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★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★ Given an array of integers A, a move consists of choosing any Return the least number of moves to make every value in Example 1: Input: [1,2,2]
Output: 1
Explanation: After 1 move, the array could be [1, 2, 3].
Example 2: Input: [3,2,1,2,1,7]
Output: 6
Explanation: After 6 moves, the array could be [3, 4, 1, 2, 5, 7].
It can be shown with 5 or less moves that it is impossible for the array to have all unique values.
Note:
给定整数数组 A,每次 move 操作将会选择任意 返回使 示例 1: 输入:[1,2,2] 输出:1 解释:经过一次 move 操作,数组将变为 [1, 2, 3]。 示例 2: 输入:[3,2,1,2,1,7] 输出:6 解释:经过 6 次 move 操作,数组将变为 [3, 4, 1, 2, 5, 7]。 可以看出 5 次或 5 次以下的 move 操作是不能让数组的每个值唯一的。 提示:
424ms 1 class Solution { 2 func minIncrementForUnique(_ A: [Int]) -> Int { 3 var dp = [Int](repeating: 0, count: 40001) 4 for a in A { 5 dp[a] += 1 6 } 7 var ans = 0 8 for i in 0 ..< 40000 { 9 let a = dp[i] 10 guard a > 1 else { 11 continue 12 } 13 dp[i] = 1 14 ans += a-1 15 dp[i+1] += a-1 16 } 17 if dp[40000] > 1 { 18 let a = dp[40000] - 1 19 for i in 1 ... a { 20 ans += i 21 } 22 } 23 return ans 24 } 25 } 440ms 1 class Solution { 2 func minIncrementForUnique(_ A: [Int]) -> Int { 3 guard A.count > 1 else { 4 return 0 5 } 6 let upperBound = 80001 7 var freqArray = Array<Int>(repeating: 0, count: upperBound) 8 for val in A { 9 freqArray[val] += 1 10 } 11 12 var ans = 0 13 var numsOfDupTaken = 0 14 for i in 0 ..< upperBound { 15 let count = freqArray[i] 16 if count > 1 { 17 numsOfDupTaken += count - 1 18 ans -= i * (count - 1) 19 } else if numsOfDupTaken > 0 && freqArray[i] == 0 { 20 ans += i 21 numsOfDupTaken -= 1 22 } 23 } 24 25 return ans 26 } 27 } 460ms 1 class Solution { 2 func minIncrementForUnique(_ A: [Int]) -> Int { 3 var count = [Int](repeating: 0, count:100000) 4 for x in A { 5 count[x] += 1 6 } 7 var ans = 0, taken = 0 8 for x in 0..<100000 { 9 if (count[x] >= 2) { 10 taken += count[x] - 1; 11 ans -= x * (count[x] - 1); 12 } 13 else if (taken > 0 && count[x] == 0) { 14 taken -= 1; 15 ans += x; 16 } 17 } 18 return ans; 19 } 20 } 476ms 1 class Solution { 2 func minIncrementForUnique(_ A: [Int]) -> Int { 3 guard A.count > 1 else { 4 return 0 5 } 6 let sortedA = A.sorted() 7 var result = 0, prev = sortedA[0] 8 for i in 1..<sortedA.count { 9 if sortedA[i] <= prev { 10 prev += 1 11 result += (prev - sortedA[i]) 12 } else { 13 prev = sortedA[i] 14 } 15 } 16 return result 17 } 18 } 484ms 1 class Solution { 2 func minIncrementForUnique(_ A: [Int]) -> Int { 3 guard A.count > 1 else { 4 return 0 5 } 6 var sortedA = A.sorted() 7 var counter = 0 8 for i in 1...sortedA.count - 1 { 9 if sortedA[i] <= sortedA[i-1] { 10 counter += sortedA[i-1] + 1 - sortedA[i] 11 sortedA[i] = sortedA[i-1] + 1 12 } 13 } 14 return counter 15 } 16 } 508ms 1 class Solution { 2 func minIncrementForUnique(_ A: [Int]) -> Int { 3 var arr:[Int] = A.sorted(by:<) 4 var r:Int = -1 5 var ret:Int = 0 6 for i in 0..