## DEV Community

Abhishek Chaudhary

Posted on

# Append K Integers With Minimal Sum

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`

SOLUTION:

``````class Solution:
def minimalKSum(self, nums: List[int], k: int) -> int:
nums.sort()
nums.insert(0, 0)
nums.append(float('inf'))
print(nums)
total = 0
n = len(nums)
for i in range(n - 1):
if k >= nums[i + 1] - nums[i] - 1:
if nums[i + 1] - nums[i] > 1:
total += nums[i + 1] * (nums[i + 1] - 1) // 2
total -= nums[i] * (nums[i] + 1) // 2
k -= nums[i + 1] - nums[i] - 1
else:
end = nums[i] + k
beg = nums[i]
total += end * (end + 1) // 2
total -= beg * (beg + 1) // 2
break