How do you solve Sum of Good Numbers on LeetCode?

Sum of Good Numbers (LeetCode #3452) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Good numbers can be at the edges of the array without neighbors.

What is the time complexity of Sum of Good Numbers?

The optimal solution for Sum of Good Numbers runs in O(n) time and uses O(1) space. We only pass through the array once, checking conditions in constant time.

What is the brute force solution for Sum of Good Numbers?

The brute force approach for Sum of Good Numbers is Brute Force, with O(n²) time complexity and O(1) space complexity. Check each element against its neighbors to determine if it's good. This method is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sum of Good Numbers?

Sum of Good Numbers uses the following concepts: Array. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for Sum of Good Numbers?

Clarify edge cases with the interviewer. Discuss time and space complexity upfront. Think aloud while coding to demonstrate your thought process.

What are common mistakes when solving Sum of Good Numbers?

Forgetting to check bounds when accessing neighbors. Confusing strict inequality with non-strict.

#3452

Sum of Good Numbers

Easy

ArrayArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

By iterating through the array once and checking conditions for good numbers, we achieve a more efficient solution.

⚙️

Algorithm

3 steps

1Step 1: Initialize a sum variable to 0.
2Step 2: Loop through each index i in nums.
3Step 3: Check if nums[i] is greater than its neighbors at i - k and i + k, adding to sum if true.

solution.py6 lines

1def sum_of_good_numbers(nums, k):
2    total = 0
3    for i in range(len(nums)):
4        if (i < k or nums[i] > nums[i - k]) and (i + k >= len(nums) or nums[i] > nums[i + k]):
5            total += nums[i]
6    return total

ℹ

Complexity note: We only pass through the array once, checking conditions in constant time.

1Good numbers can be at the edges of the array without neighbors.
2Checking both neighbors is crucial for determining goodness.

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