Day 72 of #100DaysOfCode Problem: Convert Sorted Array to Height-Balanced BST Today I learned how to efficiently convert a sorted array into a balanced Binary Search Tree using Divide & Conquer. Key Insight: Pick the middle element as the root to maintain balance. Recursively build: Left subtree from left half Right subtree from right half This ensures: Optimal height Faster search operations ⏱ Time Complexity: O(n) 📦 Space Complexity: O(log n) Consistency is the real game changer #DSA #Java #BinaryTree #LeetCode #CodingJourney
Converting Sorted Array to Balanced BST with Divide & Conquer
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Day 57 of #100DaysOfLeetCode Today’s problem focused on finding the minimum distance between a target element and a given index in an array. I solved it using a simple and efficient approach: Traversed the array once Checked for target occurrences Calculated distance using absolute difference Maintained the minimum value throughout Runtime: 0 ms (Beats 100%) Optimized space usage Key takeaway: Even straightforward iteration can lead to highly optimized solutions when combined with the right logic. #LeetCode #Java #DataStructures #Algorithms #100DaysOfCode #CodingJourney #ProblemSolving
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🚀 Day 52 of #100DaysOfLeetCode Solved: Daily Temperatures (Monotonic Stack) Today’s focus was on understanding how to optimize from a brute-force O(n²) approach to an efficient O(n) solution using a stack. Key takeaways: Learned how to identify “next greater element” patterns Understood how monotonic stacks help avoid redundant work Practiced writing clean and optimized code Consistency is starting to pay off. Onto the next one. #DSA #Java #CodingJourney #LeetCode #ProblemSolving
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Day 70 - Next Greater Node Understanding how to find the next larger value for each node using stack optimization. Approach: • Convert linked list → array for easy indexing • Use a stack to track indices • Compare current value with stack top • Update results when a greater value is found Key Insight: Monotonic stack helps avoid unnecessary comparisons Time Complexity: O(n) Space Complexity: O(n) #Day70 #LeetCode #Java #CodingPractice #TechJourney #DSA #LinkedList
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Day 90 of #100DaysOfCode – Unique Binary Search Trees II Today’s problem was all about generating all structurally unique BSTs for values from 1 to n. At first glance, it feels tricky because it's not just counting trees — we actually need to build every possible tree structure. 💡 Key Insight: Pick each number i as the root Recursively build: Left subtree from [1 ... i-1] Right subtree from [i+1 ... n] Combine every left & right subtree pair 🌳 This is a classic Recursion + Backtracking problem and is closely related to Catalan Numbers. 📌 Example: For n = 3, we get 5 unique BSTs ⚡ What I learned today: How recursion can generate combinations of structures Importance of base case (start > end → null) Combining subproblems effectively 💻 Time Complexity: Approximately O(4ⁿ / √n) Consistency is key — 90 days strong and still going 💪 Next target: 💯 #DSA #Java #LeetCode #Recursion #BinaryTree #CodingJourney #100DaysOfCode
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Day 35 of Consistency 💻🔥 Solved Longest Substring Without Repeating Characters — a classic sliding window problem that really tests optimization skills. ✨ Key Takeaways: Mastered the sliding window technique Used HashMap for efficient lookup Improved understanding of two-pointer approach ⚡ Performance: Runtime: 5 ms Beat: 87%+ submissions Every day is about getting sharper, not just solving problems. #LeetCode #DataStructures #Algorithms #CodingJourney #Consistency #Java #ProblemSolving
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𝐃𝐚𝐲 𝟕𝟕 – 𝐃𝐒𝐀 𝐉𝐨𝐮𝐫𝐧𝐞𝐲 | 𝐀𝐫𝐫𝐚𝐲𝐬 🚀 Today’s problem focused on finding all elements that appear more than n/3 times in an array. 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝐒𝐨𝐥𝐯𝐞𝐝 • Majority Element II 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡 – 𝐇𝐚𝐬𝐡𝐌𝐚𝐩 • Counted frequency of each element using a map • Calculated threshold = n / 3 • Collected elements whose frequency exceeded the threshold 𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠𝐬 • At most 2 elements can appear more than n/3 times • HashMap is straightforward for frequency counting • Understanding constraints helps reduce possibilities • This problem has an optimized Boyer-Moore Voting (extended) solution 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲 • Time: O(n) • Space: O(n) 𝐓𝐚𝐤𝐞𝐚𝐰𝐚𝐲 Constraints often reveal hidden patterns — understanding them leads to better optimizations. 77 days consistent 🚀 On to Day 78. 🔗 Problem Link: https://lnkd.in/dDwdWYJs #DSA #Arrays #HashMap #LeetCode #Java #ProblemSolving #DailyCoding #LearningInPublic #SoftwareDeveloper
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Day 80/90 Solved: Largest Rectangle in Histogram Core idea: → Use monotonic stack to find nearest smaller to left & right → Width = (right - left - 1), maximize area This is one of those problems where the pattern isn’t obvious at first. Once the left/right boundary logic clicks, the solution becomes systematic. #Consistency #DSA #LeetCode #Stack #MonotonicStack #ProblemSolving #Java #Algorithms
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Continuing my DSA journey, * Problem: Number of Atoms (726) * Approach: -I solved this using a stack-based parsing approach. -I used a stack of maps to keep track of atom counts at different levels. While iterating through the string: -On '(' → push a new map onto the stack -On ')' → pop the top map, multiply counts, and merge back -For elements → parse the name and count, then update the current map * Key Insight: -The tricky part was handling nested parentheses and applying multipliers correctly. -Using a stack helps manage different scopes efficiently. * What I Learned: This problem improved my understanding of string parsing with stacks. #LeetCode #DSA #Java #Stack #Strings #ProblemSolving
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𝐃𝐚𝐲 87/100 – 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞 𝐂𝐡𝐚𝐥𝐥𝐞𝐧𝐠𝐞 🚀 Problem: 228. 𝐒𝐮𝐦𝐦𝐚𝐫𝐲 𝐑𝐚𝐧𝐠𝐞𝐬 Today I solved a problem where we need to summarize consecutive numbers in a sorted unique array into ranges. 🔑 𝐈𝐝𝐞𝐚: Traverse the array and keep extending the range while consecutive numbers continue. Once the sequence breaks, close the range and store it. 💡 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Start with the first element as start Move forward while nums[i] + 1 == nums[i+1] If range exists → "start->end" Else → single number "start" 𝐊𝐞𝐲 𝐈𝐧𝐬𝐢𝐠𝐡𝐭: Efficient single pass solution (O(n)) by grouping consecutive elements on the fly. #LeetCode #Java #ProblemSolving #DSA #100DaysOfCode #CodingJourney
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Day 81 - Balanced Binary Tree Checked whether a binary tree is height-balanced using a bottom-up approach. Approach: • Recursively compute height of left and right subtrees • If height difference > 1 → not balanced • Use -1 as a signal to stop early Time Complexity: O(n) Space Complexity: O(h) #Day81 #LeetCode #BinaryTree #DSA #Java #CodingJourney #ProblemSolving
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