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As I prepare for upcoming technical interviews, I'm revisiting the basics to ensure I have a solid foundation. One crucial concept in TypeScript is generics, which play a vital role in creating flexible, reusable, and type-safe code. However, before delving into generics, it is easier to start with utility types. Utility types are also generics, but they are already defined and ready to use. Here are some basic utility types: — Readonly<Type>: Type can be any type or interface, such as a string, number, or a complex object. This utility constructs a type with all properties of Type set to readonly, meaning the properties cannot be reassigned. — Pick<Type, Keys>: Constructs a type by selecting the set of properties Keys (string literals or a union of string literals) from Type. — NonNullable<Type>: Constructs a type by excluding null and undefined from Type. Read more and see code examples in the official documentation.
Documentation - Utility Types
typescriptlang.org
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Day 4 of #75DaysOfCode Challenge 🌟 Continuing the journey with Day 4 of my #75DaysOfCode Challenge! 🌟 Today, I worked on generating unique subsets, finding specific combinations, and exploring letter combinations from phone numbers. These tasks deepened my understanding of recursion and backtracking. --- 1️⃣ Subset Sum II : Given an integer array `nums` that may contain duplicates, return all possible subsets (the power set) without duplicates. Return the solution in any order. Approach : - Base Case : If we reach the end of the array, add the current subset to the result. - Recursive Case : Sort the array to handle duplicates. Recursively include or exclude each element, ensuring no duplicate subsets. --- 2️⃣ Combination Sum III : Find all valid combinations of `k` numbers that sum up to `n` using numbers 1 through 9, each used at most once. Return unique combinations in any order. Approach : - Base Case : If the combination contains `k` numbers and sums to `n`, add it to the result. - Recursive Case : Iterate through numbers 1 to 9, recursively building combinations by including each number only once and ensuring the combination length and sum constraints are met. --- 3️⃣ Letter Combination of a Phone Number : Given a string containing digits from 2-9, return all possible letter combinations that the number could represent using a mapping of digits to letters (like on telephone buttons). Approach : - Base Case : If the current combination length equals the input digits length, add the combination to the result. - Recursive Case : Use a mapping of digits to letters and recursively build combinations by appending each possible letter for the current digit. --- 🔍 Learning Path : - Unique Subsets and Combinations : Practiced generating unique subsets and combinations while managing duplicates and constraints. - Recursive Depth Management : Focused on managing recursive function depth to meet all conditions without redundancy. - Mapping and Backtracking : Applied mapping techniques and backtracking to explore all possible combinations efficiently. 💡 "The future belongs to those who believe in the beauty of their dreams." - Eleanor Roosevelt 💡 🖋️Today's challenges brought new insights and valuable learning experiences. With each problem solved, I’m becoming more skilled at handling complex issues with greater confidence. Let’s keep up the pace, stay dedicated, and embrace every obstacle that comes our way! 💻Check out the problems here: https://lnkd.in/gnzNnawd https://lnkd.in/gPz7x2nG https://lnkd.in/ghH9gs-4 #Coding #ProblemSolving #DailyChallenge #StayMotivated #DSA #TechJourney #Recursion #Dedication #KeepGrowing #NeverGiveUp #striver #takeuforward #ncu #thenorthcapuniversity #leetcode #codingninjas #striverA2ZDSAsheet #75daysofcode #75dayscodechallenge
Subsets II - LeetCode
leetcode.com
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Day 2 of #75DaysOfCode Challenge 🌟 Continuing the journey with Day 2 of my #75DaysOfCode Challenge! 🌟 Today's focus was all about generating sequences and understanding their properties through recursion. Here’s what I dived into: --- 1️⃣ Generate Parentheses: Creating valid combinations of parentheses! This problem taught me how to explore all possible combinations and ensure each one is balanced. Approach: - Base Case: If the number of opening and closing parentheses used is equal to `n`, add the combination to the result. - Recursive Case: Add an opening parenthesis if it doesn't exceed `n`, and add a closing parenthesis if it doesn't exceed the number of opening parentheses used. --- 2️⃣ Print All Subsequences/Power Set: Generating all possible subsequences of a given string or array. This challenge was a great way to practice recursion by exploring each element's inclusion or exclusion. Approach: - Base Case: If we reach the end of the array/string, print the current subsequence. - Recursive Case: For each element, make two recursive calls - one including the current element and one excluding it. --- 3️⃣ Count All Subsequences with Sum K: Counting the number of subsequences that sum up to a given value `K`. This problem combined recursion with arithmetic checks to find valid subsequences. Approach: - Base Case: If we reach the end of the array, check if the current sum equals `K`. - Recursive Case: For each element, recursively include it in the current sum or exclude it, and count the valid subsequences. --- 🔍 Learning Path: - Balancing Constraints in Recursion: Focused on generating combinations with specific constraints, ensuring all conditions are met. - Backtracking Techniques: Learned how to generate all possible combinations and subsequences efficiently. - Recursive Arithmetic: Applied arithmetic checks within recursive functions to solve counting problems. 