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

<p>Return <em>the minimum number of moves to make every value in </em><code>nums</code><em> <strong>unique</strong></em>.</p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,2]
<strong>Output:</strong> 1
<strong>Explanation:</strong> After 1 move, the array could be [1, 2, 3].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,2,1,2,1,7]
<strong>Output:</strong> 6
<strong>Explanation:</strong> After 6 moves, the array could be [3, 4, 1, 2, 5, 7].
It can be shown that it is impossible for the array to have all unique values with 5 or less moves.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def minIncrementForUnique(self, nums: List[int]) -> int:
        # Через карту
        freqs = [0] * (len(nums) + max(nums) + 1)

        # Заполняем частоты
        for num in nums:
            freqs[num] += 1
        
        to_spend = 0
        #print(freqs)
        # Считаем нужные сдвиги
        for i in range(len(freqs)):
            #print(i, freqs)
            if freqs[i] <= 1:
                continue
            freqs[i+1] += freqs[i] - 1
            to_spend += freqs[i] - 1
        
        return to_spend


        # Мой способ через сортировку, продолжение вот этой изи задачи:
        # https://leetcode.com/problems/minimum-operations-to-make-the-array-increasing/
        nums.sort()
        to_spend = 0
        prev = -float('inf')
        for num in nums:
            if num <= prev:
                to_spend += prev + 1 - num
                num = prev + 1
            prev = num
        return to_spend
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