<arr.count 7 { 8 var to:Int = max(r,arr[i]) 9 ret += abs(to - arr[i]) 10 r = to + 1 11 } 12 return ret 13 } 14 } 552ms 1 class Solution { 2 func minIncrementForUnique(_ A: [Int]) -> Int { 3 4 var counts = A.reduce(into: [:]) { dict, val in 5 dict[val, default: 0] += 1 6 } 7 8 var increments = 0 9 var nextPossibleSlot = -1 10 for key in counts.keys.sorted() { 11 var keyCount = counts[key]! 12 while keyCount > 1 { 13 nextPossibleSlot = max(key, nextPossibleSlot) + 1 14 while counts[nextPossibleSlot] != nil { 15 nextPossibleSlot += 1 16 } 17 //Now nextPossibleSlot is available 18 counts[nextPossibleSlot] = 1 //not actually necessary 19 increments += nextPossibleSlot - key 20 keyCount -= 1 21 } 22 } 23 24 return increments 25 } 26 } 592ms 1 class Solution { 2 func minIncrementForUnique(_ A: [Int]) -> Int { 3 guard A.count > 0 else { return 0 } 4 guard A.count > 1 else { return 0 } 5 6 var sortA = A.sorted() 7 var mapA: [Int: Int] = [:] 8 for a in A { 9 mapA[a] = (mapA[a] ?? 0) + 1 10 } 11 var res = 0 12 var emptIter = sortA.first! - 1 13 for (index, a) in sortA.enumerated() { 14 if a > emptIter + 1 { 15 emptIter = emptIter + 1 16 break 17 } 18 emptIter = a 19 } 20 if emptIter == sortA.last! { 21 emptIter += 1 22 } 23 24 //print(sortA, emptIter) 25 var prev = -1 26 for (index, a) in sortA.enumerated() { 27 if prev == a { 28 if emptIter <= a { 29 emptIter = a 30 while mapA[emptIter] != nil { 31 emptIter += 1 32 } 33 } 34 res += (emptIter - a) 35 emptIter += 1 36 while mapA[emptIter] != nil { 37 emptIter += 1 38 } 39 //print(res, emptIter) 40 } 41 prev = a 42 } 43 44 return res 45 } 46 } 608ms 1 class Solution { 2 3 func minIncrementForUnique(_ arr: [Int]) -> Int { 4 guard arr.count > 1 else { 5 return 0 6 } 7 8 let arr = arr.sorted() 9 var nextHighestValue = arr[0] + 1 10 var result = 0 11 for i in 1..<arr.count { 12 let value = arr[i] 13 result += max(nextHighestValue - value, 0) 14 nextHighestValue = max(value, nextHighestValue) + 1 15 } 16 17 return result 18 } 19 } 612ms 1 class Solution { 2 func minIncrementForUnique(_ A: [Int]) -> Int { 3 4 if A.count <= 1 { 5 return 0 6 } 7 let a = A.sorted() 8 var ind0 = 1 9 var curMin = a[0] 10 var counter = 0 11 while ind0 < a.count { 12 if a[ind0] <= curMin { 13 curMin += 1 14 counter += curMin - a[ind0] 15 } else { 16 curMin = max(a[ind0], curMin) 17 } 18 ind0 += 1 19 } 20 return counter 21 } 22 } 772ms 1 class Solution { 2 func minIncrementForUnique(_ A: [Int]) -> Int { 3 var dict: [Int:Node] = [:] 4 var count = 0 5 6 for num in A { 7 if dict[num] == nil { 8 let node = Node(num + 1) 9 dict[num] = node 10 } else { 11 // Duplicate 12 var nextVal = num 13 let newNode = Node(nextVal + 1) 14 while dict[nextVal] != nil { 15 if let node = dict[nextVal] { 16 dict[nextVal] = newNode 17 nextVal = node.val 18 } 19 } 20 dict[nextVal] = newNode 21 newNode.val = nextVal + 1 22 count += (nextVal - num) 23 } 24 } 25 26 return count 27 } 28 29 class Node { 30 var val: Int 31 init(_ a: Int) { 32 self.val = a 33 } 34 } 35 }
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