💡 "Success is not final, failure is not fatal: It is the courage to continue that counts." - Winston Churchill 💡 Today was filled with new insights and deeper understanding. With each problem, I'm building a stronger foundation and growing more confident in my abilities. Let's keep the momentum going, stay committed, and embrace every challenge! Check out the problems here: https://lnkd.in/gTQGckT8 https://lnkd.in/gVV_5WGK https://lnkd.in/gQ6rYKbi --- #Coding #ProblemSolving #DailyChallenge #StayMotivated #DSA #TechJourney #Recursion #Dedication #KeepGrowing #NeverGiveUp #striver #takeuforward #ncu #thenorthcapuniversity #northcapuniversity #leetcode #codingninjas #striverA2ZDSAsheet #75daysofcode #75dayscodechallenge
Generate Parentheses - LeetCode
leetcode.com
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🚀 180 days Leetcode 𝐂𝐡𝐚𝐥𝐥𝐞𝐧𝐠𝐞 🚀 #leetcodechallenge 📆 18 Days Completed ✅ / 180 Days Planned 🚀 Problem name: 485. Max Consecutive Ones Problem Link: https://lnkd.in/dyjDdBkZ For your learning you can follow this link from @w3school and Striver: https://lnkd.in/dD_c2pbb The solution is an interesting one. 1. Initialization: The code defines a class named Solution. Inside the class, it creates a method named findMaxConsecutiveOnes that takes a list of integers (nums) as input. Two variables are initialized: c: This variable will be used to keep track of the current count of consecutive ones encountered so far. It's set to 0 initially. maximum: This variable will store the maximum count of consecutive ones found in the entire array. It's also set to 0 initially. 2. Iteration through the input array (nums): The code uses a for loop to iterate through each element (nums[i]) in the input list nums. i acts as the loop counter, starting from 0 and incrementing until it reaches the length of the list (len(nums)). 3. Checking for consecutive ones: Inside the loop, an if statement checks if the current element (nums[i]) is equal to 1: If it is 1, it means we've encountered a consecutive one. The c variable is incremented by 1 to keep track of the current count. 4. Handling breaks: The else block of the if statement handles cases where the current element is not 1: This indicates a break in the sequence of consecutive ones. As a result, the c variable is reset to 0, starting the count from scratch for the next sequence of ones. 5. Updating the maximum count: After checking each element, the max function is called. This function takes two arguments: the current maximum value and the current c value (count of consecutive ones). The max function returns the larger of these two values. This result is then assigned back to the maximum variable, ensuring that it always holds the highest count of consecutive ones found so far. 6. Returning the result: After iterating through the entire list, the loop completes. The maximum variable, which now holds the largest count of consecutive ones encountered in the array, is returned from the findMaxConsecutiveOnes method. Source code: https://lnkd.in/dqfTZMic Leetcode profile: https://lnkd.in/d5qkVyRp Do let me know your approach. Any suggestion is highly appreciated. Happy Learning! 💡 #codingchallenge #PlacementChallenge #consistencymatters #LearningJourney #HustleHard #DailyTasks #GoalAchievement #coding #programing #development #computervision #computerscience #internship #challenge
Max Consecutive Ones - LeetCode
leetcode.com
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🚀 180 days Leetcode 𝐂𝐡𝐚𝐥𝐥𝐞𝐧𝐠𝐞 🚀 #leetcodechallenge 📆 19 Days Completed ✅ / 180 Days Planned 🚀 Problem name: 485. Max Consecutive Ones Problem Link: https://lnkd.in/dcecjh8w For your learning you can follow this link from @w3school and Striver: https://lnkd.in/dD_c2pbb Dictionary for Counting: mpp = defaultdict(int) creates an empty dictionary named mpp using defaultdict. defaultdict ensures that for any key not yet present in the dictionary, it will automatically create a default value of 0 (an integer in this case). Counting Occurrences: The for num in nums: loop iterates through each number (num) in the input list nums. Inside the loop, mpp[num] += 1 increments the count associated with num in the dictionary mpp. If num is not present in mpp yet, it will be added with a count of 1. This effectively builds a frequency table of the numbers in the array. Checking for Majority: The outer for num, count in mpp.items(): loop iterates through each key-value pair (num and its corresponding count) in the dictionary mpp. Inside the loop, the condition if count > (len(nums) // 2): checks if the count (count) of the current number (num) is greater than half the length of the input array (len(nums) // 2). Rationale: A majority element must appear more than half the times in the array. If the count exceeds half the array size, it indicates a potential majority element. Returning the Majority Element (or -1): If the condition in step 4 is true, it means a majority element has been found. The code immediately returns the number (num) using return num. If the loop completes without finding a count exceeding half the array size, it means no majority element exists. The code returns -1 to indicate this. Key Points: This code assumes that a majority element always exists in the input array. The time complexity of this approach is O(n) due to iterating through the array once to build the frequency table and potentially once more to find the majority element. The space complexity is O(n) because the dictionary mpp can potentially store all unique elements in the array. Alternative Approaches: Boyer-Moore Voting Algorithm: This is a more efficient algorithm that uses linear time and constant extra space to find the majority element. It relies on the concept of cancellation when encountering non-majority elements. Source code: https://lnkd.in/d3VaKqCy Leetcode profile: https://lnkd.in/d5qkVyRp Do let me know your approach. Any suggestion is highly appreciated. Happy Learning! 💡 #codingchallenge #PlacementChallenge #consistencymatters #LearningJourney #HustleHard #DailyTasks #GoalAchievement #coding #programing #development #computervision #computerscience #internship #challenge
