| TECH |
Techable（テッカブル） |
企業の健康経営と利用者のQOL向上。“続けたくなる”ヘルスケアサービス、法人向け「LOOOM」が登場 |
https://techable.jp/archives/216491
|
looom |
2023-08-11 10:00:49 |
| Ruby |
Rubyタグが付けられた新着投稿 - Qiita |
プロを目指す人のためのRuby入門（チェリー本）を読んで |
https://qiita.com/KOH6/items/b33a9dac5a1167707ba9
|
通称 |
2023-08-11 19:13:28 |
| Git |
Gitタグが付けられた新着投稿 - Qiita |
【Git】空コミットの作り方 |
https://qiita.com/smartPG/items/e730d4f103df1f8b9070
|
追加 |
2023-08-11 19:22:57 |
| Ruby |
Railsタグが付けられた新着投稿 - Qiita |
プロを目指す人のためのRuby入門（チェリー本）を読んで |
https://qiita.com/KOH6/items/b33a9dac5a1167707ba9
|
通称 |
2023-08-11 19:13:28 |
| 技術ブログ |
Developers.IO |
วิธีการแจ้งเตือนค่าใช้จ่ายการใช้งาน AWS ใน Slack ทุกวัน |
https://dev.classmethod.jp/articles/how-to-notify-daily-cost-of-using-aws-in-slack/
|
วิธีการแจ้งเตือนค่าใช้จ่ายการใช้งานAWS ในSlack ทุกวันสวัสดีครับPOP จากบริษัทClassmethod Thailand ครับผู้ที่ใช้งานบริการAWS ส่วนใหญ่มีการควบคุมค่าใช้จ่ายการใ |
2023-08-11 10:21:53 |
| 海外TECH |
DEV Community |
Product of Array Except Self - LeetCode Java Solution |
https://dev.to/aswin2001barath/product-of-array-except-self-leetcode-java-solution-4j72
|
Product of Array Except Self LeetCode Java SolutionHello readers let s solve a LeetCode problem today In this blog let s solve the problem Product of Array Except Self which is one of the Blind List of LeetCode Problems This is a LeetCode Medium problem with multiple potential solutions which we will go through in this blog Understand the ProblemThe Objective of the problem is to calculate the product of all elements in an input array except for the element at each index The problem also mentions writing an algorithm that runs in O n time and without using the division operation Understand the TestcasesTo test the solution we can consider the following cases Input Output Explanation The product of all elements except the one at index is The product of all elements except the one at index is Similarly for index the product is and for index the product is Input Output Explanation The product of all elements except the one at index is The product of all elements except the one at index is Similarly for index the product is for index the product is and for index the product is Input Output Explanation Since all elements are the same the product of all elements except the one at each index will be the same element itself Input Output Explanation Since there are zero values in the input array the product of all elements except the one at each index will be zero Input Output Explanation The product of all elements except the one at index is The product of all elements except the one at index is For index the product is and for index the product is Brute Force ApproachThe brute force approach will be to calculate the product of all values in the array except the current value using two nested iterations Key Points For each value calculate the product of all other values by iterating through the array twice This approach uses two nested loops to multiply each element with all other elements excluding itself However this method is inefficient and may lead to a Time Limit Exceeded TLE error for larger test cases class Solution public int productExceptSelf int nums int size nums length int answer new int size for int i i lt size i answer i for int i i lt size i for int j j lt size j if i j continue answer i nums j return answer Time Complexity O N The time complexity of the brute force approach is quadratic as it involves nested loops that iterate through the array Space Complexity O The brute force approach doesn t require any additional space other than the answer array so the space complexity is constant Better Approachclass Solution public int productExceptSelf int nums int size nums length int answer new int size int prefix new int size int suffix new int size int temp for int i i lt size i prefix i temp temp nums i temp for int j size j gt j suffix j temp temp nums j for int i i lt size i answer i prefix i suffix i return answer Key Points The better approach uses separate iterations to calculate prefix and suffix values for each element In this approach we calculate the prefix products while traversing the array from left to right and storing them in the prefix array Then we calculate the suffix products while traversing the array from right to left and storing them in the suffix array Finally we multiply the corresponding prefix and suffix values to get the final products Time Complexity O N The better approach performs two linear passes through the array resulting in a linear time complexity Space Complexity O N The better approach requires additional space to store the prefix and suffix arrays resulting in linear space complexity Optimal Approachclass Solution public int productExceptSelf int nums int answer new int nums length int prefix for int i i lt nums length i answer i prefix prefix nums i int suffix for int j nums length j gt j answer j suffix suffix nums j return answer Key Points The optimal approach further improves efficiency by using the output array to store the prefix products In this approach we calculate the prefix products while traversing the array from left to right Then we calculate the suffix products while traversing the array from right to left Finally we multiply the suffix product with the corresponding prefix product to obtain the final answer Time Complexity O N The optimal approach also performs two linear passes through the array resulting in a linear time complexity Space Complexity O The optimal approach doesn t require any additional space other than the output array so the space complexity is constant ConclusionThe