Can you reverse only specific characters in a string without breaking the rest? 🧩 Let’s solve it smartly. Hey everyone! Day 350 of my 365-day coding journey, and today’s problem was a classic string manipulation challenge: LeetCode 345, Reverse Vowels of a String. This problem is simple on the surface but great for exploring multiple approaches and understanding trade-offs in efficiency. Let’s get into it. ✨ 🛠️ The Problem Given a string, the task is to reverse only the vowels in it while keeping all other characters in their original positions. Vowels include both lowercase and uppercase: a, e, i, o, u. Example: Input: IceCreAm Output: AceCreIm 🎯 The Approaches I solved this problem using three different methods: Solution 1: Brute Force First, collect all vowels from the string. Reverse the collected vowels. Traverse the string again and replace vowels in order. This approach is easy to understand but uses extra space. Solution 2: Two Pointers Use two pointers, one starting from the left and one from the right. Move them inward until both point to vowels, then swap. This is the most optimal approach with constant extra space. Solution 3: Stack Push all vowels into a stack. Traverse the string again and replace vowels by popping from the stack. This naturally reverses the vowels and keeps the logic clean. 🧠 Key Takeaways String problems often look easy but can be solved in many ways with different space and time trade-offs. The two-pointer technique is extremely powerful for in-place string manipulation. Understanding multiple approaches helps in choosing the right solution during interviews and real-world coding. 💡 Challenge for You! Which approach do you usually prefer for string problems — clarity first or optimal performance? Share your thoughts below. 💬 📺 Watch My Video Walkthrough I explain all three approaches step by step in my latest video: https://lnkd.in/gSZZbnx9 🔥 Join the Conversation If you’re consistently solving DSA problems or working towards mastering coding interviews, let’s connect and grow together. 🚀 #CodingJourney #DSA #LeetCode #Strings #TwoPointers #ProblemSolving #Algorithms #DataStructures #Python #DeveloperLife #CodeNewbies #Programming #365DaysOfCode #LearningEveryDay
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𝗦𝘁𝗼𝗽 𝘂𝘀𝗶𝗻𝗴 .𝘀𝗼𝗿𝘁() 𝗳𝗼𝗿 𝗺𝗲𝗿𝗴𝗶𝗻𝗴 𝗮𝗿𝗿𝗮𝘆𝘀. 𝗛𝗲𝗿𝗲 𝗶𝘀 𝘄𝗵𝘆 "𝗪𝗼𝗿𝗸𝗶𝗻𝗴 𝗕𝗮𝗰𝗸𝘄𝗮𝗿𝗱𝘀" 𝗶𝘀 𝗯𝗲𝘁𝘁𝗲𝗿. #day28 out of #day365 Solved 𝗟𝗲𝗲𝘁𝗖𝗼𝗱𝗲 𝟴𝟴: Merge Sorted Array today. As a Python developer, the temptation to just write (nums1 + nums2).sort() is real. It’s one line, readable, and "𝗣𝘆𝘁𝗵𝗼𝗻𝗶𝗰." But in an interview (and high-performance systems), that approach costs you 𝗢((𝗡+𝗠)\𝗹𝗼𝗴(𝗡+𝗠)) time. To maintain an optimal 𝗢(𝗡+𝗠) runtime and 𝗢(𝟭) space, I had to use the 𝗥𝗲𝘃𝗲𝗿𝘀𝗲 𝗧𝘄𝗼-𝗣𝗼𝗶𝗻𝘁𝗲𝗿 technique. The Challenge: Merge two sorted arrays in-place into nums1. The "Aha!" Moment: If you start merging from the front (index 0), you end up overwriting elements in nums1 that you haven't processed yet. You'd need extra space to save them. The Solution: Work backwards. Pointers: Set p1 and p2 to the end of the valid data in both arrays. Target: Set k to the absolute end of the array (where the empty buffer space is). Race: Compare nums1[p1] vs nums2[p2]. Place the larger one at k and move pointers left. This approach fills the "empty" space first, ensuring no data is ever overwritten. Optimizing for memory often means thinking in reverse. #Python #LeetCode #DataStructures #Algorithms #BackendDeveloper #Elitesecom #Optimization #Python #LeetCode #DataStructures #Algorithms #BackendDeveloper #Coding #GrowthMindset #Python #LeetCode #DailyDSA #DynamicProgramming #BackendDeveloper #Elitesecom #Coding
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Nice post! I usually go with the two-pointer approach, but the stack explanation made it really easy to understand logic .Thanks for sharing!