Formatted question description: https://leetcode.ca/all/2195.html

# 2195. Append K Integers With Minimal Sum (Medium)

You are given an integer array `nums`

and an integer `k`

. Append `k`

**unique positive** integers that do **not** appear in `nums`

to `nums`

such that the resulting total sum is **minimum**.

Return* the sum of the* `k`

*integers appended to* `nums`

.

**Example 1:**

Input:nums = [1,4,25,10,25], k = 2Output:5Explanation:The two unique positive integers that do not appear in nums which we append are 2 and 3. The resulting sum of nums is 1 + 4 + 25 + 10 + 25 + 2 + 3 = 70, which is the minimum. The sum of the two integers appended is 2 + 3 = 5, so we return 5.

**Example 2:**

Input:nums = [5,6], k = 6Output:25Explanation:The six unique positive integers that do not appear in nums which we append are 1, 2, 3, 4, 7, and 8. The resulting sum of nums is 5 + 6 + 1 + 2 + 3 + 4 + 7 + 8 = 36, which is the minimum. The sum of the six integers appended is 1 + 2 + 3 + 4 + 7 + 8 = 25, so we return 25.

**Constraints:**

`1 <= nums.length <= 10`

^{5}`1 <= nums[i], k <= 10`

^{9}

**Similar Questions**:

- Remove K Digits (Medium)
- Find All Numbers Disappeared in an Array (Easy)
- Kth Missing Positive Number (Easy)

## Solution 1. Sorting

**Intuition**: Sort the array. Traverse from left to right, sum up the missing numbers between `A[i-1]`

and `A[i]`

until we’ve used `k`

missing numbers.

**Algorithm**:

For a given `A[i]`

, the previous number `prev`

is `A[i-1]`

or `0`

if `A[i-1]`

doesn’t exist.

There are `cnt = min(k, max(0, A[i] - prev - 1))`

missing numbers inbetween, i.e. from `prev+1`

to `prev+cnt`

. The sum of these numbers is `(prev+1 + prev+cnt) * cnt / 2`

.

If there are still `k`

missing numbers after traversing the array, the rest of the missing numbers are `A[N-1]+1`

to `A[N-1]+k`

. The sum of them is `(A[N-1]+1 + A[N-1]+k) * k / 2`

.

```
// OJ: https://leetcode.com/problems/append-k-integers-with-minimal-sum/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
long long minimalKSum(vector<int>& A, int k) {
sort(begin(A), end(A));
long ans = 0, N = A.size();
for (int i = 0; i < N && k; ++i) {
long prev = i == 0 ? 0 : A[i - 1]; // the previous number
long cnt = min((long)k, max((long)0, A[i] - prev - 1)); // the count of missing numbers between `prev` and `A[i]`
k -= cnt; // use these `cnt` missing numbers
ans += (long)(prev + 1 + prev + cnt) * cnt / 2; // sum of these `cnt` missing numbers `[prev+1, prev+cnt]`.
}
if (k > 0) ans += ((long)A.back() + 1 + A.back() + k) * k / 2; // If there are still missing numbers, add the sum of numbers`[A.back()+1, A.back()+k]` to answer
return ans;
}
};
```

## Discuss

https://leetcode.com/problems/append-k-integers-with-minimal-sum/discuss/1823619/