Majority Element - LeetCode
leetcode.com
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The second chapter of my series on 'Cracking the Coding Interview' is now live! 🚀 Cracking the Coding Interview [Ch.2: Linked Lists] Check out the article here: https://lnkd.in/dwQ8yxdC
Cracking the Coding Interview [Ch.2: Linked Lists]
youssefhesham.hashnode.dev
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Senior Data and Systems Architect | Gen AI | AI and Software Architecture Consultations | Career Guidance | Ex Vice President at JP Morgan Chase | Startup Mentor | Angel Investor | Author
𝑺𝒕𝒂𝒓𝒕𝒊𝒏𝒈 𝒂 𝒔𝒆𝒓𝒊𝒆𝒔 𝒐𝒏 𝑺𝒚𝒔𝒕𝒆𝒎 𝑫𝒆𝒔𝒊𝒈𝒏 - No, not articles on how to clear system design interviews. Nothing as trivial as that. Instead, I will be writing articles on 𝗿𝗲𝗮𝗱 𝘄𝗼𝗿𝗹𝗱 𝗽𝗿𝗼𝗯𝗹𝗲𝗺𝘀 with solutions I proposed - Solutions that were suboptimal but met the Client's requirements. I encourage people to think about more viable solutions to the problem statements, it will be a good exercise to understand what sort of real world problems can present themselves when you work as a Software Architect. Article 1 in the series: https://lnkd.in/d7fhauEP #systemdesign #softwarearchitecture
System Design Series: Event Courier — Create Events dynamically
riteshshergill.medium.com
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🚀 Day 26 of #75DaysofCode Challenge 🚀 Today's coding journey focused on fundamental data structures, offering a deep dive into stacks and queues. Here’s a quick overview of what I tackled: 1. Implement Stack using Arrays : 📚 Created a stack data structure using arrays, reinforcing my understanding of stack operations and array manipulations. https://lnkd.in/gTJV96kN 2. Implement Queue using Arrays : 🚦 Developed a queue data structure using arrays, enhancing my grasp of queue operations and efficient array usage. https://lnkd.in/gNR-9-jr 3. Implement Stack using Queue : 🔄 Designed a stack utilizing two queues, a challenging problem that deepened my comprehension of how stacks and queues can be interconverted. https://lnkd.in/g4Y42pGU 4. Implement Queue using Stack : 🔁 Constructed a queue using two stacks, further exploring the relationship between these two essential data structures. https://lnkd.in/gtF7zFXV 🔍 Learning Outcomes : - Implement Stack using Arrays : Strengthened my understanding of stack operations and array manipulation techniques. - Implement Queue using Arrays : Enhanced my ability to implement queue operations and efficiently manage arrays. - Implement Stack using Queue : Gained insights into simulating stack behavior using queue operations. - Implement Queue using Stack : Mastered the techniques to simulate queue operations using stacks. 💡 "However difficult life may seem, there is always something you can do and succeed at." - Stephen Hawking Today's challenges pushed me to explore the intricate connections between stacks and queues. Each problem helped me refine my data structure skills, reinforcing my ability to tackle complex coding tasks. Embracing these challenges with determination and enthusiasm makes me more proficient and confident. Let’s keep pushing forward with passion and persistence, turning every obstacle into a stepping stone for success! #Coding #ProblemSolving #DailyChallenge #StayMotivated #DSA #TechJourney #DataStructures #Stack #Queue #Dedication #KeepGrowing #NeverGiveUp #Striver #TakeUForward #NCU #TheNorthCapUniversity #LeetCode #CodingNinjas #StriverA2ZDSASheet #75DaysOfCode #75DaysCodeChallenge
Implement Queue using Stacks - LeetCode
leetcode.com
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Level Up Your Kotlin Coding With Digma and OpenTelemetry Join us for a livestream with JetBrains where we explore applying OpenTelemetry, OSS-free tooling, and IDE features and functionality to elevate your Kotlin programming experience to the next level. Tuesday, January 23, 16:00 UTC https://lnkd.in/dRnbKgDw #kotlin #jetbrains #java #springboot #quarkus #OpenTelemetry
Live Webinar -
info.jetbrains.com
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#Design_Patterns are solutions and best practices providing reusable solutions to common problems. 🛠️ They enhance code maintainability, scalability, and readability, making it easier for developers to collaborate and maintain codebases over time. 🗄️ Understanding and applying design patterns like the Factory and Factory Method patterns is essential for leading to more efficient development processes and higher-quality software products. In Addition, it is very important for #backed #tecnicanl #interview and #interviews in general. 🚀🔧 #SoftwareDevelopment #DesignPatterns #TechInnovation #ProgrammingLanguages #SoftwareEngineering #CodeQuality #DeveloperProductivity #CodingBestPractices
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