Klika’s Post

🎉 5 Years of Excellence at Accedia! 🥳 👨🏻💻 Svetoslav Donkov, Software Consultant, is celebrating his 5-year journey at Accedia! Here is how he describes it: “In the last 5 years, I have learned too much to list. But if I had to pick one, I would say that the most important lesson is never to be afraid to ask, engage, and try. From my perspective, that's the one hurdle every new (and not-so-new) developer faces in their day-to-day work. The great team spirit at Accedia and the colleagues who are always willing to share their experiences are a great bonus to overcome exactly this challenge. The freedom of expression that the company offers you is priceless. No matter what technology you're interested in, there's always an opportunity to try something you've never seen before, share it with everyone, get feedback, and try again. Many company initiatives help with this, and every app that has come out of IDC's internal development marathon is a highlight in its own right. There are some great people working at Accedia, and regardless of who your team members are, you have a good chance of meeting someone who shares your interests inside and outside of work. The versatility of the job also brings tons of new insights and opportunities to improve every day. Pretty much all the countless team buildings, parties, and get-togethers have been special experiences, but I fondly remember the times when a few colleagues and I would regularly visit each other to chat, have a drink, and play board games. On one particular occasion, we pondered the meaning of life by the pool at night.” #accediateam #exceedingexpectations #5yearanniversary #softwareconsultant #softwaredeveloper

hey #viewers Today is my day 23, of solving a hard question everyday, for the next 75 days 🚀 Today, I solved a question named as Basic Calculator 📟 Leetcode : https://lnkd.in/d_-59cba Approach: • We can simply use a stack to store results correctly, whenever parentheses are found. · Sabse pehle, do create a counter to store final result, a value, which is used to check sign, as initially with 1 and, create a stack, that stores pair of values calculated , along their sign. · Then, check in a for loop, as, if we get an numeric value, then check for the number, and move the index, as far as the number goes. • Basically, it parses over the whole number, and updates finalAns with adding the number, multiplied by the current sign. · If a negative sign is encountered, then we update the sign value as well. · Then check if, we have encountered an open parentheses, if yes, then store the current value, and it's sign, to the stack, and assign sum and sign values to their initial values, as to calculate values inside the parentheses correctly. • Then check if, we have encountered any closing parentheses, if yes, then it can be calculated, by adding the previous value, with the current value, multiplied by the sign, to avoid incorrect results. • Also, it then pops out the previous values stored, to store a new pair. • After processing all the characters in the input string, it returns the finalAns. For further assistance, please find the code attached below 😄 In case of any queries, feel free to ask in the comments 😀 Let's see how far we can get through this challenge. #ab #mai #hashTag #use #karunga #for #good #reach #and #thats #it #75hardchallenge #connections #codechallenge #100daysofCode

This #AAPIHeritageMonth, we're shining a light on the multifaceted Workmates who make work days brighter ☀️ By day, Suman is a Senior Software Application Engineer, but outside of work, he's a passionate self-taught artist! Suman brings that same dedication and creativity to everything he does, and we're so lucky to have him on the team. What are your hidden talents? Share them in the comments below!#WDAYLife #APIDAMonth

hey #viewers Today is my day 22, of solving a hard question everyday, for the next 75 days 🚀 Today, I solved a question named as Contains Duplicates III This question was asked in many of the MNCs like Google , Adobe , Apple , Amazon 🏢 Leetcode : https://lnkd.in/gtHwitx7 Approach: • We can simply use a set to store for the window size of indexDiff (let's say k). · Sabse pehle, create a set, to store the values. · Then, iterate in a for loop, and check for the given conditions. · If i is greater than k, then, we need to shrink the window, to get the required pair only. • Then, we can find the position of lowest value, possible for the second condition. • To do so, let's have an short insight over the idea 💡: • abs(position- nums[i]) <= valueDiff ==> -valueDiff <= position - nums[i] <= valueDiff • position-nums[i] >= -valueDiff ==> position>= nums[i]-valueDiff · Then check if, a value is already present in map, then return it. • Then check if the position (index)is possible to find, and also check whether the values of that index and nums[i] follows the second condition. If yes, then return true. • If no such pair has been found yet, then insert the value of nums in the set, and move for further iterations. • Finally return false, if no such pair exist. For further assistance, please find the code attached below 😄 In case of any queries, feel free to ask in the comments 😀 Let's see how far we can get through this challenge. #ab #mai #hashTag #use #karunga #for #good #reach #and #thats #it #75hardchallenge #connections #codechallenge

