Determine the Minimum Sum of a k-avoiding Array

To construct a k-avoiding array of length ( n ) with the minimum sum, start by placing the smallest available integers in the array. Keep in mind that the sum of any two numbers in the array should not equal ( k ).

1. Initialize an empty list called result and a variable sum_result set to 0.
2. Initialize a variable current_number set to 1.
3. Loop ( n ) times to fill the result array.
4. In each iteration, check if current_number added to any other number in the result array sums up to ( k ).
5. If it does, skip this number and check the next one.
6. If it does not, append it to result and add its value to sum_result.

Here’s the code:

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from typing import List

class Solution:
    def minimumSum(self, n: int, k: int) -> int:
        # Step 1
        result = []
        sum_result = 0

        # Step 2
        current_number = 1

        # Step 3
        for _ in range(n):
            while True:
                # Step 4
                if all(current_number + x != k for x in result):
                    # Step 5 and 6
                    result.append(current_number)
                    sum_result += current_number
                    current_number += 1
                    break
                current_number += 1

        return sum_result

Explanation

• We start by setting current_number to 1 because we are looking for positive integers.
• We then loop ( n ) times to construct our result array.
• Inside each loop, we check if adding current_number to any of the numbers already in result would sum to ( k ).
• If it does, we skip this number and try the next one.
• If it doesn’t, we add it to result and move on to the next iteration of the loop.

The time complexity is ( O(n^2) ) because for each of the ( n ) numbers in result, we potentially have to check all ( n ) numbers already in the array. This is acceptable for ( n \leq 50 ).