LeetCode Challenge: Reverse String in Java

The Two-Pointer streak continues! Today’s LeetCode Problem: Is Subsequence. After using multiple pointers to sort arrays and move zeroes, I applied the exact same pattern to string manipulation today. The challenge was figuring out if a shorter string (s) is a valid subsequence of a longer string (t) without disturbing the relative order of the characters. Instead of generating all possible subsequences (which would be a massive O(2ⁿ) performance drain), the Two-Pointer approach solves this beautifully in a single pass. Knocking this out in O(n) time and O(1) space is another great reminder of how incredibly versatile this algorithmic pattern is for both arrays and strings. #DSA #Java #LeetCode #IsSebsequenceProblem

Taking the Two-Pointer technique to LeetCode! Today's challenge: Remove Duplicates from a Sorted Array. After practicing the Two-Pointer pattern for reversing arrays and swapping 0s and 1s, I applied it to a classic LeetCode problem. The challenge? Remove duplicates from an array in-place with O(1) extra memory. Because the array is already sorted, all duplicates are grouped together. • The Slow Pointer (The Writer): Keeps track of where the next unique element should be placed. • The Fast Pointer (The Reader): Scans ahead through the array to find new, unique numbers. • The Logic: If the Reader finds a duplicate, it just skips it. But the moment the Reader finds a new number, the Writer records it at the front of the array and steps forward. #DSA #Java #LeetCode #RemoveDuplicate

🚀 Day 52/100 Today’s problem was based on String Manipulation — reversing the first k characters of a string. 🧠 What I learned: - How to efficiently manipulate strings - Using "StringBuilder" for reversal in Java - Writing clean and optimized code 💡 Approach: Reversed the first k characters and appended the remaining string. ⚡ Key Insight: Even simple problems help strengthen fundamentals, which are crucial for solving complex problems later. 👨💻 Code (Java): class Solution { public String reversePrefix(String s, int k) { String first = new StringBuilder(s.substring(0, k)).reverse().toString(); String rest = s.substring(k); return first + rest; } } 📈 Consistency is the key — showing up every day matters more than perfection. #Day52 #CodingChallenge #Java #DSA #LearningJourney #Consistency #100DaysOfCode

Your switch-case multiplied again. Here's the refactoring that kills it. When your type code only affects data - not behavior - you can replace switch-case with a class. Not if-else chains, not pattern matching. A class where each type is an instance that carries its own data. This works in Java, JavaScript, Kotlin, C# - any language with classes. The idea is simple: if price, weight, and label all depend on a type code, why scatter that knowledge across three switch statements? Put it in one place. Let the constructor enforce completeness. Add a new type? Create a new instance. The compiler tells you what's missing. No grep. No "did I forget a case somewhere". (When types also change behavior, you need a different approach. That's the next post, next week.) I walked through the full refactoring with a pizza example. From int constants to enums to smart classes. Link in the first comment. #cleancode #refactoring #softwarecraft #codequality

Day 68 — LeetCode Progress Problem: Check if an array is “special” — meaning every pair of adjacent elements has different parity (one even, one odd). Required: Given an integer array, return true if no two adjacent elements have the same parity, otherwise return false. Idea: If two neighboring numbers are both even or both odd, the condition fails immediately. So just compare parity of adjacent elements throughout the array. Approach: Traverse the array starting from index 1 For each element: Compare it with the previous element If both have same parity → return false If the loop completes → return true Time Complexity: O(n) Space Complexity: O(1) Clean, simple, and a good reminder that not every problem needs something fancy — sometimes it’s just about spotting the pattern. #LeetCode #DSA #Java #Arrays #ProblemSolving #CodingJourney

🔥 𝗗𝗮𝘆 𝟵𝟯/𝟭𝟬𝟬 — 𝗟𝗲𝗲𝘁𝗖𝗼𝗱𝗲 𝗖𝗵𝗮𝗹𝗹𝗲𝗻𝗴𝗲 𝟭𝟲𝟬𝟴. 𝗦𝗽𝗲𝗰𝗶𝗮𝗹 𝗔𝗿𝗿𝗮𝘆 𝗪𝗶𝘁𝗵 𝗫 𝗘𝗹𝗲𝗺𝗲𝗻𝘁𝘀 𝗚𝗿𝗲𝗮𝘁𝗲𝗿 𝗧𝗵𝗮𝗻 𝗼𝗿 𝗘𝗾𝘂𝗮𝗹 𝗫 | 🟢 Easy | Java A self-referential condition — x elements must be ≥ x. Elegant problem, elegant solution. 🎯 🔍 𝗧𝗵𝗲 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 Find x such that exactly x elements in the array are ≥ x. Return -1 if no such x exists. ⚡ 𝗔𝗽𝗽𝗿𝗼𝗮𝗰𝗵 — 𝗙𝗿𝗲𝗾𝘂𝗲𝗻𝗰𝘆 𝗖𝗼𝘂𝗻𝘁 + 𝗦𝘂𝗳𝗳𝗶𝘅 𝗦𝘂𝗺 ✅ Cap all values at n (array length) — anything larger contributes the same way ✅ Build a frequency count array of size n+1 ✅ Traverse from right to left, accumulating a running suffix sum ✅ When suffix sum == current index i → x = i is the answer! 💡 𝗪𝗵𝘆 𝗰𝗮𝗽 𝗮𝘁 𝗻? x can never exceed n (can't have more elements than the array size). So values above n are equivalent — capping them avoids index overflow and keeps the logic clean. 📊 𝗖𝗼𝗺𝗽𝗹𝗲𝘅𝗶𝘁𝘆 ⏱ Time: O(n) — two passes 📦 Space: O(n) — frequency array No sorting. No binary search. Just a clever frequency count + suffix accumulation. Sometimes the cleanest approach is right under your nose. 🧠 📂 𝗙𝘂𝗹𝗹 𝘀𝗼𝗹𝘂𝘁𝗶𝗼𝗻 𝗼𝗻 𝗚𝗶𝘁𝗛𝘂𝗯: https://lnkd.in/gm2c4-6x 𝟳 𝗺𝗼𝗿𝗲 𝗱𝗮𝘆𝘀. 𝗦𝗼 𝗰𝗹𝗼𝘀𝗲 𝘁𝗼 𝟭𝟬𝟬! 💪 #LeetCode #Day93of100 #100DaysOfCode #Java #DSA #Arrays #FrequencyCount #CodingChallenge #Programming

