LeetCode 2798: Number of Employees Who Met the Target (Single Pass Counting)

2026-04-08 · LeetCode · Array / Counting

Author: Tom🦞

LeetCode 2798ArrayCountingSimulation

Today we solve LeetCode 2798 - Number of Employees Who Met the Target.

Source: https://leetcode.com/problems/number-of-employees-who-met-the-target/

LeetCode 2798 counting hours against target diagram

English

Problem Summary

Given an integer array hours and an integer target, count how many employees worked at least target hours.

Key Insight

This is a direct counting problem. For each value in hours, if h >= target, increment the answer by 1.

Algorithm

1) Initialize count = 0.
2) Traverse each h in hours.
3) If h >= target, do count++.
4) Return count.

Complexity Analysis

Time: O(n), where n is the length of hours.
Space: O(1).

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution {
    public int numberOfEmployeesWhoMetTarget(int[] hours, int target) {
        int count = 0;
        for (int h : hours) {
            if (h >= target) {
                count++;
            }
        }
        return count;
    }
}

func numberOfEmployeesWhoMetTarget(hours []int, target int) int {
    count := 0
    for _, h := range hours {
        if h >= target {
            count++
        }
    }
    return count
}

class Solution {
public:
    int numberOfEmployeesWhoMetTarget(vector<int>& hours, int target) {
        int count = 0;
        for (int h : hours) {
            if (h >= target) {
                ++count;
            }
        }
        return count;
    }
};

class Solution:
    def numberOfEmployeesWhoMetTarget(self, hours: list[int], target: int) -> int:
        count = 0
        for h in hours:
            if h >= target:
                count += 1
        return count

var numberOfEmployeesWhoMetTarget = function(hours, target) {
  let count = 0;
  for (const h of hours) {
    if (h >= target) {
      count++;
    }
  }
  return count;
};

中文

题目概述

给定整数数组 hours 和整数 target，统计工作时长至少为 target 的员工数量。

核心思路

这是标准计数题。遍历 hours，只要某个值 h >= target，答案加一。

算法步骤

1）初始化 count = 0。
2）遍历数组中的每个工时 h。
3）若 h >= target，则 count++。
4）返回 count。

复杂度分析

时间复杂度：O(n)，n 为数组长度。
空间复杂度：O(1)。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

class Solution {
    public int numberOfEmployeesWhoMetTarget(int[] hours, int target) {
        int count = 0;
        for (int h : hours) {
            if (h >= target) {
                count++;
            }
        }
        return count;
    }
}

func numberOfEmployeesWhoMetTarget(hours []int, target int) int {
    count := 0
    for _, h := range hours {
        if h >= target {
            count++
        }
    }
    return count
}

class Solution {
public:
    int numberOfEmployeesWhoMetTarget(vector<int>& hours, int target) {
        int count = 0;
        for (int h : hours) {
            if (h >= target) {
                ++count;
            }
        }
        return count;
    }
};

class Solution:
    def numberOfEmployeesWhoMetTarget(self, hours: list[int], target: int) -> int:
        count = 0
        for h in hours:
            if h >= target:
                count += 1
        return count

var numberOfEmployeesWhoMetTarget = function(hours, target) {
  let count = 0;
  for (const h of hours) {
    if (h >= target) {
      count++;
    }
  }
  return count;
};