hey #viewers Today marks my day 30, of solving a hard question everyday, for the next 75 days 🚀 Also, as today's POTD was an hard problem, you can check out my approach Link : https://lnkd.in/gi9BFXve Today, I solved a question named as Count of small numbers after self 🤏 Leetcode : https://lnkd.in/ggHx5FvY This question was previously asked in many of the MNCs like Amazon, Adobe, Google 🏢 Approach: • We can simply use the merge sort algorithm by leveraging it, so that it counts for both left and right halves. · Sabse pehle, create a vector, which stores elements, along their indices. · Then, create a vector, initially with zero, to store the correct counts, for every value. · Then, call for the 'solve' function with left as zero and right as end value's index and the array, and the counting array. · Then check if, left is greater than or equal to right or not. If yes, then return. • Then, check for the both halves of the array, and call the 'merge' function, to get the count of smaller numbers in a sorted manner. • Then, first create a temporary vector, to store the merged elements temporarily. • Then assign three values as i, j, k as I deals with the left half, j deals with the right half, and k for the temporary vector. • Then, check in a while loop, that, if the next value is smaller or not. If yes, then, move to the next value, to preserve sorting. • Else, just assign the value, as the remaining elements in the right half. • Also, remember to copy all the leftover values to the temporary vector. • Finally, the temporary vector is copied back to the temporary vector. • At last, return the counting vector. For further assistance, please find the code attached below 😀 In case of any queries, feel free to ask in the comments 🙂 Let's see how far we can get through this challenge. #ab #mai #hashTag #use #karunga #for #good #reach #and #thats #it #75hardchallenge #connections #codechallenge

hey #viewers Today marks my day 43, of solving a hard question everyday, for the next 75 days 🚀 Today, I solved a question named as Largest Palindromic Product ✖️ Leetcode : https://lnkd.in/gfBktV_G This question was asked in Yahoo Approach: • We can simply use maths, and some logic to solve the question. The key is to find a value such that its value must have factors greater than square root. · Sabse pehle, check if n is 1. If yes, return 9, as it will be the highest possible palindromic product for it. · Then, assign the lowest value, as pow(10, n-1) and highest value, below which a number can be accessed, as pow(10,n)-1. For example, in case of 2, the lowest value possible will be 10, and the highest is 100, exclusively. · Then, iterate in a for loop, from highest value -1, to the lowest value, inclusive, and convert the current value, into string. • Now, copy the string into another, reverse it, and concatenate it to the original string, as we have to check for palindrome. · Then, convert the value of the original string, to a long integer. • Then, iterate inside the outer loop, from highest value - 1, to the square root of the long integer, as to check factors efficiently. • Now, check in the loop, if the value is divisible by j or not. If yes, it means, it does have factors, so just return the final value, MOD 1337. • Finally, return 0, if nothing has been found. For further assistance, please find the code attached below ⬇️ In case of any queries, feel free to ask in the comments 😄 Let's see how far we can get through this challenge. #ab #mai #hashTag #use #karunga #for #good #reach #and #thats #it #75hardchallenge #connections #codechallenge

