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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

• Java
• C++
• Python
• Go

• 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)
  }
  

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