<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:
        deq = Deque([])
        nums_set = set(nums)
        # freq = Counter(nums)
        #1, 1, 2, 2, 3, 3, 7    

        #1, 2, 3    
        nums.sort()
        # print(nums)
        ans = 0
        prev_v = nums[0]
        for i in range(1, len(nums)):
            # print(deq)
            c_v = nums[i]
            if c_v == prev_v:
                deq.append(c_v)
            elif c_v - prev_v > 1:
                n_prev = prev_v + 1
                while len(deq) > 0 and c_v - n_prev >= 1:
                    pop_v = deq.popleft()
                    ans += n_prev - pop_v
                    n_prev += 1
            prev_v = c_v
        
        n_prev = nums[-1] + 1
        while len(deq) > 0:
            pop_v = deq.popleft()
            ans += n_prev - pop_v
            n_prev += 1
        return ans



        # if curr_v is same as prev
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