LeetCode Challenge Day 92: Final Value of Variable After Operations

✅ Day 90 of 100 Days LeetCode Challenge Problem: 🔹 #476 – Number Complement 🔗 https://lnkd.in/gzE6gM7d Learning Journey: 🔹 Today’s problem focused on finding the complement of a number by flipping its binary bits. 🔹 I first converted the integer to its binary representation using bin(num)[2:]. 🔹 Then, I created a helper function to flip each bit: • '0' → '1' • '1' → '0' 🔹 After generating the flipped binary string, I converted it back to an integer using int(..., 2). 🔹 Returned the final complemented value. Concepts Used: 🔹 Binary Representation 🔹 Bit Manipulation 🔹 String Traversal 🔹 Base Conversion Key Insight: 🔹 The complement operation is essentially a bitwise NOT, but only within the significant bits of the number (ignoring leading zeros). 🔹 Converting to binary simplifies the flipping logic for beginners. Complexity: 🔹 Time: O(log n) 🔹 Space: O(log n) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #Python #ProblemSolving #LearningInPublic #TechCareers

🚀 Just solved the “Valid Number” problem on LeetCode! This problem looks simple at first glance—but handling edge cases like decimals, signs, and exponents makes it a great test of attention to detail and logical thinking. ✅ Key takeaways: Careful handling of edge cases is crucial Validating input step-by-step can simplify complex parsing problems Writing clean, readable logic beats overcomplicated solutions 💡 Performance: ⚡ Runtime: 3 ms 🧠 Efficient space usage ✅ All test cases passed Problems like this remind me that consistency in practice is what builds strong problem-solving skills. On to the next one! 🔥 #LeetCode #Coding #Python #ProblemSolving #SoftwareEngineering #AIEngineerJourney link of #Solution :- https://lnkd.in/ga9b5pVb

🚀 Day 83 of #100DaysOfCode 📌 Problem Solved: Letter Combinations of a Phone Number (LeetCode 17) Today’s challenge was all about exploring backtracking and how recursive decision-making builds combinations efficiently. 🔍 Given a string of digits (2–9), the goal is to generate all possible letter combinations based on the classic phone keypad mapping. 💡 Key Learning: Instead of trying every possible string blindly, backtracking helps us build combinations step-by-step and explore only valid paths — making the solution clean and efficient. ⚡ Highlights: ✔️ Used recursion to explore all possibilities ✔️ Implemented a digit-to-letter mapping ✔️ Built combinations incrementally ✔️ Achieved optimal runtime performance 📈 Result: ✅ All test cases passed ⚡ 0 ms runtime (Beats 100%) Consistency is starting to compound — one problem at a time. 💪 #LeetCode #DSA #Python #CodingJourney #Backtracking #100DaysOfCode

✅ Day 98 of 100 Days LeetCode Challenge Problem: 🔹 #338 – Counting Bits 🔗 https://lnkd.in/gXdNxX66 Learning Journey: 🔹 Today’s problem focused on counting the number of 1s in the binary representation of numbers from 0 to n. 🔹 I initialized an array ans of size n+1. 🔹 For each number i, I converted it to binary using bin(i)[2:]. 🔹 Counted the number of '1' bits by iterating through the binary string. 🔹 Stored the count in ans[i] and returned the final array. Concepts Used: 🔹 Bit Manipulation (Binary Representation) 🔹 Array Traversal 🔹 String Processing 🔹 Brute Force Approach Key Insight: 🔹 Each number’s bit count can be computed independently. 🔹 Converting to binary and counting '1' works, but can be optimized further using DP or bit tricks. Complexity: 🔹 Time: O(n log n) (binary conversion for each number) 🔹 Space: O(n) #LeetCode #Algorithms #DataStructures #100DaysOfCode #Python #CodingJourney #ProblemSolving #LearningInPublic

✅ Day 95 of 100 Days LeetCode Challenge Problem: 🔹 #869 – Reordered Power of 2 🔗 https://lnkd.in/gkNXaSFM Learning Journey: 🔹 Today’s problem focused on checking whether the digits of a number can be rearranged to form a power of 2. 🔹 I created a helper function to extract and store the digits of a number. 🔹 Then I sorted the digits of the input number for comparison. 🔹 Next, I generated powers of 2 iteratively and compared their sorted digit lists with the input. 🔹 If any match was found, I returned True. Otherwise, continued until the digit length exceeded the input. Concepts Used: 🔹 Digit Extraction 🔹 Sorting 🔹 Simulation of Powers of 2 🔹 Brute Force Optimization Key Insight: 🔹 Instead of generating permutations (which is expensive), sorting digits allows quick comparison. 🔹 Any valid rearrangement must have the same digit frequency as some power of 2. Complexity: 🔹 Time: O(log n * d log d) 🔹 Space: O(d) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #Python #ProblemSolving #LearningInPublic #TechCareers

