Minimum increment to make array unique

class Solution:
    def minIncrementForUnique(self, nums: List[int]) -> int:
        count = [0] * int(2e5)
        for x in nums:
            count[x] += 1

        ans = taken = 0
        for x in range(int(2e5)):
            if count[x] >= 2:
                taken += count[x] - 1
                ans -= x * (count[x] - 1)
            elif taken > 0 and count[x] == 0:
                taken -= 1
                ans += x

        return ans

Minimum Increment to Make Array Unique

You are given an integer array nums. In one move, you can pick an index i where 0 <= i < nums.length and increment nums[i] by 1.

Return the minimum number of moves to make every value in nums unique.

The test cases are generated so that the answer fits in a 32-bit integer.

Example 1:

Input: nums = [1,2,2]
Output: 1
Explanation: After 1 move, the array could be [1, 2, 3].

Example 2:

Input: nums = [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.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁵