README.md

2195. 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

Solutions (Rust)

1. Solution

impl Solution {
    pub fn minimal_k_sum(nums: Vec<i32>, k: i32) -> i64 {
        let mut nums = nums;
        let mut k = k;
        let mut ret = 0;

        nums.push(0);
        nums.push(i32::MAX);
        nums.sort_unstable();

        for i in 1..nums.len() {
            if nums[i] == nums[i - 1] {
                continue;
            } else if nums[i] - nums[i - 1] - 1 < k {
                k -= nums[i] - nums[i - 1] - 1;
                ret +=
                    (nums[i - 1] as i64 + nums[i] as i64) * (nums[i] - nums[i - 1] - 1) as i64 / 2;
            } else {
                ret += (nums[i - 1] as i64 * 2 + 1 + k as i64) * k as i64 / 2;
                break;
            }
        }

        ret
    }
}