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Formatted question description: https://leetcode.ca/all/2195.html

2195. Append K Integers With Minimal Sum (Medium)

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], k <= 109

Similar Questions:

• Remove K Digits (Medium)
• Find All Numbers Disappeared in an Array (Easy)
• Kth Missing Positive Number (Easy)

Solution 1. Sorting

Intuition: Sort the array. Traverse from left to right, sum up the missing numbers between A[i-1] and A[i] until we’ve used k missing numbers.

Algorithm:

For a given A[i], the previous number prev is A[i-1] or 0 if A[i-1] doesn’t exist.

There are cnt = min(k, max(0, A[i] - prev - 1)) missing numbers inbetween, i.e. from prev+1 to prev+cnt. The sum of these numbers is (prev+1 + prev+cnt) * cnt / 2.

If there are still k missing numbers after traversing the array, the rest of the missing numbers are A[N-1]+1 to A[N-1]+k. The sum of them is (A[N-1]+1 + A[N-1]+k) * k / 2.

• C++
• Python
• Java
• Go

• // OJ: https://leetcode.com/problems/append-k-integers-with-minimal-sum/
  // Time: O(NlogN)
  // Space: O(1)
  class Solution {
  public:
      long long minimalKSum(vector<int>& A, int k) {
          sort(begin(A), end(A));
          long ans = 0, N = A.size();
          for (int i = 0; i < N && k; ++i) {
              long prev = i == 0 ? 0 : A[i - 1]; // the previous number
              long cnt = min((long)k, max((long)0, A[i] - prev - 1)); // the count of missing numbers between `prev` and `A[i]`
              k -= cnt; // use these `cnt` missing numbers
              ans += (long)(prev + 1 + prev + cnt) * cnt / 2; // sum of these `cnt` missing numbers `[prev+1, prev+cnt]`.
          }
          if (k > 0) ans += ((long)A.back() + 1 + A.back() + k) * k / 2; // If there are still missing numbers, add the sum of numbers`[A.back()+1, A.back()+k]` to answer
          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
  
  ############
  
  # 2195. Append K Integers With Minimal Sum
  # https://leetcode.com/problems/append-k-integers-with-minimal-sum/
  
  class Solution:
      def minimalKSum(self, nums: List[int], k: int) -> int:
          res = 0
          nums.sort()
          prev = 0
          
          for i, x in enumerate(nums):
              if (i == 0 and x == 1) or (i > 0 and x == nums[i - 1]): 
                  prev = x
                  continue
              
              xx = x - 1
              currentDelta = min(k, xx - prev)
              if k <= currentDelta:
                  for j in range(prev + 1, prev + k + 1):
                      res += j
                  
                  return res
              else:
                  delta = xx
                  
              if currentDelta == 0: 
                  prev = x
                  continue
                  
              upper = delta * (delta + 1) // 2
              lower = prev * (prev + 1) // 2
              
              res += upper - lower
              
              if currentDelta >= k:
                  return res
              k -= currentDelta
              prev = x
  
          if k > 0:
              delta = prev + k
              upper = delta * (delta + 1) // 2
              lower = prev * (prev + 1) // 2
              res += upper - lower
          
          return res
  
  

• 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;
      }
  }
  

• 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)
  }
  
  func min(a, b int) int {
  	if a < b {
  		return a
  	}
  	return b
  }
  

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https://leetcode.com/problems/append-k-integers-with-minimal-sum/discuss/1823619/

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