<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 getSum(int num) {
        int ans = 0;
        if(num&1) {
            ans = ((num+1)/2) * num;
        }
        else {
            ans = (num+1) * (num/2);
        }
        return ans;
    }

    int minIncrementForUnique(vector<int>& nums) {
        map<int,int> mp;
        for(int num: nums) {
            mp[num]++;
        }
        int ans = 0;
        int temp = 0;
        int prev = -1;
        int gap = 0;
        for(auto x=mp.begin();x!=mp.end();x++) {
            if(prev != -1) {
                gap = x->first - prev-1;
                if(gap > 0) {
                    int numb = min(gap,temp);
                    ans += getSum(numb);
                    temp -= numb;
                }
            }
            ans += ((gap+1)*temp);
            temp += x->second-1;
            prev = x->first;
        }
        ans += getSum(temp);
        return ans;
    }
};
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