How do you solve Determine the Minimum Sum of a k-avoiding Array on LeetCode?

Determine the Minimum Sum of a k-avoiding Array (LeetCode #2829) can be solved using Brute Force. The optimal approach is Brute Force with O(n²) time complexity and O(n) space complexity. Key insight: The brute-force approach involves generating all possible arrays of length n and checking if they are k-avoiding. This method is straightforward but inefficient due to the large number of combinations.

What is the time complexity of Determine the Minimum Sum of a k-avoiding Array?

The optimal solution for Determine the Minimum Sum of a k-avoiding Array runs in O(n²) time and uses O(n) space. The time complexity is O(n²) because for each number added, we check if its complement (k - num) is already in the array, which can take O(n) time in the worst case.

What algorithm or data structure is used to solve Determine the Minimum Sum of a k-avoiding Array?

Determine the Minimum Sum of a k-avoiding Array uses the following concepts: Math, Greedy. The recommended patterns to study are: .

Determine the Minimum Sum of a k-avoiding Array — LeetCode #2829 (Medium)

Tags: Math, Greedy

Related patterns:

Brute Force approach

Time complexity: O(n²). Space complexity: O(n).

The brute-force approach involves generating all possible arrays of length n and checking if they are k-avoiding. This method is straightforward but inefficient due to the large number of combinations.

The time complexity is O(n²) because for each number added, we check if its complement (k - num) is already in the array, which can take O(n) time in the worst case.

Step 1: Start with an empty array and initialize a sum variable to 0.
Step 2: Iterate through the smallest positive integers, adding them to the array if they do not create a pair that sums to k.
Step 3: Continue until the array reaches the desired length n.
Step 4: Return the sum of the elements in the array.

1. Start with num = 1, result = [], current_sum = 0. 2. num = 1: k - num = 3, not in result, add 1. result = [1], current_sum = 1. 3. num = 2: k - num = 2, not in result, add 2. result = [1, 2], current_sum = 3. 4. num = 3: k - num = 1, in result, skip. 5. num = 4: k - num = 0, not in result, add 4. result = [1, 2, 4], current_sum = 7. 6. num = 5: k - num = -1, not in result, add 5. result = [1, 2, 4, 5], current_sum = 12. 7. num = 6: k - num = -2, not in result, add 6. result = [1, 2, 4, 5, 6], current_sum = 18.

#2829

Determine the Minimum Sum of a k-avoiding Array

Medium

Math↗Greedy↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(n)

⚙️

Algorithm

4 steps

1Step 1: Start with an empty array and initialize a sum variable to 0.
2Step 2: Iterate through the smallest positive integers, adding them to the array if they do not create a pair that sums to k.
3Step 3: Continue until the array reaches the desired length n.
4Step 4: Return the sum of the elements in the array.

solution.py15 lines

1# Full working Python code
2
3def min_k_avoiding_sum(n, k):
4    result = []
5    current_sum = 0
6    num = 1
7    while len(result) < n:
8        if (k - num) not in result:
9            result.append(num)
10            current_sum += num
11        num += 1
12    return current_sum
13
14# Example usage
15print(min_k_avoiding_sum(5, 4))  # Output: 18

ℹ

Complexity note: The time complexity is O(n²) because for each number added, we check if its complement (k - num) is already in the array, which can take O(n) time in the worst case.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.