#2195

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Append K Integers With Minimal Sum

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In this problem, you need to append k unique positive integers to a given array nums such that the sum is minimized.

Problem Statement

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.

Examples

Example 1

Input: nums = [1,4,25,10,25], k = 2

Output: 5

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

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

Solution Approach

Greedy Approach

To minimize the sum, we can use a greedy approach where we append the smallest integers that are not in nums. Start with 1 and keep adding the next smallest integer not in nums.

Efficient Set Lookup

To efficiently track which integers have already been used, maintain a set of numbers present in nums. This allows you to quickly check if a number is missing and append it.

Sum Calculation

The final sum is the sum of the k smallest numbers added to nums. Keep track of the sum while appending to minimize computation overhead.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n + k) where n is the length of nums and k is the number of integers to be appended. Space complexity is O(n) due to the storage of nums in a set for fast lookup.

What Interviewers Usually Probe

Check if the candidate understands how to apply greedy strategies to minimize sums.
Evaluate the candidate's ability to optimize using set-based lookups.
Look for knowledge of efficient ways to append elements while minimizing time complexity.

Common Pitfalls or Variants

Common pitfalls

Not considering the uniqueness constraint of the appended integers.
Incorrectly calculating the sum, failing to track only the sum of the appended numbers.
Inefficient lookups or appending methods that degrade performance.

Follow-up variants

Appending a larger number of integers, testing how the algorithm scales with larger k.
Variation where nums already contains most of the integers, requiring careful selection of missing integers.
Approach using sorting or other methods instead of greedy choice to check optimization.

FAQ

How does the greedy approach work in 'Append K Integers With Minimal Sum'?

The greedy approach works by always selecting the smallest integers that are missing from nums, ensuring the sum of appended integers is minimized.

What is the time complexity of the solution?

The time complexity is O(n + k), where n is the length of nums and k is the number of integers to append.

How do you efficiently track which integers are missing from nums?

You can use a set to store the elements of nums, allowing for quick lookup to determine missing integers.

Can this problem be solved using sorting?

Sorting can be used, but a greedy approach with set-based lookups is more efficient for minimizing the sum.

What is the main pattern used in this problem?

The main pattern is greedy choice with invariant validation, ensuring the minimum sum by always choosing the smallest missing integers.

terminal

Solution

Solution 1: Sorting + Greedy + Mathematics

We can add two sentinel nodes to the array, which are $0$ and $2 \times 10^9$.

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class Solution:
    def minimalKSum(self, nums: List[int], k: int) -> int:
        nums.extend([0, 2 * 10**9])
        nums.sort()
        ans = 0
        for a, b in pairwise(nums):
            m = max(0, min(k, b - a - 1))
            ans += (a + 1 + a + m) * m // 2
            k -= m
        return ans

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