🚀 Day 80 of #100DaysOfCode Solved LeetCode Problem #1653 – Minimum Deletions to Make String Balanced ✅ A clean greedy + DP-style problem that looks simple but really tests decision-making at each step. The trick is choosing whether to delete a character now or rely on previous counts to minimize total deletions. Key Takeaways: -> Greedy decisions can be optimized with running state -> Tracking counts (bCount) simplifies future choices -> Sometimes the best DP is just two variables -> Elegant logic beats complex data structures Language: Java -> Runtime: 19 ms (Beats 93.09%) ⚡ -> Memory: 47.80 MB Showing up daily, sharpening logic, and trusting the process. 💻🔥 #LeetCode #Java #Greedy #DynamicProgramming #Strings #ProblemSolving #100DaysOfCode
LeetCode 1653: Minimum Deletions to Balance String
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🚀 Day 88 of #100DaysOfCode | LeetCode Daily Solved LeetCode Problem #67 – Add Binary ➕💻✅ A clean fundamentals problem that reinforces how low-level operations really work under the hood. Simulating binary addition digit by digit with carry is simple, efficient, and elegant. Key Takeaways: -> Binary addition using carry logic -> Traversing strings from right to left -> Using StringBuilder for efficient string construction -> Handling edge cases when carry remains at the end Language: Java -> Runtime: 1 ms (Beats 99.65%) -> Memory: 43.90 MB Strong basics make complex problems easier. One day, one win. 🔥💻 #LeetCode #Java #Binary #Strings #ProblemSolving #100DaysOfCode
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🚀 Day 81 of #100DaysOfCode Solved LeetCode Problem #110 – Balanced Binary Tree ✅ A classic tree problem that rewards thinking bottom-up. Instead of recalculating heights repeatedly, combining height computation with balance checking makes the solution both clean and efficient. Key Takeaways: -> Bottom-up recursion simplifies tree problems -> Early termination saves unnecessary computation -> Returning sentinel values (-1) is a powerful pattern -> One DFS can solve both height & balance checks Language: Java -> Runtime: 0 ms (Beats 100%) ⚡ -> Memory: 45.76 MB Staying consistent, one tree at a time. 🌳💻🔥 #LeetCode #Java #BinaryTree #Recursion #DFS #ProblemSolving #100DaysOfCode
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🚀 Day 82 of #100DaysOfCode Solved LeetCode Problem #1382 – Balance a Binary Search Tree ✅ This problem is a great reminder that sometimes rebuilding is better than fixing. By leveraging the sorted nature of BSTs, converting it to a balanced tree becomes clean and intuitive. Key Takeaways: -> Inorder traversal gives a sorted sequence in BST -> Divide & conquer helps rebuild a height-balanced tree -> Choosing the middle element ensures balance -> Clean recursion beats complex rotations here Language: Java -> Runtime: 2 ms (Beats 97.82%) ⚡ -> Memory: 48.44 MB Consistency compounds. Trees today, forests tomorrow. 🌳💻🔥 #LeetCode #Java #BinarySearchTree #Recursion #DivideAndConquer #ProblemSolving #100DaysOfCode
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🚀 Day 91 of #100DaysOfCode | LeetCode Daily Solved LeetCode Problem #693 – Binary Number with Alternating Bits 🔢💡✅ A clean and elegant bit manipulation problem. The trick is to XOR the number with its right shift and check if the result becomes all 1s — simple, fast, and effective. Key Takeaways: -> Using n ^ (n >> 1) to detect alternating patterns -> Leveraging highestOneBit() for boundary checks -> Writing concise, branch-free logic with bitwise ops -> Sometimes the smartest solutions are the shortest Language: Java -> Runtime: 0 ms (Beats 100%) -> Memory: 41.81 MB (Beats 97.08%) One more day, one more concept sharpened. 🔥 Onward to Day 92 💻🚀 #LeetCode #Java #BitManipulation #BinaryNumbers #ProblemSolving #100DaysOfCode
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🚀 Day 90 of #100DaysOfCode | LeetCode Daily Solved LeetCode Problem #401 – Binary Watch ⌚💻✅ A fun problem that blends bit manipulation with simple iteration. The idea is to count set bits for hours and minutes and generate all valid time combinations. Key Takeaways: -> Smart use of Integer.bitCount() for counting LEDs -> Separating hours (0–11) and minutes (0–59) cleanly -> Formatting time output correctly (leading zeros matter!) -> Brute force done right with clear constraints Language: Java -> Runtime: 3 ms (Beats 91.19%) -> Memory: 43.45 MB (Beats 94.31%) Consistency builds confidence. 90 days in — still pushing forward. 🔥💻 #LeetCode #Java #BitManipulation #BinaryWatch #ProblemSolving #100DaysOfCode
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Day 52/100 – LeetCode Challenge ✅ Problem: #1653 Minimum Deletions to Make String Balanced Difficulty: Medium Language: Java Approach: Greedy with Counter Time Complexity: O(n) Space Complexity: O(1) Key Insight: String must be of form aa...abb...b (all 'a's before 'b's). For each 'a' encountered after a 'b', we must delete either this 'a' or a previous 'b'. Greedily delete 'a' when it appears out of order. Solution Brief: Tracked count of 'b's seen so far. When encountering 'a': If no preceding 'b's, it's in correct position If there are preceding 'b's, delete this 'a' (increment result) and reduce 'b' counter Final result = minimum deletions needed. #LeetCode #Day52 #100DaysOfCode #Greedy #Java #Algorithm #CodingChallenge #ProblemSolving #StringBalance #MediumProblem #Counter #Optimization #DSA
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🚀 Day 85 of #100DaysOfCode Solved LeetCode Problem #3713 – Longest Balanced Substring I ✅ A classic string + frequency-based problem that rewards careful iteration and validation. The goal was to find the longest substring where all present characters appear the same number of times—simple idea, but requires disciplined checking. Key Takeaways: -> Brute-force with pruning can still pass when constraints allow -> Frequency arrays are powerful for substring analysis -> Early balance checks save unnecessary computation -> Always align solution strategy with input limits Language: Java -> Runtime: 83 ms (Beats 88.95%) ⚡ -> Memory: 47.29 MB Small wins matter. Staying consistent, one problem at a time. 💻🔥 #LeetCode #Java #Strings #ProblemSolving #Algorithms #100DaysOfCode
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🚀 Day 13 of #LeetCode Challenge 🔢 Problem: 1758 – Minimum Changes To Make Alternating Binary String Today’s problem was about converting a binary string into an alternating pattern with minimum changes. 💡 Key Insight: An alternating string can only start with either ‘0’ or ‘1’. So instead of trying many possibilities, I compared both patterns and counted mismatches — then returned the minimum. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) ✨ This problem improved my thinking in: Pattern observation Optimization approach Writing clean and efficient Java code Consistency > Motivation 💪 #Day13 #Java #DSA #LeetCode #CodingJourney #FutureFullStackDeveloper
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🚀 Day 98 of #100DaysOfCode Solved LeetCode #1356 – Sort Integers by The Number of 1 Bits ✅ A clean bit manipulation + sorting problem that rewards thinking beyond plain comparisons. Key Takeaways: -> Using Integer.bitCount() to count set bits efficiently -> Encoding sort priority directly into values for simplicity -> Leveraging built-in sorting for clean and fast solutions -> Small tricks can lead to elegant code ✨ Language: Java -> Runtime: 6 ms (Beats 93.70%) ⚡ -> Memory: 47.42 MB Almost there. Staying consistent till the end 💻🔥 #LeetCode #Java #BitManipulation #Sorting #ProblemSolving #DSA #100DaysOfCode
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LeetCode Problem Solved | #1209. Remove All Adjacent Duplicates in String II Today I worked on LeetCode 1209, which is a great extension of the classic adjacent-duplicate problem and a perfect example of the stack pattern in DSA. 🧠 Problem Summary Given a string s and an integer k, repeatedly remove k adjacent identical characters until no more removals are possible. Example: Input: s = "deeedbbcccbdaa", k = 3 Output: "aa" 💡 Approach (Using Stack) 1. Traverse the string character by character 2. If the current character matches the stack top -> increment count 3. If count reaches k -> pop from stack 4. Otherwise -> push a new Pair 5 .Finally, rebuild the string from the stack #LeetCode #DSA #Java #Stack #ProblemSolving #CodingPractice
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