Day 96 - LeetCode Journey Solved LeetCode 901: Online Stock Span in Java ✅ This problem is all about recognizing the pattern and using the right data structure. Instead of checking previous prices one by one, I used a Monotonic Stack to efficiently calculate spans. Every element is pushed and popped at most once → super optimized 🔥 Key idea: Keep removing smaller or equal previous prices and accumulate their spans. Key takeaways: • Monotonic Stack concept (very important) • Avoiding nested loops using stack optimization • Efficient span calculation • Thinking in patterns, not brute force ✅ All test cases passed ⚡ O(n) time and O(n) space This is one of those problems that truly levels up your stack game 💯 #LeetCode #DSA #Java #Stack #MonotonicStack #ProblemSolving #CodingJourney #InterviewPrep #Consistency #100DaysOfCode

Today’s Java DSA challenge: Sort Colors. After conquering Two-Pointer problems like moving zeroes and reversing arrays, I tackled the classic Dutch National Flag algorithm on LeetCode to sort an array of 0s, 1s, and 2s. You could easily solve this with a counting sort approach (count the 0s, 1s, and 2s, then overwrite the array in a second pass). But we can optimize this to a single pass with O(1) space using three pointers! Here is how the magic works: • low pointer: Tracks where the next 0 should go. • mid pointer: The explorer scanning through the array. • high pointer: Tracks where the next 2 should go. The Logic: • If mid finds a 0: Swap it with low, then move both low and mid forward. • If mid finds a 1: It's already in the safe middle zone, so just move mid forward. • If mid finds a 2: Swap it with high, and move high backward. (The catch? Don't move mid yet, because you still need to check the newly swapped number!) Devs: The Dutch National Flag algorithm is an absolute masterpiece. What other algorithm completely blew your mind the first time? 👇 #DSA #Java #LeetCode #SortColors

🚀 Day 41/100 – LeetCode DSA Practice 📌 Problem: 561. Array Partition Today’s challenge was about grouping elements into pairs such that the sum of the minimum values of each pair is maximized. 💡 My Approach: Initially thought about checking all possible pair combinations ❌ (too complex) Then realized this is a problem Sorted the array in ascending order Paired adjacent elements Added elements at even indices (0, 2, 4...) to get the maximum sum 🧠 Key Insight: After sorting, pairing neighbors ensures that the minimum in each pair is as large as possible, which maximizes the total sum. 💻 Code (Java): import java.util.Arrays; class Solution { public int arrayPairSum(int[] nums) { Arrays.sort(nums); int sum = 0; for(int i = 0; i < nums.length; i += 2) { sum += nums[i]; } return sum; } } 📚 What I Learned Today: Importance of recognizing patterns Sorting can simplify complex pairing problems Always look for pattern instead of brute force 🔥 Small insight, big impact! #Day41 #100DaysOfCode #LeetCode #DSA #GreedyAlgorithm #CodingJourney

Day 93 - LeetCode Journey Solved LeetCode 9: Palindrome Number in Java ✅ At first glance, it feels like a string problem… but the real challenge is solving it without converting to string. Instead of reversing the whole number, I reversed only half of it and compared both parts. This avoids overflow and keeps it efficient. Smart approach > brute force 💡 Key takeaways: • Handling edge cases (negative numbers, trailing zeroes) • Reversing only half of the number • Avoiding extra space (no string conversion) • Writing optimized mathematical logic ✅ All test cases passed ⚡ O(log n) time and O(1) space Sometimes the best solutions are the simplest ones, just a different way of thinking 🔥 #LeetCode #DSA #Java #Math #ProblemSolving #CodingJourney #InterviewPrep #Consistency #100DaysOfCode