#3452

Easy

auto_awesomeArray-driven solution strategy

LeetCode Problem Workspace

Sum of Good Numbers

The problem asks for the sum of all good numbers in an array based on specific conditions of neighboring elements.

Problem Statement

You are given an array of integers nums and an integer k. An element nums[i] is considered good if it is strictly greater than the elements at indices i - k and i + k (if those indices exist). If neither of these indices exists, nums[i] is still considered good.

Your task is to return the sum of all the good elements in the array. For example, for the array nums = [1,3,2,1,5,4] and k = 2, the good elements are 3, 5, and 4, resulting in the sum of 12.

Examples

Example 1

Input: nums = [1,3,2,1,5,4], k = 2

Output: 12

The good numbers are nums[1] = 3 , nums[4] = 5 , and nums[5] = 4 because they are strictly greater than the numbers at indices i - k and i + k .

Example 2

Input: nums = [2,1], k = 1

Output: 2

The only good number is nums[0] = 2 because it is strictly greater than nums[1] .

Constraints

2 <= nums.length <= 100
1 <= nums[i] <= 1000
1 <= k <= floor(nums.length / 2)

Solution Approach

Iterate Through the Array

Start by iterating through the array, checking for each element if it meets the condition of being greater than its surrounding elements. For indices within the bounds of the array, compare nums[i] to nums[i - k] and nums[i + k].

Edge Case Handling

Ensure proper handling of boundary elements. For elements at the beginning or end of the array, check only the valid neighboring indices that exist, ensuring that no out-of-bounds access occurs.

Summing the Good Elements

Sum the elements that satisfy the condition of being good. As you iterate through the array, maintain a running total of the good elements and return the final sum.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach used to iterate through the array and check conditions for each element. A straightforward approach would be O(n) where n is the length of the array, but optimizing for space and edge cases might alter this. Space complexity is O(1) since only a few variables are needed for iteration and checking conditions.

What Interviewers Usually Probe

Candidate demonstrates a solid understanding of array manipulation and boundary condition handling.
Candidate efficiently checks the conditions without unnecessary recomputation.
Candidate demonstrates the ability to handle edge cases like the array boundaries correctly.

Common Pitfalls or Variants

Common pitfalls

Not properly handling boundary elements, causing out-of-bounds errors.
Inefficient checking of neighbors for each element, leading to unnecessary complexity.
Not considering the array length limits or invalid inputs which might cause runtime issues.

Follow-up variants

Increase the complexity by adding more complex boundary conditions or larger values for k.
Optimize the approach for larger arrays with higher values of k.
Modify the problem to include elements with different types of neighboring conditions (e.g., divisibility instead of strict greater).

FAQ

How do I handle boundary elements in the Sum of Good Numbers problem?

To handle boundary elements, only check the neighboring elements that exist within the array's bounds. If the element is at the beginning or end, adjust accordingly.

What is the optimal approach for solving the Sum of Good Numbers problem?

The optimal approach is to iterate through the array, check the conditions for each element, and sum the good numbers. The time complexity can be O(n) with proper boundary handling.

Can I use additional data structures to optimize this problem?

Since the problem involves checking neighboring elements, additional data structures are not necessary. A single pass through the array with O(1) space should suffice.

What happens if k is larger than half the array's length?

If k is larger than half the array's length, it will still work, but boundary conditions become more important, especially when checking for elements at the extremes.

How do I optimize the solution for larger inputs in the Sum of Good Numbers problem?

For larger inputs, ensure the solution iterates only once through the array and efficiently checks the conditions without recomputing unnecessary comparisons.

terminal

Solution

Solution 1: Traversal

We can traverse the array $\textit{nums}$ and check each element $\textit{nums}[i]$ to see if it meets the conditions:

1

2

3

4

5

6

7

8

9

10

class Solution:
    def sumOfGoodNumbers(self, nums: List[int], k: int) -> int:
        ans = 0
        for i, x in enumerate(nums):
            if i >= k and x <= nums[i - k]:
                continue
            if i + k < len(nums) and x <= nums[i + k]:
                continue
            ans += x
        return ans

Continue Practicing

#3453 separate squares i #3454 separate squares ii #3449 maximize the minimum game score