Append K Integers With Minimal Sum

// Accepted solution for LeetCode #2195: Append K Integers With Minimal Sum
class Solution {
    public long minimalKSum(int[] nums, int k) {
        int n = nums.length;
        int[] arr = new int[n + 2];
        arr[1] = 2 * 1000000000;
        System.arraycopy(nums, 0, arr, 2, n);
        Arrays.sort(arr);
        long ans = 0;
        for (int i = 0; i < n + 1 && k > 0; ++i) {
            int m = Math.max(0, Math.min(k, arr[i + 1] - arr[i] - 1));
            ans += (arr[i] + 1L + arr[i] + m) * m / 2;
            k -= m;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2195: Append K Integers With Minimal Sum
func minimalKSum(nums []int, k int) (ans int64) {
	nums = append(nums, []int{0, 2e9}...)
	sort.Ints(nums)
	for i, b := range nums[1:] {
		a := nums[i]
		m := max(0, min(k, b-a-1))
		ans += int64(a+1+a+m) * int64(m) / 2
		k -= m
	}
	return ans
}

# Accepted solution for LeetCode #2195: Append K Integers With Minimal Sum
class Solution:
    def minimalKSum(self, nums: List[int], k: int) -> int:
        nums.extend([0, 2 * 10**9])
        nums.sort()
        ans = 0
        for a, b in pairwise(nums):
            m = max(0, min(k, b - a - 1))
            ans += (a + 1 + a + m) * m // 2
            k -= m
        return ans

// Accepted solution for LeetCode #2195: Append K Integers With Minimal Sum
/**
 * [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 <= 10^5
 * 	1 <= nums[i] <= 10^9
 * 	1 <= k <= 10^8
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/append-k-integers-with-minimal-sum/
// discuss: https://leetcode.com/problems/append-k-integers-with-minimal-sum/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn minimal_k_sum(nums: Vec<i32>, k: i32) -> i64 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2195_example_1() {
        let nums = vec![1, 4, 25, 10, 25];
        let k = 2;

        let result = 5;

        assert_eq!(Solution::minimal_k_sum(nums, k), result);
    }

    #[test]
    #[ignore]
    fn test_2195_example_2() {
        let nums = vec![1, 4, 25, 10, 25];
        let k = 2;

        let result = 5;

        assert_eq!(Solution::minimal_k_sum(nums, k), result);
    }
}

// Accepted solution for LeetCode #2195: Append K Integers With Minimal Sum
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;
}