🚀 Day 28 of Problem Solving Journey Today, I worked on an interesting problem: Group Anagrams 🔍 Problem Statement: Given an array of strings, group the anagrams together. Anagrams are words that have the same characters but arranged differently. 💡 Approaches I explored: ✅ Approach 1: Character Frequency Count (Optimized) Used a fixed-size array (26 letters) to count character occurrences Converted the count into a tuple to use as a dictionary key Achieved an efficient time complexity of O(n * k) ✅ Approach 2: Sorting Strings Sorted each string and used it as a key Simple and intuitive approach Time complexity: O(n * k log k) 📌 Key Learning: Understanding how hashing works with different representations (frequency vs sorted string) helps in optimizing solutions. ⚡ Takeaway: There are always multiple ways to solve a problem, but choosing the most efficient one makes a difference in real-world applications and interviews. 💻 Tech Used: Python | HashMap | Arrays #Day28 #ProblemSolving #Python #DataStructures #Algorithms #CodingJourney #100DaysOfCode

🚀 Day 36 – LeetCode Journey Today’s problem: String to Integer (atoi) ✔️ Handled leading spaces and signs (+/-) ✔️ Processed numeric characters step by step ✔️ Managed overflow conditions within 32-bit integer range 💡 Key Insight: Carefully handling edge cases (like spaces, signs, and overflow) is just as important as the core logic. Small conditions can make a big difference in correctness. This problem strengthened my understanding of string parsing, edge cases, and boundary conditions. Learning to write robust and reliable code every day 💪🔥 #LeetCode #Day36 #Strings #EdgeCases #Python #ProblemSolving #CodingJourney #100DaysOfCode

My thesis was stuck. A matrix had the wrong shape and I had no idea why. I could have printed the entire dataset to find the error. I did not. Instead I used Python's debugger. One breakpoint. One look at the intermediate state. Wrong dimensions. Found in seconds. That moment changed how I work. Not because debugging saved my thesis. But because it taught me something I still use every day: You do not need to see all the data to understand what is wrong. You just need to see the right data at the right moment. Since then, every time a pipeline breaks or a model behaves unexpectedly, I reach for the debugger first. Not print statements. Not guesswork. A breakpoint. An intermediate result. A clear answer. Debugging is not a last resort. It is the fastest way to understand what your code is actually doing. What is your go-to strategy when something breaks unexpectedly? #Python #Debugging #DataScience #MachineLearning #FreelanceDataScientist

🚀 Day 75 of #100DaysOfCode 🔥 LeetCode 179 – Largest Number 💡 Problem: Given a list of non-negative integers, arrange them such that they form the largest possible number. 🧠 Key Insight: Normal sorting won't work here ❌ We need a custom comparator based on string concatenation. 👉 Compare: - ""a + b"" vs ""b + a"" - Whichever gives a larger value should come first. ⚙️ Approach: 1. Convert numbers to strings 2. Sort using custom comparison logic 3. Join the result 4. Handle edge case (like "[0,0] → "0"") ⚡ Complexity: - Time: O(n log n) - Space: O(n) 🎯 Result: ✅ Accepted ⚡ Runtime: 0 ms (100%) 📌 Lesson Learned: Sometimes sorting logic depends on combination, not value. #LeetCode #Python #CodingJourney #DSA #100DaysOfCode #Sorting #ProblemSolving

🚀 Solved a great problem today: “Consecutive 1’s Not Allowed” At first glance, it looked like a simple binary string problem… but it quickly turned into a lesson in pattern recognition and dynamic thinking. 📌 What the problem was about: Count all binary strings of length n such that no two 1’s are consecutive. 💡 What I learned: Instead of brute forcing all combinations (which would be exponential), the key was to observe a pattern: If a string ends with 0 → we can add 0 or 1 If it ends with 1 → we can only add 0 This leads to a recurrence: 👉 dp[n] = dp[n-1] + dp[n-2] Which is basically the Fibonacci pattern in disguise. 🧠 Big takeaway: Many problems are not about coding harder… they’re about seeing the hidden pattern behind the problem. This was a reminder that: Brute force is rarely the answer Thinking in terms of state transitions is powerful Optimization often comes from observation, not syntax 📷 Sharing my solution screenshot below 👇 #DataStructures #DynamicProgramming #ProblemSolving #Python #LearningInPublic #DataAnalyticsJourney