## Algorithm

Problem Name: 945. Minimum Increment to Make Array Unique

You are given an integer array `nums`. In one move, you can pick an index `i` where `0 <= i < nums.length` and increment `nums[i]` by `1`.

Return the minimum number of moves to make every value in `nums` unique.

The test cases are generated so that the answer fits in a 32-bit integer.

Example 1:

```Input: nums = [1,2,2]
Output: 1
Explanation: After 1 move, the array could be [1, 2, 3].
```

Example 2:

```Input: nums = [3,2,1,2,1,7]
Output: 6
Explanation: After 6 moves, the array could be [3, 4, 1, 2, 5, 7].
It can be shown with 5 or less moves that it is impossible for the array to have all unique values.
```

Constraints:

• `1 <= nums.length <= 105`
• `0 <= nums[i] <= 105`

## Code Examples

### #1 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const minIncrementForUnique = function(nums) {
const seen = new Set()
const queue = []
let res = 0
for(const e of nums) {
else queue.push(e)
}
queue.sort((a, b) => b - a)
for(let i = 0; i <= 1e5 || queue.length; i++) {
if(!seen.has(i> && i > last(queue)) {
res += i - queue.pop()
}
}

return res

function last(arr) {
return arr[arr.length - 1]
}
};
``````
Copy The Code &

Input

cmd
nums = [1,2,2]

Output

cmd
1

### #2 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def minIncrementForUnique(self, A):
st, used, move = set(A), set(), 0
heapq.heapify(A)
empty = [i for i in range(80000) if i not in st][::-1] if A else []
while A:
num = heapq.heappop(A)
if num not in used:
else:
while empty[-1] < num:
empty.pop()
move += empty[-1] - num
heapq.heappush(A, empty.pop())
return move
``````
Copy The Code &

Input

cmd
nums = [1,2,2]

Output

cmd
1