Minimum Increment to Make Array Unique LT945

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;
     }
 }