LeetCode 2829 - Determine the Minimum Sum of a k-avoiding Array

The problem asks us to construct an array of n distinct positive integers such that no two elements in the array sum to a given integer k. This type of array is called k-avoiding.

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LeetCode Problem 2829

Difficulty: 🟡 Medium
Topics: Math, Greedy

Solution

Problem Understanding

The problem asks us to construct an array of n distinct positive integers such that no two elements in the array sum to a given integer k. This type of array is called k-avoiding. Among all possible k-avoiding arrays of length n, we are asked to find the minimum possible sum of its elements.

The input consists of two integers: n represents the length of the array, and k is the forbidden sum for any two distinct elements. The output is a single integer representing the minimum sum.

Constraints tell us that both n and k are relatively small (1 <= n, k <= 50), so solutions that examine numbers in a straightforward way are feasible. We need to ensure all array elements are positive integers and distinct, and that no pair adds up to k.

Important edge cases include:

  • n = 1, where the array has only one element and any number is valid.
  • k being very small relative to n, which may force us to skip some small integers to avoid creating forbidden pairs.
  • Arrays where n is large and k is just above n, which may require careful selection of numbers to maintain minimal sum.

Approaches

Brute Force

A brute-force approach would generate all possible arrays of length n with distinct positive integers, check if they are k-avoiding, and track the minimum sum. This approach is correct because it explores every possible valid configuration. However, it is extremely inefficient because the number of possible arrays grows combinatorially. Even for n = 10, the number of candidate arrays becomes astronomical.

Optimal Approach

The key insight is that to minimize the sum, we want to pick the smallest positive integers possible while avoiding pairs that sum to k. If we start from 1 and go upwards, we can skip integers that would pair with a previously selected number to form k. This guarantees minimal sum because any larger numbers would only increase the total sum.

Specifically, if x is already in the array, we cannot include k - x in the array. Therefore, we iterate from 1 upward, adding numbers to the array if k - x is not already chosen. We continue until we have n elements.

Approach Time Complexity Space Complexity Notes
Brute Force O((max_val choose n) * n) O(n) Generate all possible arrays and filter k-avoiding ones
Optimal O(n*k) O(k) Greedily choose smallest numbers, skip numbers that would form k

Algorithm Walkthrough

  1. Initialize an empty set to keep track of numbers already included in the array.
  2. Start from integer 1 and iterate upwards.
  3. For each integer i, check if k - i is already in the set. If it is, skip i because including it would form a forbidden pair.
  4. If k - i is not in the set, add i to the set and increment a counter of chosen elements.
  5. Repeat steps 2-4 until we have chosen n numbers.
  6. Compute the sum of all numbers in the set and return it.

Why it works: The algorithm ensures that the smallest integers are always chosen unless they would violate the k-avoiding property. Since we process numbers in ascending order and skip only those that form forbidden pairs, the resulting sum is guaranteed to be minimal.

Python Solution

class Solution:
    def minimumSum(self, n: int, k: int) -> int:
        chosen = set()
        current = 1
        while len(chosen) < n:
            if k - current not in chosen:
                chosen.add(current)
            current += 1
        return sum(chosen)

Explanation: We maintain a set of chosen numbers. We iterate from 1 upwards and only add a number if it does not form a sum of k with any already chosen number. Once we have n numbers, we compute the sum. The algorithm is direct and uses a greedy approach to minimize the sum while respecting the k-avoiding condition.

Go Solution

func minimumSum(n int, k int) int {
    chosen := make(map[int]bool)
    sum := 0
    current := 1
    count := 0
    for count < n {
        if !chosen[k-current] {
            chosen[current] = true
            sum += current
            count++
        }
        current++
    }
    return sum
}

Explanation: The Go version mirrors the Python logic. A map tracks chosen numbers. We iterate from 1 upwards, skip numbers that would violate the k-avoiding property, and accumulate the sum directly. Go requires explicit counting and map usage instead of a set.

Worked Examples

Example 1: n = 5, k = 4

Step Current Chosen Set Comment
1 1 {1} 1 is added
2 2 {1,2} 2 does not pair with 1 to form 4
3 3 {1,2} 3 is skipped because 1 + 3 = 4
4 4 {1,2,4} 4 is added
5 5 {1,2,4,5} 5 is added
6 6 {1,2,4,5,6} 6 is added, count = 5

Sum = 1+2+4+5+6 = 18

Example 2: n = 2, k = 6

Step Current Chosen Set Comment
1 1 {1} 1 is added
2 2 {1,2} 2 is added, count = 2

Sum = 1+2 = 3

Complexity Analysis

Measure Complexity Explanation
Time O(n*k) In the worst case, we may iterate up to n*k to collect n numbers because some numbers are skipped to avoid pairs
Space O(k) We use a set or map to track numbers already chosen up to k

The time complexity is acceptable for constraints n, k <= 50, as n*k is at most 2500 operations.

Test Cases

# Provided examples
assert Solution().minimumSum(5, 4) == 18  # Example 1
assert Solution().minimumSum(2, 6) == 3   # Example 2

# Edge cases
assert Solution().minimumSum(1, 1) == 1   # Single element, trivial
assert Solution().minimumSum(3, 3) == 8   # Must skip 2 because 1+2=3, picks 1,3,4
assert Solution().minimumSum(4, 7) == 10  # Avoid pairs that sum to 7
assert Solution().minimumSum(50, 50) > 0  # Max input size, should complete
Test Why
n=5, k=4 Validates normal case with small n and k
n=2, k=6 Small array, sum minimal
n=1, k=1 Edge case with single element
n=3, k=3 Checks correct skipping of numbers that form k
n=4, k=7 Checks multiple skips needed
n=50, k=50 Stress test on max input size

Edge Cases

Single element array: When n = 1, the array can contain any positive integer. The solution correctly chooses 1 because it is the smallest positive integer and automatically k-avoiding.

k smaller than n: If k is smaller than some of the first n numbers, the algorithm may need to skip certain numbers to avoid forming k. By iterating upwards and checking k - current in the set, it ensures no pair violates the condition.

Large n close to k: When n is large relative to k, multiple small numbers may be blocked due to forbidden sums. The greedy selection ensures that we always choose the next smallest valid number, guaranteeing a minimal sum without violating the k-avoiding property.