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Determine the Minimum Sum of a k-avoiding Array

Number: 2811

Difficulty: Medium

Paid? No

Companies: Infosys

Problem Description

Given two integers n and k, construct an array of n distinct positive integers such that no two distinct elements in the array add up to k. Return the minimum possible sum of any valid k-avoiding array.

Key Insights

  • The array must use distinct positive integers.
  • By choosing smaller numbers first, we minimize the overall sum.
  • A candidate number x must be skipped if its complementary value (k - x) is already selected (since x + (k - x) = k).
  • When x equals k/2 (only possible if k is even), ensure that x is only chosen once because using it twice (if allowed) would form a pair that sums to k.
  • A greedy approach works since we can iterate from 1 upward and select valid candidates until reaching n elements.

Space and Time Complexity

Time Complexity: O(m) where m is the candidate number at which we stop; m will be relatively small given n, k <= 50. Space Complexity: O(n) to store the selected elements.

Solution

The solution uses a greedy algorithm implemented as follows:

  1. Initialize an empty set or list for chosen numbers.
  2. Start iterating candidate numbers from 1 upwards.
  3. For each candidate, check if its complement (k - candidate) is already part of the chosen set.
    • If it is, skip the candidate.
    • Otherwise, add the candidate.
  4. Stop once n numbers have been chosen.
  5. Finally, compute the sum of the selected numbers which is the minimum sum of a valid k-avoiding array.

This greedy algorithm works because by always picking the smallest valid candidate, we guarantee the minimal sum under the given constraints.

Code Solutions

# Python Solution
def minimum_sum_k_avoiding(n: int, k: int) -> int:
    selected = set()  # to quickly check if complement exists
    current = 1
    while len(selected) < n:
        # Check if current's complement has been added.
        if (k - current) in selected:
            current += 1
            continue
        # If current equals its complement and it is already in the set (only possible when current * 2 == k), skip.
        if current * 2 == k and current in selected:
            current += 1
            continue
        selected.add(current)
        current += 1
    return sum(selected)

# Example usage:
print(minimum_sum_k_avoiding(5, 4))  # Output: 18
print(minimum_sum_k_avoiding(2, 6))  # Output: 3
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