brute force solution has a time complexity of O N² but no extra memory usage Using two extra arrays improves the time complexity to O N but also requires extra memory with the complexity of O N Using the linear traversals and answer array offers the best time complexity of O N but doesn t require any additional space other than the output array so the space complexity is constant O NeetCode Solution Video Recommended Resources to Learn Data Structures and AlgorithmsBasics of DS Algo Blogs Hashing in JavaRecommended YouTubers for LeetCode Problems NeetCode Take U ForwardFree Resources for Learning Data Structures and Algorithms NeetCode Roadmap Striver s SDE SheetRecommended Courses for Learning Data Structures and Algorithms NeetCode CoursesZTM Mastering the Coding Interview Big Tech Available on Udemy and ZTM AcademyZTM Mastering the Coding Interview Available on Udemy and ZTM AcademyData Structures amp Algorithms Level up for Coding Interviews CourseStriver s AZ Free CourseTop Coursera Courses for Learning Data Structures and Algorithms Coding Interview Preparation Meta Algorithms Course Part I Princeton University Algorithms Course Part II Princeton University Data Structures and Algorithms Specialization UC San Diego Algorithms Specialization Stanford Note The Coursera courses can be audited to get free access to the lectures Disclosure Please note that some of the links mentioned on this page may be affiliate links This means that if you click on one of these links and make a purchase I may earn a small commission from the sale Who Am I I m Aswin Barath a Software Engineering Nerd who loves building Web Applications now sharing my knowledge through Blogging during the busy time of my freelancing work life Here s the link to all of my craziness categorized by platforms under one place Keep LearningNow I guess this is where I say goodbye But hey it s time for you to start learning with your newfound Knowledge Power Good Job that you made it this far amp Thank you so much for reading my Blog |
2023-08-11 10:48:03 |
| 海外TECH |
DEV Community |
Two Sum — LeetCode Java Solution |
https://dev.to/aswin2001barath/two-sum-leetcode-java-solution-5545
|
Two Sum ーLeetCode Java SolutionHello readers let s solve a LeetCode problem today In this blog let s solve Two Sum which is one of the Blind List of LeetCode Problems The Two Sum problem is a classic coding challenge and the No problem on LeetCode that asks us to find two numbers in an array that add up to a given target In this blog post we will explore two different approaches to solving this problem the Brute Force approach and the Optimal approach We will examine the problem statement understand the test cases and dive into the code implementations for both approaches Additionally we will analyze the time and space complexities of each solution Understanding the Problem Given an array of integers nums and a target integer target the task is to find two numbers in the array such that their sum equals the target Our objective is to return the indices of these two numbers The problem also mentions that each input will have exactly one solution and you cannot use the same element twice Understanding the Test Cases To better understand the problem let s consider a few simple test cases Test Case Input nums target Output Explanation The sum of the numbers at indices and is equal to the target value of Test Case Input nums target Output Explanation The sum of the numbers at indices and is equal to the target value of Test Case Input nums target Output Explanation The sum of the numbers at indices and is equal to the target value of Brute Force Approach Nested Loopsclass Solution public int twoSum int nums int target int result new int for int i i lt nums length i for int j i j lt nums length j if nums i nums j target result i result j return result return result Key Points We go through each number in the array and add it to every other number to see if their sum equals the target If we find two numbers that add up to the target we return their indices Time Complexity O N The brute force approach uses nested loops to check every pair of numbers resulting in a time complexity of O N where n is the size of the input array Space Complexity O The space complexity of this approach is O because we only need a fixed size result array to store the indices Optimal Approach Hashmapclass Solution public int twoSum int nums int target int result new int HashMap lt Integer Integer gt map new HashMap lt gt for int i i lt nums length i int numberToFind target nums i if map containsKey numberToFind result map get numberToFind result i map put nums i i return result Key Points For each number in the array we calculate the difference between the target and that number We check if this difference exists in the HashMap which means we have found a pair of numbers that add up to the target If we find such a pair we return their indices Otherwise we keep adding the numbers and their indices to the HashMap Time Complexity O N The optimal approach has a time complexity of O N because we iterate through the input array nums only once and each lookup operation in the HashMap takes constant time Space Complexity O N The space complexity of the optimal approach is O N because in the worst case we may need to store all the elements of the array in the HashMap Google Engineer Coding Interview Style Video SolutionAlso checkout NeetCode Video Solution Recommended Resources to Learn Data Structures and AlgorithmsBasics of DS Algo Blogs Hashing in JavaRecommended YouTubers for LeetCode Problems NeetCode Take U ForwardFree