🌟 Introducing ECGiT's All Employee Featuring Month! 🌟 At ECGiT, we're more than just a team; we're a family! To celebrate our incredible employees and their unique contributions, we're excited to announce the ECGiT All Employee Featuring Month, starting October 19, 2023. 🔦 What's the Spotlight Month? 🔦 Throughout this month, we'll showcase one exceptional ECGiT team member every day. We believe in celebrating the talent, dedication, and diverse skills that make our organization extraordinary. 🌐 Why Are We Doing This? 🌐 Our team is the heart and soul of ECGiT, and we want to share their stories, achievements, and experiences. This initiative is not just about recognizing individuals; it's about inspiring others, fostering connections, and building a stronger ECGiT community. 🤝 How Can You Get Involved? 🤝 Follow our LinkedIn page to stay updated. Engage with our daily posts - show your appreciation, ask questions, and connect with your colleagues. Share the spotlight with your network! Tag your friends who might be interested in our journey. 🎉 Join Us in Celebrating Our Team! 🎉  We can't wait to showcase the incredible people who drive ECGiT's success. Stay tuned for daily employee spotlights, insights, and behind-the-scenes glimpses into the amazing work happening here. Let's make this month memorable together. 🙌 #ECGiTTeamSpotlight #EmployeeRecognition #CompanyCulture

🚀 The #30DaysOfSwiftUI Challenge has officially begun! Are you ready to level up your SwiftUI skills? 💻 Check out the kickoff post and full table of contents here https://lnkd.in/gU2zPTjA #SwiftUI #30DaysOfCode

hey #viewers Today marks my day 38, of solving a hard question everyday, for the next 75 days 🚀 Today, I solved a question named as Perfect Rectangle 👌 Leetcode : https://lnkd.in/g4CYDb2E This question was previously asked in many of the MNCs like Apple, Google 🏢 Approach: • We can simply use an map to keep track of the count of each corner of the the rectangles. · To do so, sabse pehle, create a map to keep track of the count of corners, and the key is a pair of integers, represents the x and y coordinates of a corner point. · The value of the key in the map, represents the count of how many rectangles have that corner point. · Then, iterate over every rectangle, and perform the following steps : • Increment the count of the bottom-left corner. • Decrement the count of the bottom-right corner. • Decrement the count of the top-left corner. • Increment the count of the top-right corner. · Then, iterate through the map, and check if each corner point has a count of 1 or -1. If it doesn't, then return false, as it indicates overlapping between two rectangles. • Else, increment the count for each corner point. • Finally, it checks if exactly 4 corner points have a count of either 1 or -1, which would indicate that these points are the corners of the outer rectangle formed by all the input rectangles. • If this condition is true, it returns true, otherwise, it returns false. For further assistance, please find the code attached below 🙂 In case of any queries, feel free to ask in the comments 😀 Let's see how far we can get through this challenge. #ab #mai #hashTag #use #karunga #for #good #reach #and #thats #it #75hardchallenge #connections #codechallenge

21 Ways In 21 Day #18 Our National Day is Postcode Pride Day. What if we reframed Australia’s National Day to “Postcode Pride Day”. We’d spend our energy and focus on learning to embrace and belong with all the people in our postcode. We would have street parties, community events, sports competitions, arts and food festivals in our neighbourhoods to strengthen our sense of community. We would acknowledge and pay respect to the history of our postcode area. We would tell stories about how it has changed over time. We would create fun competitions that reflect our best values of belonging together. ‘Postcode Pride Day’ is an alternative story to our postcodes being a determining factor of access, privilege and success in life. ‘Postcode Pride Day’ is about being proud of your own community and it can build a strong bond of belonging to flip the script on negative stereotypes. We could develop happiness and belonging indicators to tell a different narrative about our people in our Postcodes. The ‘Postcode Pride Day’ is a chance to build real connections with local people in your place. It is a way of acknowledging that we are the community that call this postcode home. Be proud and keep learning about the place you call home. Be encouraged and encourage others. Coming up next: Our National Day is Generations Day.