<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) {
        int n = nums.size();
        sort(nums.begin(), nums.end());
        unordered_set<int> st(nums.begin(), nums.end());

        // Create a Min heap which stores element from start to end
        priority_queue<int, vector<int>, greater<int>> minHeap;
        for (int i=nums[0]; i<=nums[n-1]; i++) {
            if (st.find(i) == st.end()) {
                minHeap.push(i);
            }
        }

        int ele = nums[n-1] + 1, cost = 0;
        for (int i=1; i<n; i++) {
            if (nums[i] == nums[i-1]) {
                while (minHeap.size() > 0 && minHeap.top() < nums[i]) {
                    minHeap.pop();
                }
                if (minHeap.size() <= 0) {
                    minHeap.push(ele);
                    ele++;
                }
                cost += minHeap.top() - nums[i];
                minHeap.pop();
            }
        }

        return cost;
    }
};
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