What is the most efficient way to solve Number of Employees Who Met the Target?

Iterate through the hours array and count each employee whose hours are greater than or equal to target. This single-pass array approach is optimal.

Can I use built-in array functions to solve this problem?

Yes, functions like filter or reduce work, but a simple for-loop is clearer and avoids unnecessary overhead, keeping the solution O(n).

Does the array need to be sorted to count employees meeting the target?

No, sorting is unnecessary since the pattern only requires checking each element independently.

What happens if all employees have hours below target?

The function should return 0 because no employee meets or exceeds the target value.

How does the array-driven solution handle large target values?

Each employee's hours are compared directly to target, so even if target is large, the solution remains O(n) with minimal memory.

#2798

Easy

auto_awesome数组·driven

LeetCode 题解工作台

满足目标工作时长的员工数目

公司里共有 n 名员工，按从 0 到 n - 1 编号。每个员工 i 已经在公司工作了 hours[i] 小时。公司要求每位员工工作至少 target 小时。给你一个下标从 0 开始、长度为 n 的非负整数数组 hours 和一个非负整数 target 。请你用整数表示并返回工作至少 tar…

数组

题目描述

公司里共有 n 名员工，按从 0 到 n - 1 编号。每个员工 i 已经在公司工作了 hours[i] 小时。

公司要求每位员工工作至少 target 小时。

给你一个下标从 0 开始、长度为 n 的非负整数数组 hours 和一个非负整数 target 。

请你用整数表示并返回工作至少 target 小时的员工数。

示例 1：

输入：hours = [0,1,2,3,4], target = 2
输出：3
解释：公司要求每位员工工作至少 2 小时。
- 员工 0 工作 0 小时，不满足要求。
- 员工 1 工作 1 小时，不满足要求。
- 员工 2 工作 2 小时，满足要求。
- 员工 3 工作 3 小时，满足要求。
- 员工 4 工作 4 小时，满足要求。
共有 3 位满足要求的员工。

示例 2：

输入：hours = [5,1,4,2,2], target = 6
输出：0
解释：公司要求每位员工工作至少 6 小时。
共有 0 位满足要求的员工。

提示：

1 <= n == hours.length <= 50
0 <= hours[i], target <= 10⁵

lightbulb

解题思路

方法一：遍历计数

我们可以遍历数组 $hours$ ，对于每个员工，如果其工作时长 $x$ 大于等于 $target$ ，则将计数器 $ans$ 加一。

遍历结束后，返回答案即可。

时间复杂度 $O(n)$ ，其中 $n$ 是数组 $hours$ 的长度。空间复杂度 $O(1)$ 。

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class Solution:
    def numberOfEmployeesWhoMetTarget(self, hours: List[int], target: int) -> int:
        return sum(x >= target for x in hours)

speed

复杂度分析

指标	值
时间	complexity is O(n) because each employee is checked exactly once. Space complexity is O(1) since only a counter is maintained, making it very efficient for arrays up to length 50.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Check if you can solve this using only a simple array scan without additional data structures.
question_mark
They may hint at counting instead of filtering or mapping to emphasize direct array operations.
question_mark
Expect questions on handling edge cases where no employee meets the target or all employees meet it.

warning

常见陷阱

外企场景

error
Forgetting to include employees whose hours exactly equal the target.
error
Using extra arrays or structures unnecessarily instead of a single counter.
error
Misinterpreting 0-based indexing when iterating over hours array.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Return a list of employee indices who met the target instead of the count.
arrow_right_alt
Given a dynamic list of hours, continuously update the count of employees meeting changing targets.
arrow_right_alt
Compute percentage of employees meeting the target instead of absolute count.

help

常见问题

外企场景

继续练习

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