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
Solution 1: Sorting + Greedy + Mathematics
We can add two sentinel nodes to the array, which are $0$ and $2 \times 10^9$.
Then we sort the array. For any two adjacent elements $a$ and $b$ in the array, the integers in the interval $[a+1, b-1]$ do not appear in the array, and we can add these integers to the array.
Therefore, we traverse the adjacent element pairs $(a, b)$ in the array from small to large. For each adjacent element pair, we calculate the number of integers $m$ in the interval $[a+1, b-1]$. The sum of these $m$ integers is $\frac{m \times (a+1 + a+m)}{2}$. We add this sum to the answer and subtract $m$ from $k$. If $k$ is reduced to $0$, we can stop the traversal and return the answer.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Where $n$ is the length of the array.
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