LeetCode Day 9: Container With Most Water Optimization

🚀 Day 17/60 — LeetCode Discipline Problem Solved: 4Sum (Revision) Difficulty: Medium Today’s session revisited the classic k-Sum family problem — 4Sum. After solving 2Sum and 3Sum patterns previously, this problem naturally extends the idea by combining sorting with nested iteration and the two-pointer technique to efficiently identify unique quadruplets. The key challenge lies in managing duplicates carefully while keeping the solution efficient and structured. 💡 Focus Areas: • Strengthened k-Sum pattern understanding • Practiced sorting + two-pointer combination • Improved duplicate handling in multi-pointer problems • Reinforced nested iteration optimization • Focused on writing structured and readable code Problems like these highlight how mastering one pattern allows you to scale the same idea to more complex variations. #LeetCode #60DaysOfCode #100DaysOfCode #DSA #TwoPointers #Arrays #Algorithms #DataStructures #ProblemSolving #CodingJourney #SoftwareEngineering #Programming #Developers #TechCareers

Day 47 on LeetCode — Partition Labels ✂️✅ Today’s problem was a great example of greedy strategy with smart indexing. 🔹 Approach Used in My Solution The goal was to split the string into maximum number of partitions such that each character appears in only one part. Key idea in the solution: • First, store the last occurrence of each character in an array • Traverse the string while maintaining a current partition range • Continuously update the end of the partition using the last index of characters encountered • When the current index reaches end, it means the partition is complete • Store the partition size and start a new one This greedy approach ensures each partition is as small as possible while still valid. ⚡ Complexity: • Time Complexity: O(n) • Space Complexity: O(1) 💡 Key Takeaways: • Learned how preprocessing (last occurrence tracking) simplifies problems • Practiced greedy partitioning techniques • Strengthened understanding of interval expansion logic #LeetCode #DSA #Algorithms #DataStructures #GreedyAlgorithm #Strings #ProblemSolving #Coding #Programming #Cpp #STL #SoftwareEngineering #ComputerScience #CodingPractice #DeveloperLife #TechJourney #CodingDaily #100DaysOfCode #BuildInPublic #AlgorithmPractice #CodingSkills #Developers #TechCommunity #SoftwareDeveloper #EngineeringJourney

🚀 Day 13/60 — LeetCode Discipline Problem Solved: Contains Duplicate II (Revision) Difficulty: Easy Today’s practice focused on revisiting a classic array problem that blends hashing with a sliding window concept. The goal was to efficiently determine whether duplicate values appear within a specific index range. Instead of checking every possible pair, the approach maintains a moving window of elements using a hash structure, allowing constant-time lookups while traversing the array. 💡 Focus Areas: • Reinforced hash-based lookup techniques • Practiced sliding window with bounded size • Strengthened index-distance reasoning • Improved understanding of time-efficient duplicate detection • Focused on clean and readable implementation Even simple problems can reinforce powerful ideas when approached with discipline and clarity. #LeetCode #60DaysOfCode #100DaysOfCode #DSA #SlidingWindow #HashSet #Arrays #Algorithms #DataStructures #ProblemSolving #CodingJourney #SoftwareEngineering #Programming #Developers

🚀 Day 79 of My Coding Challenge Today I solved the Super Reduced String problem. The challenge was to repeatedly remove adjacent matching characters from a string until no such pairs remain. If the final string becomes empty, we return "Empty String". 💡 Key Idea: I used a stack-based approach to efficiently remove adjacent duplicates in O(n) time complexity. 📌 Example: Input: aaabccddd Output: abd Steps: aaabccddd → abccddd → abddd → abd 🧠 What I practiced today: Stack-based string processing Efficient character comparison Problem-solving with linear time complexity Consistency is the key — every day a little progress toward becoming a better developer. 💻🔥 #Day79 #CodingChallenge #LeetCode #Programming #ProblemSolving #DataStructures #100DaysOfCode #DeveloperJourney

Day 50 on LeetCode — Maximum Average Subarray I 📊✅ Half-century milestone! 🚀 Today’s problem was a classic and clean application of the Sliding Window technique. 🔹 Approach Used in My Solution The goal was to find the maximum average of a subarray of size k. Key idea in the solution: • First, compute the sum of the first window of size k • Then slide the window forward by:   – Adding the next element   – Removing the element leaving the window • Continuously update the maximum sum encountered • Finally, divide the maximum sum by k to get the result This avoids recomputing sums and ensures an efficient solution. ⚡ Complexity: • Time Complexity: O(n) • Space Complexity: O(1) 💡 Key Takeaways: • Mastered the fixed-size sliding window pattern • Learned how to optimize subarray sum problems • Reinforced thinking in terms of window reuse instead of recomputation #LeetCode #DSA #Algorithms #DataStructures #SlidingWindow #Arrays #ProblemSolving #Coding #Programming #Cpp #STL #SoftwareEngineering #ComputerScience #CodingPractice #DeveloperLife #TechJourney #CodingDaily #100DaysOfCode #BuildInPublic #AlgorithmPractice #CodingSkills #Developers #TechCommunity #SoftwareDeveloper #EngineeringJourney

