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.

#3452

Easy

auto_awesome数组·driven

LeetCode 题解工作台

好数字之和

给定一个整数数组 nums 和一个整数 k ，如果元素 nums[i] 严格大于下标 i - k 和 i + k 处的元素（如果这些元素存在），则该元素 nums[i] 被认为是好的。如果这两个下标都不存在，那么 nums[i] 仍然被认为是好的。返回数组中所有好元素的和。示例…

数组

题目描述

给定一个整数数组 nums 和一个整数 k，如果元素 nums[i] 严格大于下标 i - k 和 i + k 处的元素（如果这些元素存在），则该元素 nums[i] 被认为是好的。如果这两个下标都不存在，那么 nums[i] 仍然被认为是好的。

返回数组中所有好元素的和。

示例 1：

输入： nums = [1,3,2,1,5,4], k = 2

输出： 12

解释：

好的数字包括 nums[1] = 3，nums[4] = 5 和 nums[5] = 4，因为它们严格大于下标 i - k 和 i + k 处的数字。

示例 2：

输入： nums = [2,1], k = 1

输出： 2

解释：

唯一的好数字是 nums[0] = 2，因为它严格大于 nums[1]。

提示：

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

lightbulb

解题思路

方法一：遍历

我们可以遍历数组 $\textit{nums}$ ，对于每个元素 $\textit{nums}[i]$ ，检查是否满足条件：

如果 $i \ge k$ 且 $\textit{nums}[i] \le \textit{nums}[i - k]$ ，则 $\textit{nums}[i]$ 不是好数字；
如果 $i + k < \textit{len}(\textit{nums})$ 且 $\textit{nums}[i] \le \textit{nums}[i + k]$ ，则 $\textit{nums}[i]$ 不是好数字。
否则， $\textit{nums}[i]$ 是好数字，我们将其累加到答案中。

遍历结束后，返回答案即可。

时间复杂度 $O(n)$ ，其中 $n$ 是数组 $\textit{nums}$ 的长度。空间复杂度 $O(1)$ 。

1

2

3

4

5

6

7

8

9

10

11

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

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

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

warning

常见陷阱

外企场景

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

swap_horiz

进阶变体

外企场景

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

help

常见问题

外企场景

继续练习

#3453 分割正方形 I #3454 分割正方形 II #3449 最大化游戏分数的最小值