2195. Append K Integers With Minimal Sum

Problem Description

In this problem, you are given an array of integers named nums and an integer k. The goal is to append k unique positive integers to the nums array, which are not already present in nums, so that the total sum after appending these new integers is the smallest possible. The question requires you to find the sum of the k integers that you append to the array.

For example, if nums = [1,4,25] and k = 2, the two unique integers to append could be 2 and 3, as they are the smallest positive integers not present in nums. The sum we are looking for would be 2 + 3 = 5.

It's essential to note that the integers appended to nums must be positive and non-repeating.

Intuition

The key intuition behind the solution is to find the smallest gaps in the sorted nums array where the missing positive integers can be placed. To do this efficiently, we first add two boundaries to the nums array: 0 at the beginning and 2*10^9 at the end (which is sufficiently large to ensure that we can find k unique positive integers).

The array is then sorted to facilitate the process of finding missing numbers between the contiguous elements of nums.

After sorting, we look into the gaps between consecutive elements. Specifically, we check the difference between each pair of neighbouring elements to calculate how many unique positive integers could fit into each gap. We prioritize smaller numbers and only insert as many integers as we need (up to k).

This process is repeated iteratively, reducing k by the number of integers inserted each time until k becomes 0, which means no more numbers need to be appended. The sum of the numbers added within each gap is calculated using the arithmetic progression formula: sum = n/2 * (first_term + last_term), where n is the number of terms to add, first_term is the start of the sequence to append, and last_term is the end of the sequence to append.

The loop breaks immediately once we have appended k integers, and we return the cumulative sum of all the appended integers.

Learn more about Greedy, Math and Sorting patterns.

Solution Approach

The implementation of the solution follows these steps:

1. Append Boundaries to nums: We start by appending the integer 0 at the beginning and a very large integer (here, 2 * 10^9) at the end of the nums array. These boundaries help us handle edge cases where the smallest or largest possible integers must be appended.

2. Sort the nums Array: We sort the nums array to make it easier to go through it sequentially and find gaps between the existing numbers where new integers can be added.

3. Iterate Over Pairs of Elements: Utilizing the pairwise function (new in Python 3.10), we iterate over the nums array in pairs to check for the difference (gap) between each pair of neighbouring elements.

4. Calculate the Number to Append: For each pair (a, b), we calculate the number of unique positive integers that can be appended in the gap (b - a - 1). We limit this to the minimum of the gap size and k as we are only interested in appending up to k numbers.

5. Sum Calculation: Using the formula for the sum of an arithmetic sequence, we calculate the sum of the consecutive integers from (a + 1) to (a + n), where n is the number of integers to include in the particular gap. The sum is calculated by the formula (a + 1 + a + n) * n // 2, taking advantage of the fact that the sum of a sequence can be found by multiplying the average of the first and last term by the number of terms.

6. Update k and the Total Sum: We subtract n from k, reducing the number of integers left to append, and add the sum of the appended integers to ans, which holds the cumulative sum of integers appended so far.

7. Break Condition: As soon as k reaches 0, we break out of the loop since we have appended enough integers to meet the requirement.

The solution approach effectively uses sorting to organize the data and then a sweep algorithm with an arithmetic sequence sum calculation to quickly derive the sum without an explicit enumeration of each number that needs to be appended.

1class Solution:
2    def minimalKSum(self, nums: List[int], k: int) -> int:
3        nums.append(0)
4        nums.append(2 * 10**9)
5        nums.sort()
6        ans = 0
7        for a, b in pairwise(nums):
8            n = min(k, b - a - 1)
9            if n <= 0:
10                continue
11            k -= n
12            ans += (a + 1 + a + n) * n // 2
13            if k == 0:
14                break
15        return ans

The beauty of this solution lies in its efficiency, as it makes a single pass over the sorted array and calculates the sums in a constant number of operations per gap, thus achieving a time complexity that is linearithmic (O(n*log(n))) due to sorting, where n is the number of elements in the initial nums array.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach. Suppose we have nums = [3, 5, 7] and k = 3. We want to append k = 3 unique positive integers to nums such that the total sum is minimized.

1. Append Boundaries to nums: We first add the boundaries, so nums becomes [0, 3, 5, 7, 2000000000].

2. Sort the nums Array: nums is already in sorted order after appending the boundaries.

3. Iterate Over Pairs of Elements: We now consider each pair of neighbouring elements and the gaps between them:

   • Between 0 and 3, the gap is 2 ([1, 2]). We can append two numbers here.
   • Between 3 and 5, no gap exists.
   • Between 5 and 7, no gap exists.
   • Between 7 and 2000000000, there's a large gap, but we will only fill as many numbers as we need to reach k.

4. Calculate the Number to Append:

   • For the pair (0, 3), we calculate the gap is 2 and k = 3. We can use this entire gap, so we append 1 and 2.

5. Sum Calculation:

   • The sum for the gap between (0, 3) will be (1 + 2), which equals 3.

6. Update k and the Total Sum:

   • After filling the gap between (0, 3) with 1 and 2, k reduces from 3 to 1 (k = k - 2), and our running total (the answer ans) becomes 3.

7. Continue Iterating Until k is 0:

   • Since k is not yet 0, we continue to the next gap. However, there are no gaps between 3 and 5, nor between 5 and 7.
   • Finally, we look at the gap between 7 and 2000000000.
   • We have 1 left to append (k = 1).
   • The next number we can append is 8 (the first number after 7).

8. Sum Calculation:

   • We append 8, and our sum (ans) becomes 3 + 8 = 11.

9. Update k and the Total Sum:

   • k is now 0, and we've achieved our goal of appending 3 unique positive integers that were not already in nums.

Therefore, the sum of the appended numbers is 11. This corresponds to appending the numbers 1, 2, and 8 to the array.

This walkthrough demonstrates how the algorithm efficiently calculates the sum of the minimal k unique positive integers that can be appended to nums to achieve the smallest possible total sum.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the minimalKSum method can be split into a few distinct parts:

1. nums.append(0) and nums.append(2 * 10**9) - These operations are constant-time, i.e., O(1).
2. nums.sort() - Sorting the array nums takes O(n log n) time, where n is the length of the array.
3. The for loop and pairwise(nums) - Iterating over the array takes O(n) time. Inside the loop, operations are constant time, such as arithmetic operations and comparison. The pairwise function iterates over the list, generating tuples of adjacent elements. The complexity of generating each pair is O(1), thus pairwise itself does not add to the asymptotic complexity of the loop.
4. Note that the inner operations that include arithmetic and the if condition are all O(1) operations.

Combining these, the time complexity is dominated by the sorting step, which results in an overall time complexity of O(n log n).

Space Complexity

For space complexity:

1. An additional 0 and 2 * 10**9 are appended to the existing list nums, this is O(1) additional space.
2. pairwise(nums) function - This function generates an iterator, which does not create a full copy of nums, and thus takes O(1) space.
3. No additional data structures are used that depend on the size of the input.

Thus, the space complexity of the method is O(1) — constant space, aside from the input list itself.

Learn more about how to find time and space complexity quickly using problem constraints.