Resources for Learning Data Structures and Algorithms NeetCode Roadmap Striver s SDE SheetRecommended Courses for Learning Data Structures and Algorithms NeetCode CoursesZTM Mastering the Coding Interview Big Tech Available on Udemy and ZTM AcademyZTM Mastering the Coding Interview Available on Udemy and ZTM AcademyData Structures amp Algorithms Level up for Coding Interviews CourseBecome a Job Ready Programmer Java Striver s AZ Free CourseTop Coursera Courses for Learning Data Structures and Algorithms Coding Interview Preparation Meta Algorithms Course Part I Princeton University Algorithms Course Part II Princeton University Data Structures and Algorithms Specialization UC San Diego Algorithms Specialization Stanford Note The Coursera courses can be audited to get free access to the lectures Disclosure Please note that some of the links mentioned on this page may be affiliate links This means that if you click on one of these links and make a purchase I may earn a small commission from the sale Who Am I I m Aswin Barath a Software Engineering Nerd who loves building Web Applications now sharing my knowledge through Blogging during the busy time of my freelancing work life Here s the link to all of my craziness categorized by platforms under one place Keep LearningNow I guess this is where I say goodbye But hey it s time for you to start learning with your newfound Knowledge Power Good Job that you made it this far amp Thank you so much for reading my Blog |
2023-08-11 10:45:56 |
| 海外TECH |
DEV Community |
Valid Anagram - LeetCode Java Solution |
https://dev.to/aswin2001barath/valid-anagram-leetcode-java-solution-48nm
|
Valid Anagram LeetCode Java SolutionHello readers let s solve a LeetCode problem today In this blog let s solve Valid Anagram which is one of the Blind List of LeetCode Problems This is a very good LeetCode Easy problem for beginners There are multiple potential solutions for this problem which we will go through in this blog Understand the problemAn Anagram is a word or phrase formed by rearranging the letters of a different word or phrase typically using all the original letters exactly once Given two strings s and t return true if t is an anagram of sreturn false otherwise Understand the test casesExample Input s anagram t nagaram Output trueThe input strings “anagram and nagaram exactly contain a g m n and r So nagaram can be formed by rearranging the letters of “anagram And hence they are both anagrams of each other Example Input s rat t car Output falseThe string “car does not contain the character t from the string s “rat so these are not anagrams Brute force Approach Sortingclass Solution public boolean isAnagram String s String t char sc s toCharArray char tc t toCharArray Arrays sort sc Arrays sort tc if new String sc equals new String tc return true return false Key Points The above code checks if two strings s and t are anagrams by converting them into character arrays sorting the arrays in ascending order and then comparing the sorted arrays for equality If the sorted arrays of characters are equal it means that the strings have the same characters in the same frequencies indicating that they are anagrams of each other The code returns true if the strings are anagrams and false otherwise Time complexity O N log N The time complexity of the above code is O N log N where N is the length of the longer string between s and t This is because the code uses Arrays sort to sort the character arrays sc and tc The time complexity of the Arrays sort method is O n log n in the average case Therefore the overall time complexity is dominated by the sorting operation Space complexity O N The space complexity of the code is O N where N is the length of the longer string between s and t This is because the code creates two character arrays sc and tc which store the characters of the input strings The length of each character array is equal to the length of the respective input string Therefore the space required by the character arrays is proportional to the length of the longer string Better Approach Count using two HashMapsclass Solution public boolean isAnagram String s String t HashMap lt Character Integer gt count new HashMap lt gt HashMap lt Character Integer gt count new HashMap lt gt count count count s count count count t return count equals count true false public HashMap lt Character Integer gt count HashMap lt Character Integer gt count String str for int i i lt str length i Character c str charAt i if count containsKey c count put c count get c continue count put c return count Key Points HashMaps count and count are used to count the number of characters in the given two strings s and t respectively The function count is used to traverse the given string calculate the number of occurrences of a character and store them in the count HashMap and return it Finally count and count are compared for equality to verify of the given two strings are anagrams of each other or not Time Complexity O N The time complexity of the given code is O N where N is the total number of characters in both strings combined This is because the code iterates over each character in the strings once while building the frequency count in the count method Since the count method is called twice once for each input string the total time complexity is O x y where x and y are the lengths of the input strings s and t respectively However since we typically ignore constant factors and focus on the dominant term the time complexity is simplified to O N Space Complexity O or ConstantThe space complexity of the code is O k where k