Problem Description

You are given an integer array nums and an integer k. Your task is to append k unique positive integers to nums that do not already appear in nums, such that the sum of these k appended integers is minimized.

The goal is to find and return the minimum possible sum of the k integers that need to be appended.

For example, if nums = [1, 4, 25, 10, 25] and k = 2, the smallest positive integers not in nums are 2 and 3. So you would append these two numbers, and return their sum which is 2 + 3 = 5.

The key constraints are:

• The appended integers must be positive (greater than 0)
• The appended integers must be unique (no duplicates among the k integers)
• The appended integers must not already exist in nums
• You want to minimize the total sum of the appended integers

The problem essentially asks you to find the k smallest positive integers that are not present in the given array and return their sum.

Intuition

To minimize the sum of k integers we need to append, we should choose the smallest possible positive integers that don't exist in nums. Think of it like filling gaps - we want to start from 1 and pick the smallest available numbers.

If nums didn't exist, we'd simply pick 1, 2, 3, ..., k and the sum would be k * (k + 1) / 2. But since some numbers already exist in nums, we need to skip those and find the next smallest available ones.

The key insight is that after sorting nums, we can identify "gaps" between consecutive elements. For any two adjacent elements a and b in the sorted array, all integers in the range [a+1, b-1] are missing from the original array and can be used.

For example, if we have consecutive elements 3 and 7 in the sorted array, then 4, 5, 6 are all missing and available to use. These are exactly the kind of small numbers we want to pick first.

By adding sentinel values 0 at the beginning and a very large number like 2 * 10^9 at the end, we ensure that:

• We can consider positive integers starting from 1 (gap after 0)
• We have enough room to find all k integers even in worst case

As we traverse through pairs of consecutive elements, we greedily take as many integers as possible from each gap (up to k remaining), starting from the smallest gaps first. The sum of consecutive integers from a+1 to a+m can be calculated using the arithmetic series formula: m * (first + last) / 2 = m * (a+1 + a+m) / 2.

This greedy approach works because we're always picking the smallest available integers first due to the sorted order, which guarantees the minimum possible sum.

Learn more about Greedy, Math and Sorting patterns.

Solution Approach

Let's walk through the implementation step by step:

1. Add Sentinel Nodes: We first extend the array with two sentinel values - 0 at the beginning and 2 * 10^9 at the end. This ensures we can consider all positive integers starting from 1, and have enough upper bound to find all k integers.

   nums.extend([0, 2 * 10**9])

2. Sort the Array: We sort the modified array to identify gaps between consecutive elements where missing integers can be found.

   nums.sort()

3. Process Adjacent Pairs: We iterate through each pair of adjacent elements (a, b) in the sorted array using pairwise(). For each pair:

   • Calculate Available Integers: The number of missing integers between a and b is b - a - 1. For example, between 3 and 7, there are 7 - 3 - 1 = 3 missing integers: 4, 5, 6.

   • Determine How Many to Take: We take the minimum of:

     • k (remaining integers we need)
     • b - a - 1 (available integers in this gap)
     • We also ensure it's non-negative using max(0, ...) to handle edge cases

     m = max(0, min(k, b - a - 1))

   • Calculate Sum Using Arithmetic Series: The integers we're taking are a+1, a+2, ..., a+m. This is an arithmetic sequence with:

     • First term: a + 1
     • Last term: a + m
     • Number of terms: m
     • Sum formula: m * (first + last) / 2 = m * (a+1 + a+m) / 2

     ans += (a + 1 + a + m) * m // 2

   • Update Remaining Count: Decrease k by the number of integers we just took.

     k -= m

4. Return Result: Once we've processed all pairs (or k reaches 0), we return the accumulated sum.

The algorithm efficiently finds the k smallest missing positive integers by processing gaps in sorted order, ensuring we always pick the smallest available numbers first, which guarantees the minimum possible sum.

Example Walkthrough

Let's trace through the algorithm with nums = [1, 4, 25, 10, 25] and k = 2.

