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:

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.

  • // 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;
        }
    };
    
  • // Todo
    
  • # 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
    
    

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