is the number of unique characters across both strings In the worst case scenario each unique character in the strings will be stored as a key in the count and count HashMaps Therefore the space required by HashMaps is proportional to the number of unique characters Since the total number of unique characters is typically much smaller than the length of the strings we can consider the space complexity to be O or constant Optimal Approach Count using a frequency arraypublic class Solution public boolean isAnagram String s String t int alphabet new int for int i i lt s length i alphabet s charAt i a for int i i lt t length i alphabet t charAt i a for int i alphabet if i return false return true Key Points In the above code a size int array alphabet is created as a frequency array for each letter in the Alphabet The indices of the alphabet represent the letters of the Alphabet So represents a represents b so on until represents z Then we calculate the frequency of the characters We increment the alphabet array values for each of the characters of String s in the first iteration And then decrement the values of the same alphabet array with string t in the second iteration In both iterations we perform the operations s charAt i a and t charAt i a Here the characters of the string s and string t are subtracted with the character a to get an integer value of the letters that correspond to the alphabet array If they are anagrams the alphabet bucket should remain with the initial value of the array which is zero So the final iteration just checks if all values are zero if any value is not zero we return false and return trueTime complexity O N The time complexity of the above code is O n where n is the length of the longer string between s and t This is because the code iterates over each character in both strings only once using two separate loops The first loop counts the occurrences of characters in string s and the second loop subtracts the occurrences of characters in string t Finally there is a third loop that checks if any count in the alphabet array is nonzero indicating a mismatch in character counts between the two strings Space complexity O or ConstantThe space complexity of the code is O or constant because uses an integer array alphabet of size to store the counts of characters in the English alphabet Regardless of the length of the input strings the alphabet array remains a fixed size because it represents a fixed set of characters lowercase English letters Therefore the space required by the array is constant and does not depend on the input size NeetCode Solution VideoRecommended Resources to Learn Data Structures and AlgorithmsBasics of DS Algo Blogs Hashing in JavaRecommended YouTubers for LeetCode Problems NeetCode Take U ForwardFree Resources for Learning Data Structures and Algorithms NeetCode Roadmap Striver s SDE SheetRecommended Courses for Learning Data Structures and Algorithms NeetCode CoursesZTM Mastering the Coding Interview Big Tech Available on Udemy and ZTM AcademyZTM Mastering the Coding Interview Available on Udemy and ZTM AcademyData Structures amp Algorithms Level up for Coding Interviews CourseStriver s AZ Free CourseTop Coursera Courses for Learning Data Structures and Algorithms Coding Interview Preparation Meta Algorithms Course Part I Princeton University Algorithms Course Part II Princeton University Data Structures and Algorithms Specialization UC San Diego Algorithms Specialization Stanford Note The Coursera courses can be audited to get free access to the lectures Disclosure Please note that some of the links mentioned on this page may be affiliate links This means that if you click on one of these links and make a purchase I may earn a small commission from the sale Who Am I I m Aswin Barath a Software Engineering Nerd who loves building Web Applications now sharing my knowledge through Blogging during the busy time of my freelancing work life Here s the link to all of my craziness categorized by platforms under one place Keep LearningNow I guess this is where I say goodbye But hey it s time for you to start learning with your newfound Knowledge Power Good Job that you made it this far amp Thank you so much for reading my Blog |
2023-08-11 10:43:00 |
| 海外TECH |
DEV Community |
Getting started with react native storybook |
https://dev.to/dannyhw/getting-started-with-react-native-storybook-96c
|
Getting started with react native storybookThis guide is a written version of a video I recently made and will show you how to get started with react native storybook on an existing project This assumes that you already have a react native project to start with For my purposes I m starting with the expo typescript template but you can use react native cli if you prefer Heres the video if you want to watch me do it and talk through the processThe first thing you ll want to do is open the terminal in your project folder and runnpx sb latest init type react nativeThis will create all the files and install all the packages you need to get started As you can see from the message here the setup isn t fully automated and the next step is to get storybook to render by replacing your entry point with the Storybook component Lets now take a look at what got added to our projectThere were some new packages installed two new scripts and a storybook folder which contains all the storybook config The storybook index js file exports the Storybook component which we need to render to get storybook working Now in your editor open up App tsx or js jsx or wherever your app entrypoint and import and export the component