Given an array of integers A, a move consists of choosing any A[i], and incrementing it by 1.

Return the least number of moves to make every value in A unique.

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:

  1. 0 <= A.length <= 40000
  2. 0 <= A[i] < 40000
 

Idea 1. Once the array is sorted, the next available number is at least a big as the previous + 1,

Time complexity: O(nlogn)

Space complexity: O(1), unless counting the space needed for sorting

 class Solution {
public int minIncrementForUnique(int[] A) {
if(A.length == 0) {
return 0;
} Arrays.sort(A);
int prev = A[0];
int cnt = 0;
for(int i = 1; i < A.length; ++i) {
if(A[i] <= prev) {
cnt += prev - A[i] + 1;
prev += 1;
}
else {
prev = A[i];
}
}
return cnt;
}
}

网上更简洁的方法,亮瞎眼啊

 class Solution {
public int minIncrementForUnique(int[] A) {
if(A.length == 0) {
return 0;
} Arrays.sort(A);
int nextAvailable = A[0] + 1;
int cnt = 0;
for(int i = 1; i < A.length; ++i) {
cnt += Math.max(nextAvailable - A[i], 0);
nextAvailable = Math.max(nextAvailable, A[i]) + 1;
}
return cnt;
}
}

Idea 2. Map + count, didn't get the limit of 10000 at first, then imagin n= 4, instead of 10000, [4, 4, 4, 4], the max would be max(A) + len(A) -1

Time complexity: O(n + max(A)), n = A.length

Space complexity: O(n + max(A))

 class Solution {
public int minIncrementForUnique(int[] A) {
int n = A.length;
int[] count = new int[80000]; for(int a: A) {
++count[a];
} int result = 0;
int needed = 0;
for(int i = 0; i < 80000; ++i) {
if(count[i] >= 2) {
result -= (count[i]-1) * i;
needed += count[i] - 1;
}
else if(needed > 0 && count[i] == 0) {
--needed;
result += i;
}
} return result;
}
}

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