Day 39 DSA Streak: Number Complement Solution

🚀 𝐃𝐚𝐲 89/100 – 𝐈𝐬 𝐒𝐮𝐛𝐬𝐞𝐪𝐮𝐞𝐧𝐜𝐞 Today’s problem was 𝐈𝐬 𝐒𝐮𝐛𝐬𝐞𝐪𝐮𝐞𝐧𝐜𝐞 — a simple yet fundamental problem to understand two-pointer technique. 🔍 𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠: We can efficiently check if one string is a subsequence of another using two pointers. 💡 𝐂𝐨𝐫𝐞 𝐈𝐝𝐞𝐚: Traverse both strings Move pointer of s only when characters match Always move pointer of t  𝐖𝐡𝐲 𝐢𝐭 𝐰𝐨𝐫𝐤𝐬? We try to match characters of s in order within t without disturbing sequence. ⚡ 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Initialize two pointers i (for s) and j (for t) If characters match → move both Else → move only j If i reaches end of s → subsequence exists ⏱️ 𝐓𝐢𝐦𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: 𝐎(𝐧) 📦 𝐒𝐩𝐚𝐜𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: 𝐎(1) #Day89 #100DaysOfCode #Java #DSA #LeetCode #TwoPointers #CodingJourney

Day 33/50 🚀 — Valid Palindrome (Two Pointer Approach) Today’s problem was a great mix of string manipulation + two pointers. 🔹 Ignored non-alphanumeric characters 🔹 Handled case-insensitivity 🔹 Compared characters from both ends efficiently Key insight: Instead of preprocessing the string, we can optimize in-place using two pointers, skipping unwanted characters on the go. 💡 This improves both readability and performance. Performance: ⚡ Runtime: 2 ms (99%+) 📦 Memory: Efficient #Day33 #LeetCode #TwoPointers #DSA #Java #CodingJourney #50DaysOfCode #ProblemSolving

Can you predict the output? 🤔 int x = 5; System.out.println(x++ + ++x); Drop your answer below 👇 I’ll share the explanation in a few hours ⏳ #Java #CodingChallenge #DSA

Day 25/30 – LeetCode streak Problem: Complement of Base 10 Integer  You need to flip all bits in 'n'’s binary representation, but only up to its most significant '1' (no leading zeros). Core idea * The complement should only flip bits within the bit-length of 'n'. For example, '5' is '101₂'; its complement is '010₂' = '2', not some infinite stream of ones. * Build a mask of the form '111...1' that has the same length as 'n' in binary. Then 'n ^ mask' flips exactly those bits. * Special case: if 'n == 0', binary is "0" and complement is "1", so return '1'. Day 25 takeaway: This is a clean bitmask + XOR pattern: instead of flipping bits one-by-one, construct a full '111...1' mask up to the MSB and XOR once to get the complement. #leetcode #dsa #java #bitmanipulation #consistency

🚀 Day #82/100 – 𝐑𝐞𝐦𝐨𝐯𝐞 𝐄𝐥𝐞𝐦𝐞𝐧𝐭 (LeetCode) Today’s problem was Remove Element — a simple yet important question to understand in-place array manipulation. 🔍 𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠: We don’t need extra space! We can solve this using a two-pointer approach efficiently.  𝐖𝐡𝐲 𝐢𝐭 𝐰𝐨𝐫𝐤𝐬? We overwrite unwanted elements and keep only the valid ones at the beginning of the array. ⚡ 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Initialize k = 0 Traverse array: If element ≠ val → place it at index k Increment k Return k (new length) ⏱️ 𝐓𝐢𝐦𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: O(n) 📦 𝐒𝐩𝐚𝐜𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: O(1) #Day82 #100DaysOfCode #Java #DSA #LeetCode #TwoPointers #CodingJourney

🚀100 Days of Code Day - 17 Problem Solved: Letter Combinations of a Phone Number Today I worked on an interesting backtracking problem where I generated all possible letter combinations from a given digit string (2–9), based on the classic phone keypad mapping. 🔍 Key Learnings: • Understood how recursion helps explore all possible combinations • Strengthened my knowledge of backtracking techniques • Improved problem-solving and logical thinking 💡Approach: Used a recursive backtracking method to build combinations step by step, exploring all possible letter mappings for each digit. 📌 Example: Input: "23" Output: ["ad","ae","af","bd","be","bf","cd","ce","cf"] This problem is a great example of how powerful recursion can be when dealing with combinations. #Java #DSA #Backtracking #ProblemSolving #CodingJourney #LeetCode

Day: 72/365 📌 LeetCode POTD: Count Submatrices With Equal Frequency of X and Y Medium Key takeaways/Learnings from this problem: 1. The key idea is converting the problem into numbers (like +1 and -1) so equal frequency turns into a sum = 0 problem. 2. 2D prefix sums really help here—you can quickly get submatrix sums without recalculating every time. 3. It’s a nice extension of 1D subarray problems to 2D, so knowing those basics pays off big time. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

Solved a problem where we need to check if two strings can be made equal using a special operation. The rule is: you can swap characters only if the distance between their positions is even. So basically, characters at even indices can only swap among themselves, and same for odd indices. Idea: Instead of actually swapping, I just counted characters separately for even and odd positions in both strings. If both match, then it’s possible — otherwise not. Simple concept, but interesting twist! 😊 #LeetCode #DSA #Java #ProblemSolving #CodingJourney

Day 3 #SDE Binary Search based problems Solved: • Pow(x, n) • Find Peak Element Key Learnings: • Pow(x, n): Implemented fast exponentiation using Binary Exponentiation to reduce time complexity from O(n) to O(log n). • Find Peak Element: Used binary search to identify a peak efficiently instead of scanning the entire array. #LeetCode #DSA #BinarySearch #Java #SoftwareEngineering

🚀 Day 47/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 448. Find All Numbers Disappeared in an Array Used an in-place marking technique by treating indices as a hash map and marking visited numbers as negative to identify missing elements. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) (excluding output list) Strengthening understanding of array manipulation and in-place hashing tricks. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency