LeetCode Challenge Day 87: Check Strings with Operations II

✅ Day 92 of 100 Days LeetCode Challenge Problem: 🔹 #2011 – Final Value of Variable After Performing Operations 🔗 https://lnkd.in/gX-JQNUJ Learning Journey: 🔹 Today’s problem was about evaluating a sequence of increment and decrement operations. 🔹 I initialized a variable ans = 0 to track the value. 🔹 Used a hashmap to map each operation to its effect: • "++X" and "X++" → +1 • "--X" and "X--" → -1 🔹 Iterated through the operations and updated ans accordingly. 🔹 Returned the final computed value. Concepts Used: 🔹 HashMap / Dictionary 🔹 String Matching 🔹 Simple Simulation Key Insight: 🔹 Instead of using multiple condition checks, mapping operations to values simplifies logic and improves readability. Complexity: 🔹 Time: O(n) 🔹 Space: O(1) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #Python #ProblemSolving #LearningInPublic #TechCareers

✅ 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 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

✅ 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 86 of 100 Days LeetCode Challenge Problem: 🔹 #2839 – Check if Strings Can be Made Equal With Operations I 🔗 https://lnkd.in/gG4CaJpf Learning Journey: 🔹 Today’s problem involved determining if two strings can be made equal using swaps where indices differ by 2. 🔹 I observed that only certain positions can swap among themselves: • Even indices (0, 2) • Odd indices (1, 3) 🔹 Instead of simulating swaps, I directly generated possible permutations of s2 using these allowed swaps. 🔹 Then, I checked if s1 matches any of these valid transformed versions of s2. 🔹 If a match is found, return True; otherwise, False. Concepts Used: 🔹 String Manipulation 🔹 Permutations (restricted swaps) 🔹 Pattern Observation Key Insight: 🔹 Swaps are limited to positions with the same parity, meaning even and odd indices form independent groups. 🔹 This reduces the problem to checking a small set of possible rearrangements instead of brute-force simulation. Complexity: 🔹 Time: O(1) (fixed string length = 4) 🔹 Space: O(1) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

🚀 Day 39/60 — LeetCode Discipline Problem Solved: Sqrt(x) Difficulty: Easy Today’s challenge was to compute the square root of a number without using built-in functions. Instead of brute force, I used Binary Search — a classic, elegant approach that narrows down the answer efficiently. 💡 Key Learnings: • Binary Search application beyond arrays • Handling edge cases (x < 2) • Avoiding overflow using conditions carefully • Finding floor value of square root • Optimized thinking over brute force ⚡ Performance: Runtime: 4 ms Like walking in a foggy path, I didn’t see the answer directly… But step by step— cutting the search space in half— the truth revealed itself. That’s the beauty of algorithms. #LeetCode #60DaysOfCode #DSA #BinarySearch #ProblemSolving #CodingJourney #Python #Consistency #TechGrowth

🚀 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

🚀 Day 42/100 – LeetCode Challenge Solved 206. Reverse Linked List today! 🔍 Problem Summary: Given the head of a singly linked list, reverse the list and return the new head. 💡 Approach: Implemented an iterative solution using three pointers (prev, curr, next) to reverse the links efficiently in-place. Also explored the recursive approach to strengthen understanding of linked list manipulation. ⚡ Complexity: • Time Complexity: O(n) • Space Complexity: O(1) (Iterative) 📌 Key Takeaways: Importance of pointer manipulation Handling edge cases like empty list Difference between iterative vs recursive approaches Consistency is key — one step closer to mastering Data Structures & Algorithms! 💪 #LeetCode #100DaysOfCode #DataStructures #Algorithms #CodingJourney #Python #ProblemSolving #LinkedList

🚀 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 10/100 – DSA Challenge Today’s problem: Move Zeroes (LeetCode 283) What I Learned: The goal was to move all zeroes to the end of an array while maintaining the relative order of non-zero elements — and importantly, doing it in-place. Key Idea: Two-Pointer Technique I used a two-pointer approach: One pointer (fast) iterates through the array Another pointer (slow) tracks where the next non-zero element should go Whenever a non-zero element is found, it is swapped with the element at the slow pointer, ensuring all non-zero elements are shifted forward while zeroes naturally move to the end. Why this approach? Maintains order of elements Works in O(n) time complexity Uses O(1) extra space (in-place) Takeaway: This problem reinforced how powerful the two-pointer technique is for array manipulation problems, especially when constraints require in-place operations. Looking forward to tackling more problems and improving consistency! #Day10 #100DaysOfCode #DSA #Python #CodingJourney #LeetCode #ProblemSolving