filter({$0 == tile}).count, B.filter({$0 == tile}).count, m)
14     }
15     m = A.count - m
16     return m
17   }
18 }
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  • 原文地址：https://www.cnblogs.com/strengthen/p/10504894.html
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