Determine the Minimum Sum of a k-avoiding Array - Solution & Explanation
Problem Statement
You are given two integers, n and k.
An array of distinct positive integers is called a k-avoiding array if there does not exist any pair of distinct elements that sum to k.
Return the minimum possible sum of a k-avoiding array of length n.
Example 1:
Input: n = 5, k = 4 Output: 18 Explanation: Consider the k-avoiding array [1,2,4,5,6], which has a sum of 18. It can be proven that there is no k-avoiding array with a sum less than 18.
Example 2:
Input: n = 2, k = 6 Output: 3 Explanation: We can construct the array [1,2], which has a sum of 3. It can be proven that there is no k-avoiding array with a sum less than 3.
Constraints:
1 <= n, k <= 50
Approach Overview
Problem Overview: You need to build an array of n positive integers such that no two elements add up to k. Among all valid arrays, return the minimum possible sum of its elements.
Approach 1: Incremental Element Selection (Greedy with Set) (Time: O(n), Space: O(n))
Build the array one number at a time starting from 1. For every candidate number x, check whether its complement k - x already exists in the chosen set. If it does, adding x would create a pair whose sum equals k, so skip it. Otherwise include x and continue until you collect n numbers.
This works because choosing the smallest valid numbers always minimizes the total sum. A hash set allows constant‑time complement checks while iterating through natural numbers. The algorithm performs at most a few extra skips, so it remains linear in practice. This approach directly models the greedy rule and is easy to reason about during interviews involving greedy decisions and simple math observations.
Approach 2: Direct Number Generation (Math + Greedy Insight) (Time: O(n), Space: O(1))
Observe the structure of forbidden pairs: x conflicts with k - x. To minimize the sum, you want the smallest possible integers. All numbers from 1 up to k / 2 can safely be chosen because their complements lie in the larger range near k. After selecting those values, the next safe numbers must start from k, since any number between k/2 and k-1 would pair with something already chosen.
So the construction becomes deterministic: first take numbers from 1 upward until reaching k/2 (or until the array already has n elements). If more elements are needed, continue from k, k+1, and so on. This avoids complement conflicts entirely and removes the need for a hash set. The result is a clean arithmetic construction based purely on math reasoning.
Recommended for interviews: The incremental greedy approach demonstrates clear understanding of the constraint by explicitly checking complements. Interviewers often accept it first because it is intuitive and easy to verify. The direct generation method shows deeper pattern recognition and stronger greedy reasoning, making it the most polished final solution.
Approach 1: Approach with Incremental Element Selection
This approach selects the smallest natural numbers step by step until the desired length n is achieved ensuring no pair sums to k. If a number would create a pair summing to k with a number already in the array, it is skipped.
The solution initializes the current number at 1 and constructs the sum while ensuring no two numbers sum to k. If a potential complement exists that already achieves sum k, it skips to the next number.
Complexity
The time complexity is O(n) since we potentially look at up to n numbers, and the space complexity is O(1) as only variables are used.
Approach 2: Approach with Direct Number Generation
This approach leverages mathematical deduction to directly determine the number set providing the minimum sum by considering consecutive numbers unless they have a complement already in the set adding up to k. If a conflict arises, it skips directly to numbers above k.
The strategy involves calculating elements based on their position relative to k/2, directly adjusting if they begin to overlap with complements in the resultant numbering schema without iteration.
Complexity
Time complexity is O(n) due to a single loop through n numbers. Space complexity remains O(1).
Approach 3: Greedy + Simulation
Starting from the positive integer i = 1, we sequentially determine if i can be added to the array. If it can be added, we add i to the array, accumulate it to the answer, and then mark k - i as visited, indicating that k-i cannot be added to the array. We continue this process until the array's length reaches n.
The time complexity is O(n + k), and the space complexity is O(n + k). Where n is the length of the array.
Complexity Comparison
| Approach | Complexity |
|---|---|
| Approach with Incremental Element Selection | The time complexity is |
| Approach with Direct Number Generation | Time complexity is |
| Greedy + Simulation | — |
Detailed Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Incremental Element Selection (Greedy + Set) | O(n) | O(n) | When explaining the logic step‑by‑step or demonstrating greedy reasoning with complement checks |
| Direct Number Generation (Math Insight) | O(n) | O(1) | Best optimized solution once the pattern of forbidden pairs around k is recognized |
Video Solution
Leetcode Weekly contest 359 - Medium- Determine the Minimum Sum of a k-avoiding Array • Prakhar Agrawal • 603 views views
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