How hard is LeetCode 2798. Number of Employees Who Met the Target?

LeetCode 2798. Number of Employees Who Met the Target is rated Easy on LeetCode.

What is the time complexity of LeetCode 2798. Number of Employees Who Met the Target?

The Python solution on this page runs in O(n), where n is the length of the array hours.

What is the space complexity of LeetCode 2798. Number of Employees Who Met the Target?

The Python solution on this page uses O(1) auxiliary space.

What topics does LeetCode 2798. Number of Employees Who Met the Target cover?

LeetCode 2798. Number of Employees Who Met the Target is tagged Array on LeetCode.

EasyNo. 2798

Number of Employees Who Met the Target — LeetCode 2798 Python Solution

Topics: Array
Reading time: 2 min
Source: leetcode.com

The problem

There are n employees in a company, numbered from 0 to n - 1. Each employee i has worked for hours[i] hours in the company.

Example

Input: hours = [0,1,2,3,4], target = 2
Output: 3
Explanation: The company wants each employee to work for at least 2 hours.

Python solution

PythonSolution

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

Complexity

Time: O(n)where n is the length of the array hours
Space · auxiliary: O(1)

Related problems
No.	Problem	Difficulty
0057	Insert Interval	Medium
0228	Summary Ranges	Easy
0485	Max Consecutive Ones	Easy
0665	Non-decreasing Array	Medium
0674	Longest Continuous Increasing Subsequence	Easy
0699	Falling Squares	Hard

Frequently asked questions

How hard is LeetCode 2798. Number of Employees Who Met the Target?: LeetCode 2798. Number of Employees Who Met the Target is rated Easy on LeetCode.
What is the time complexity of LeetCode 2798. Number of Employees Who Met the Target?: The Python solution on this page runs in O(n), where n is the length of the array hours.
What is the space complexity of LeetCode 2798. Number of Employees Who Met the Target?: The Python solution on this page uses O(1) auxiliary space.
What topics does LeetCode 2798. Number of Employees Who Met the Target cover?: LeetCode 2798. Number of Employees Who Met the Target is tagged Array on LeetCode.

Stuck on problems like this in a live interview?

Stealth Interview is a desktop app for macOS and Windows. It reads the problem off your screen and returns a working solution with a step-by-step explanation and its time and space complexity, invisible to screen sharing.

Download for Mac Download for Windows

Free to download. Check it runs on your machine, then pick a plan.

90%Subscription

python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

Technical interview · 45 minsolution.pyShared editor · Python 3Longest substring without repeatsMediumGiven a string s, return the length of the longest contiguous substring in which every character is unique.s = "abcabcbb"
→ 31def length_of_longest_substring(s: str) -> int:2 last_seen = {}3 left = best = 04 for right, ch in enumerate(s):5 if ch in last_seen and last_seen[ch] >= left:

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 Terra

Transcribing...

90%Subscription

python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 TerraClear chat

Transcribing...

QuitEscHide+QCapture+PCrop+OSubmit+EnterPress to get answerLeft/Right+Up/Down+

90%Subscription

python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 TerraClear chat

Transcribing...

Primary display

QuitEscHide+QCapture+PCrop+OSubmit+EnterPress to get answerLeft/Right+Up/Down+

The problem

Example

Python solution

Complexity

Related problems

Frequently asked questions

Stuck on problems like this in a live interview?

Approach

Approach

Approach