# 2195. Append K Integers With Minimal Sum

## Description

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 = 2
Output: 5
Explanation: 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 = 6
Output: 25
Explanation: 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 <= 105
• 1 <= nums[i] <= 109
• 1 <= k <= 108

## Solutions

• class Solution {
public long minimalKSum(int[] nums, int k) {
int[] arr = new int[nums.length + 2];
arr[arr.length - 1] = (int) 2e9;
for (int i = 0; i < nums.length; ++i) {
arr[i + 1] = nums[i];
}
Arrays.sort(arr);
long ans = 0;
for (int i = 1; i < arr.length; ++i) {
int a = arr[i - 1], b = arr[i];
int n = Math.min(k, b - a - 1);
if (n <= 0) {
continue;
}
k -= n;
ans += (long) (a + 1 + a + n) * n / 2;
if (k == 0) {
break;
}
}
return ans;
}
}

• class Solution {
public:
long long minimalKSum(vector<int>& nums, int k) {
nums.push_back(0);
nums.push_back(2e9);
sort(nums.begin(), nums.end());
long long ans = 0;
for (int i = 1; i < nums.size(); ++i) {
int a = nums[i - 1], b = nums[i];
int n = min(k, b - a - 1);
if (n <= 0) continue;
k -= n;
ans += 1ll * (a + 1 + a + n) * n / 2;
if (k == 0) break;
}
return ans;
}
};

• class Solution:
def minimalKSum(self, nums: List[int], k: int) -> int:
nums.append(0)
nums.append(2 * 10**9)
nums.sort()
ans = 0
for a, b in pairwise(nums):
n = min(k, b - a - 1)
if n <= 0:
continue
k -= n
ans += (a + 1 + a + n) * n // 2
if k == 0:
break
return ans

• func minimalKSum(nums []int, k int) int64 {
nums = append(nums, 0, 2e9)
sort.Ints(nums)
ans := 0
for i := 1; i < len(nums); i++ {
a, b := nums[i-1], nums[i]
n := min(k, b-a-1)
if n <= 0 {
continue
}
k -= n
ans += (a + 1 + a + n) * n / 2
if k == 0 {
break
}
}
return int64(ans)
}

• function minimalKSum(nums: number[], k: number): number {
nums.push(...[0, 2 * 10 ** 9]);
nums.sort((a, b) => a - b);
let ans = 0;
for (let i = 0; i < nums.length - 1; ++i) {
const m = Math.max(0, Math.min(k, nums[i + 1] - nums[i] - 1));
ans += Number((BigInt(nums[i] + 1 + nums[i] + m) * BigInt(m)) / BigInt(2));
k -= m;
}
return ans;
}