storybook index js like you would normally export the App component For now you can comment out your existing default export but later you can setup a way to swap between storybook and your app import StorybookUI from storybook export default StorybookUI Next we ll make one quick change to our metro config If you re on expo you ll need to generate that file withnpx expo customize metro config jsThen in that file add sb modern to the start of the resolver resolverMainFields array In expos generated file you can do it like this Learn more const getDefaultConfig require expo metro config type import expo metro config MetroConfig const config getDefaultConfig dirname config resolver resolverMainFields unshift sbmodern module exports config note you ll be happy to know sbmodern won t be needed in v of storybookNow you can run yarn ios or yarn android and when the app opens up you should see the storybook UI Why aren t my new stories showing up When you add new stories you might notice that they don t immediately show up thats because currently you need to run yarn storybook generate each time you add a new story I like to add it to the start command like this so that I never forget start yarn storybook generate amp amp expo start This is needed because until recently react native didn t support dynamic imports and we generate a list of story imports based on your main js config You can see this in the storybook requires js file This is something that we ll be changing in the future but for now make sure that you run storybook generate whenever your new stories aren t showing up Thats it folksThanks for reading this post if you re wondering where to go from here I recommend taking a look at some of my other posts or following the tutorial I wrote for react native storybook You can find me on twitter here Danny H W You can find my work on github here dannyhwMy DM s are open on twitter in case you want to get in touch |
2023-08-11 10:41:36 |
| 海外TECH |
DEV Community |
Contains Duplicate - LeetCode Java Solution |
https://dev.to/aswin2001barath/contains-duplicate-leetcode-java-solution-128d
|
Contains Duplicate LeetCode Java SolutionHello readers let s solve a LeetCode problem today In this blog let s solve Contains Duplicate which is one of the Blind List of LeetCode Problems This is a very good LeetCode Easy problem for beginners There are multiple potential solutions for this problem which we will go through in this blog Understand the ProblemThe Objective of the problem is to Determine if an array contains any duplicate valuesIn the Problem Description it is given Given an array of numbers numsReturn true if there are any values in the array that appear at least twiceReturn false if all values in the array are distinct Understand the TestcasesTo understand the problem better three examples are given In Example it is given Input nums Output trueHere we can spot there are two ones at the first and last index and hence is the duplicate value in this array So we can return the output true In Example it is given Input nums Output falseHere we can see that all four numbers are distinct values in this array So we can return the output false In Example it is given Input nums Output trueHere we can see many duplicate values of the numbers and in this array So we can return the output true Brute Force ApproachThe brute force approach will be to compare each number with every other number in the array until we find the duplicate number class Solution public boolean containsDuplicate int nums for int i i lt nums length i for int j i j lt nums length j if nums i nums j found the duplicate return true return false Here are the key points to understand the code This approach follows a brute force method of comparing each number with every other number in the array to find duplicates We use nested loops to traverse through all possible pairs of elements in the array The outer loop iterates over each element in the array and the inner loop compares that element with the remaining elements If any two elements are found to be equal it means we have found a duplicate value and we return true After comparing all possible pairs and not finding any duplicates we return false Time complexity The time complexity of this approach is O N where N is the number of elements in the nums array This is because we have nested loops causing us to compare each element with every other element in the worst case Space complexity The space complexity of this approach is O because it does not require any additional space that grows with the input size Better Approach class Solution public boolean containsDuplicate int nums Arrays sort nums for int i i lt nums length i if nums i nums i return true return false Here are the key points to understand the code This approach takes advantage of sorting the array first to simplify the duplicate checking process The array is sorted in ascending order using the Arrays sort method By sorting the array duplicate elements become adjacent to each other making it easier to identify duplicates by comparing adjacent elements We traverse the sorted array and compare each element with its next adjacent element If any adjacent elements are equal it means we have found a duplicate value and we return true If no duplicates are found after checking all adjacent pairs we return false Time complexity The time complexity of this approach is O N log N where N is the number of elements in the nums array This is because sorting the array takes O N log N time using efficient sorting algorithms like Merge Sort or Quick Sort Space complexity The