<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 {
public:
    int minIncrementForUnique(vector<int>& nums) {
        map<int, int> counts;
        for (const int num : nums) {
            if (auto it = counts.find(num); it == counts.end()) {
                counts[num] = 0;
            }
            counts[num]++;
        }
        vector<int> keys;
        for(const auto& pair : counts) {
            keys.push_back(pair.first);
        }
        sort(keys.begin(), keys.end());
        int minIncrement = 0;
        for (int i = 0; i < keys.size(); i++) {
            int max = 1 << 30;
            if (i < keys.size() - 1) {
                max = keys[i+1] - keys[i];
            }
            int count = counts[keys[i]];
            if (count <= max) {
                minIncrement += ((count - 1) * count / 2);
            } else {
                minIncrement += ((max - 1) * max / 2);
                minIncrement += (max * (count - max));
                counts[keys[i+1]] += (count - max);
            }
        }
        return minIncrement;
    }
};

// cost: 6
// (1, 1); (2, 1); (3, 3); (7; 1)

89095