Make Unique - Greedy Approach

Sort the array. Scan left to right. Each element must be at least one more than the previous.

If $arr[i] \leq arr[i-1]$, increment $arr[i]$ to $arr[i-1] + 1$. The cost is the difference.

Here's greedy because you fix each violation with the minimum possible increment.

You cannot do better: the element must be unique and at least as large as it was. Why this is optimal: each element must be at least its original value.

If $arr[i] \leq arr[i-1]$, the smallest valid value is $arr[i-1] + 1$. Choosing anything larger would waste increments. The greedy choice minimizes cost at each step.