space complexity of this approach is O log N where N is the number of elements in the nums array This space is used for the recursive calls in the sorting algorithm Java Solution Using HashSet We can use a HashSet to efficiently check for duplicates in the given array HashSet is a data structure that stores unique elements It allows for constant time O operations like adding and searching for elements class Solution public boolean containsDuplicate int nums HashSet lt Integer gt dups new HashSet lt Integer gt for int n nums if dups contains n return true dups add n return false Here are the key points to understand the code We traverse the nums array and for each element n we check if it is already present in the HashSet If the value is present it means we have found a duplicate so we return true If the value is not present we add it to the HashSet to keep track of unique elements After traversing the entire array and not finding any duplicates we return false Time complexity The time complexity of this solution is O N where N is the number of elements in the nums array This is because we traverse the entire array once Space complexity The space complexity of this solution is O N where N is the number of elements in the nums array This is because in the worst case all elements in the array could be unique and stored in the HashSet Java Solution Using HashMap We can use a HashMap to efficiently check for duplicates in the given array HashMap is a data structure that stores key value pairs where each key is unique class Solution public boolean containsDuplicate int nums HashMap lt Integer Boolean gt dups new HashMap lt gt for int n nums if dups containsKey n return true dups put n true return false Here are the key points to understand the code We traverse the nums array and for each element n we check if it is already a key in the HashMap If it is a key it means we have found a duplicate value so we return true If the element is not present as a key we add it to the HashMap with a value of true to keep track of its presence After traversing the entire array and not finding any duplicates we return false Time complexity The time complexity of this solution is O N where N is the number of elements in the nums array This is because we traverse the entire array once Space complexity The space complexity of this solution is O N where N is the number of elements in the nums array This is because in the worst case all elements in the array could be unique and stored as keys in the HashMap ConclusionThe brute force solution has a time complexity of O N but no extra memory usage Sorting the array improves the time complexity to O N log N but still requires no extra memory Using a HashMap amp HashSet offers the best time complexity of O N but requires additional memory NeetCode Solution Video Recommended Resources to Learn Data Structures and AlgorithmsBasics of DS Algo Blogs Hashing in JavaRecommended YouTubers for LeetCode Problems NeetCodeTake U ForwardFree Resources for Learning Data Structures and Algorithms NeetCode RoadmapStriver s SDE SheetRecommended Courses for Learning Data Structures and Algorithms NeetCode CoursesZTM Mastering the Coding Interview Big Tech Available on Udemy and ZTM AcademyZTM Mastering the Coding Interview Available on Udemy and ZTM AcademyData Structures amp Algorithms Level up for Coding Interviews CourseStriver s AZ Free CourseTop Coursera Courses for Learning Data Structures and Algorithms Coding Interview Preparation Meta Algorithms Course Part I Princeton University Algorithms Course Part II Princeton University Data Structures and Algorithms Specialization UC San Diego Algorithms Specialization Stanford Note The Coursera courses can be audited to get free access to the lectures Disclosure Please note that some of the links mentioned on this page may be affiliate links This means that if you click on one of these links and make a purchase I may earn a small commission from the sale Who Am I I m Aswin Barath a Software Engineering Nerd who loves building Web Applications now sharing my knowledge through Blogging during the busy time of my freelancing work life Here s the link to all of my craziness categorized by platforms under one place Keep LearningNow I guess this is where I say goodbye But hey it s time for you to start learning with your newfound Knowledge Power Good Job that you made it this far amp Thank you so much for reading my Blog |
2023-08-11 10:36:07 |
| 海外TECH |
DEV Community |
4-Day Workweek, Works! |
https://dev.to/crabnebula/4-day-workweek-works-1eie
|
Day Workweek Works NEWSFLASH Attracting top talent is not all about financial compensation So how did we get some crazy talented engineers and team members to join us To be perfectly transparent maximizing their financial compensation packages was not our primary focus Among the plethora of tools available to attract talent I am always amazed how much emphasis is placed on financial compensation whether it be salary equity or performance bonuses I am not saying that fair and transparent financial compensation is not needed In fact it should be expected But I wonder how the working world missed the adage that financial rewards neither foster commitment nor promote positive creativity Financial rewards have shown to only temporarily change behavior to meet short term goals but do not help companies build agile thriving teams Pay compensation usually ranks th or th on lists of what matters most to employees During interviews with potential candidates for our company nearly all were seeking something different and believed that our ethos