Step 1: Add Sentinel Values

• Add 0 at the beginning and 2 * 10^9 at the end
• Array becomes: [1, 4, 25, 10, 25, 0, 2000000000]

Step 2: Sort the Array

• After sorting: [0, 1, 4, 10, 25, 25, 2000000000]

Step 3: Process Adjacent Pairs

Initialize: ans = 0, k = 2

Pair 1: (0, 1)

• Gap between 0 and 1: 1 - 0 - 1 = 0 integers available
• Take m = min(2, 0) = 0 integers
• No integers to add, ans = 0, k = 2

Pair 2: (1, 4)

• Gap between 1 and 4: 4 - 1 - 1 = 2 integers available (namely 2, 3)
• Take m = min(2, 2) = 2 integers
• We're taking integers 2 and 3:
  • First integer: 1 + 1 = 2
  • Last integer: 1 + 2 = 3
  • Sum: (2 + 3) * 2 / 2 = 5
• Update: ans = 0 + 5 = 5, k = 2 - 2 = 0

Since k = 0, we've found all needed integers. The algorithm returns 5.

Verification: We appended integers 2 and 3 (which don't exist in the original array), and their sum is 2 + 3 = 5, which is indeed the minimum possible sum.

Solution Implementation

Time and Space Complexity

The time complexity is O(n × log n), where n is the length of the array nums. This is dominated by the sorting operation nums.sort(), which takes O((n+2) × log(n+2)) time. Since we add 2 constant elements to the array before sorting, the complexity simplifies to O(n × log n). The subsequent iteration through pairs using pairwise() takes O(n) time, which doesn't affect the overall complexity.

The space complexity is O(log n). While the nums.extend([0, 2 * 10**9]) operation modifies the list in place and adds only 2 elements (constant space), the sorting algorithm typically uses O(log n) space for its recursive call stack in Python's Timsort implementation. The pairwise() function creates an iterator that uses O(1) additional space.

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

Common Pitfalls

1. Handling Duplicate Elements in the Original Array

The most common pitfall is forgetting that the original array might contain duplicate elements. If we don't handle duplicates properly, we might incorrectly calculate the gaps between numbers.

Pitfall Example:

# Incorrect approach - processes duplicates multiple times
nums = [1, 4, 4, 4, 10]
k = 2
# Without removing duplicates, we might process the gap between 4 and 4 multiple times

Solution: Convert the array to a set before processing to eliminate duplicates:

def minimalKSum(self, nums: List[int], k: int) -> int:
    # Remove duplicates first
    nums = list(set(nums))
    nums.extend([0, 2 * 10**9])
    nums.sort()
    # ... rest of the code remains the same

2. Integer Overflow in Sum Calculation

When calculating the sum of an arithmetic sequence, the intermediate calculations can overflow for large values of k or when dealing with large gaps.

Pitfall Example:

# This could overflow in languages with fixed integer sizes
# or cause precision issues with very large numbers
total_sum += (first_number + last_number) * numbers_to_take // 2

Solution: While Python handles large integers automatically, for better practice and portability:

# Alternative calculation to avoid potential overflow
# Sum = m * a + m * (m + 1) / 2
total_sum += numbers_to_take * current_num + numbers_to_take * (numbers_to_take + 1) // 2

3. Incorrect Boundary Values

Using an insufficient upper bound sentinel value could cause the algorithm to miss valid numbers if k is large and the array has large gaps.

Pitfall Example:

# If nums = [1, 1000000] and k = 1000000
# Using a small upper bound might not provide enough range
nums.extend([0, 10**6])  # Too small upper bound

Solution: Ensure the upper bound is large enough to accommodate all possible k values:

# Use a sufficiently large upper bound based on constraints
# If k can be up to 10^8, ensure upper bound > max(nums) + k
nums.extend([0, max(nums) + k + 1])
# Or use a very large value like 2 * 10^9 as in the original solution

4. Not Checking if k Becomes Zero

Continuing to process pairs after finding all k numbers wastes computation time.

Pitfall Example:

for current_num, next_num in pairwise(nums):
    # Processing continues even after k reaches 0
    numbers_to_take = max(0, min(k, next_num - current_num - 1))
    # ...

Solution: Add an early termination condition:

for current_num, next_num in pairwise(nums):
    if k == 0:
        break
    numbers_to_take = max(0, min(k, next_num - current_num - 1))
    # ... rest of the calculation
    k -= numbers_to_take