🚀 Day 14/60 — LeetCode Discipline Problem Solved: Maximum Number of Vowels in a Substring of Given Length (Revision) Difficulty: Medium Today’s practice revisited another elegant application of the sliding window technique. The task was to determine the maximum number of vowels present in any substring of fixed length k. Instead of recalculating the count for every possible substring, the sliding window approach allows the window to move forward while updating the vowel count efficiently. This pattern once again highlights how maintaining a running state can transform a brute-force idea into a clean and optimal solution. 💡 Focus Areas: • Strengthened fixed-size sliding window intuition • Practiced efficient character counting • Improved substring traversal logic • Reinforced constant-time window updates • Focused on writing clean and readable code ⚡ Performance Highlight: Achieved solid runtime efficiency on submission. Each day of deliberate practice adds another layer of clarity to fundamental algorithmic patterns. #LeetCode #60DaysOfCode #100DaysOfCode #DSA #SlidingWindow #Strings #Algorithms #DataStructures #ProblemSolving #CodingJourney #SoftwareEngineering #Programming #Developers #TechCareers

🚀 Day 6 of 100 Days LeetCode Challenge Problem: Check if Binary String Has at Most One Segment of Ones Today’s problem is all about pattern observation in strings—simple, but easy to overthink. 💡 Key Insight: The string should contain only one continuous block of '1's. 👉 That means: Once a 0 appears after a 1, There should be no more '1's later 🔍 Simplest Trick: Just check if the pattern "01" appears more than once OR even better → check if "10" appears followed by another "1" 💡 Cleaner approach: Traverse the string Count transitions from 1 → 0 If you ever see 1 again after that → ❌ Invalid 🔥 What I Learned Today: Many problems are just pattern validation Clean logic beats complex conditions Always try to reduce the problem to a simple rule 📈 Challenge Progress: Day 6/100 ✅ Consistency building strong! LeetCode, Strings, Pattern Recognition, Greedy, DSA Practice, Coding Challenge, Problem Solving, Algorithm Thinking, Programming #100DaysOfCode #LeetCode #DSA #CodingChallenge #Strings #ProblemSolving #TechJourney #ProgrammerLife #SoftwareDeveloper #CodingLife #LearnToCode #Developers #Consistency #GrowthMindset #InterviewPrep

🚀 Day 181 of #200DaysOfCoding Today I solved “Minimum Changes To Make Alternating Binary String” on LeetCode. 🔹 Problem: We are given a binary string consisting of 0 and 1. In one operation, we can flip any character (0 → 1 or 1 → 0). The goal is to make the string alternating, meaning no two adjacent characters are the same. Example of alternating strings: 0101, 1010 🔹 Key Insight: An alternating binary string can only have two possible patterns: 1️⃣ 010101... 2️⃣ 101010... So the idea is simple: Count how many changes are needed to convert the string to pattern 1. Count how many changes are needed to convert it to pattern 2. The minimum of the two will be the answer. ⏱ Time Complexity: O(n) 📦 Space Complexity: O(1) Consistent practice is helping me improve my problem-solving and pattern recognition skills every day. #leetcode #coding #programming #cpp #100daysofcode #200daysofcoding #softwaredevelopment

Today I worked on leetcode problem no. 1415 The k-th Lexicographical Happy String of Length n.” First, let's understand the concept. A happy string is a string that: Uses only the characters 'a', 'b', 'c' Does not allow two adjacent characters to be the same Concepts practiced: Backtracking Lexicographical ordering Recursion Start with an empty string Try adding 'a', 'b', 'c' Skip a character if it is the same as the previous one When the string length becomes n, we found one valid happy string Since we generate them in lexicographical order, the moment we reach the k-th string, that is our answer. #leetcode #leetcode1415 #dsa #datastructures #algorithms #backtracking #recursion #coding #programming #problemSolving #codingpractice #softwareengineering #developer #codinglife #computerscience #codingjourney #learncoding #codingcommunity #buildinpublic #100daysofcode #competitiveprogramming #cpp #techskills

Day 42 on LeetCode — Merge Sorted Array (Two Pointer Approach) ✅ Today’s problem focused on efficient in-place array manipulation using the two-pointer technique. 🔹 Approach Used in My Solution The key insight was to compare elements from the back of both arrays instead of the front. Since nums1 already has extra space at the end to accommodate elements from nums2, we can fill the array from the last index backwards. Key points in the logic: • Initialize pointers at the end of the valid elements of nums1 and nums2 • Compare the elements and place the larger one at the back of nums1 • Move the pointers accordingly until all elements are merged • This allows the final sorted array to be stored directly in nums1 This strategy avoids unnecessary shifting of elements and keeps the solution efficient. ⚡ Complexity: • Time Complexity: O(m + n) • Space Complexity: O(1) (in-place merge) 💡 Key Takeaways: • Learned how working from the back can simplify in-place merges • Strengthened understanding of the two-pointer technique • Practiced optimizing array operations without using extra space #LeetCode #DSA #Algorithms #DataStructures #TwoPointers #Arrays #ProblemSolving #Coding #Programming #Cpp #STL #SoftwareEngineering #ComputerScience #CodingPractice #DeveloperLife #TechJourney #CodingDaily #100DaysOfCode #BuildInPublic #AlgorithmPractice #CodingSkills #Developers #TechCommunity #SoftwareDeveloper #EngineeringJourney