is precisely what they were looking for People come to our company craving workplace flexibility and “work life balance not because they want to take advantage of it but because they would like to contribute to it A place to work productively with integrity and their well being at the center of what they do and why they do it Why a Day Workweek We implemented a day workweek hours at full time from the very beginning Why Because burnout is real and our employees NEED to have tangible time outside of work Many of our team have family responsibilities in addition to their own lives We believe that one tool to help them achieve balance in their lives is hours of continuous free time to accomplish what they need to and still have time for their own well being We do NOT want our team available to us That is NOT okay ProductivityThe loudest argument against the day hr week is that productivity will never reach that of a day hr week Please let me digress for a moment as we take this opportunity to jump down the Parkinson s Law rabbit hole This observation can be boiled down to “work expands to fill the time available So if work expands to fill the time we have for it time is not a constant that is continually “refilled with more new work as soon as a task is completed Rather we expand the time needed for a given task based on the time available If there is less time available the task can still be completed as it will not expand as far to fill the less time available Okay…now returning to the research the day week has shown not to reduce company productivity Workplace FlexibilityThe next argument against the day week can be boiled down to this cannot work for all industries I even read on a LinkedIn post a commentator said that the day week is only for the privileged desk workers or bougie tech engineers working remotely People in various service industries retail or even healthcare could not possibly accommodate it The commenter completed his argument with that he hates vacations and only likes to work but I digress again My question is Why not The hospitality and retail industries have been relying on flexible employees and staggered scheduling for decades Perhaps it is time that these industries remind the corporate world how workplace flexibility is still possible A related argument concludes that the day week is impossible due to the widespread staffing shortages experienced across nearly all service industries I would answer this concern with a question why do you think certain professions read here healthcare have staffing shortages The burnout rate for healthcare employees is among the highest When we start being reasonable about what people should be expected to accomplish during their working hours and start truly taking care of the people who take care of business industries will attract the staff they need Staffing CostsA final understandable argument is accommodating a day week will most certainly increase staffing costs My answer what if the day week was an actual magic bullet to help companies significantly reduce one of their larger expenses voluntary turnover people quitting Gallup estimated that voluntary turnover costs US businesses trillion USD per year To replace a single employee can cost the company from to times their annual salary Not only are companies losing that individual s productivity but also their knowledge and experience That is a wildly expensive price to pay and appears to be completely self inflicted If the day week boosts employee satisfaction voluntary turnover will decline MRL consulting experienced a retention rate during its trial with the day week If a company could reinvest what would normally be spent on replacing employees and instead invest the same money on engaged and motivated employees wouldn t that be a great investment in productivity see above about productivity Granted a day workweek is not going to solve all the staffing problems facing companies and their teams today The business world has at least acknowledged that employees across all industries are struggling with burnout and it is hurting their bottom lines Now is the time for companies to reinvent the norms of the workplace and start putting their greatest assets their employees first The day workweek is a studied tool for this task by improving employee happiness health and therefore retention rate Author Elizabeth Duck Chief Operating Officer |
2023-08-11 10:23:25 |
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Japan Sweden World Cup semi finals beckon as Arsenal and Man City stars scoreSweden produce a magnificent performance to book a semi final date with Spain and end Japan s dreams of winning the Women s World Cup |
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MP Angus MacNeil expelled by SNP after chief whip row |
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westminster |
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Irn-Bru drivers walk out on first day of strike |
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walkouts |
2023-08-11 10:50:12 |
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Women's World Cup 2023: Sweden beat Japan to progress to semis - highlights |
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Women x s World Cup Sweden beat Japan to progress to semis highlightsWatch highlights as Sweden produce an excellent display to beat Japan and book their place in the Women s World Cup semi finals |
2023-08-11 10:36:08 |
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hawaiian |
2023-08-11 10:19:41 |
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