LeetCode 2976: Minimum Cost String Conversion with Graph Algorithms

🚀 Day 72 of #100DaysOfCode Solved LeetCode Problem #2977 – Minimum Cost to Convert String II ✅ This was a more advanced follow-up that required handling string-level transformations efficiently. The challenge was to minimize the total cost by choosing optimal substring conversions rather than single-character changes. Key Learnings: -> Using a Trie to store and index valid string transformations -> Modeling transformation costs with dynamic programming -> Combining Trie traversal + DP for efficient substring matching -> Carefully managing state transitions to avoid unnecessary recomputation Language Used: Java -> Runtime: 209 ms (Beats 34.88%) -> Memory: 48.68 MB (Beats 18.60%) Step by step, leveling up problem-solving depth and string algorithm intuition 🚀 #LeetCode #DynamicProgramming #Trie #Java #ProblemSolving #100DaysOfCode

🚀 Day 78 of #100DaysOfCode Solved LeetCode Problem #3379 – Transformed Array ✅ A clean and elegant problem focused on index manipulation and modular arithmetic. The key was handling circular indexing correctly, especially with negative shifts—simple logic, but precision matters. Key Takeaways: -> Using modulo smartly to handle circular arrays -> Dealing with negative indices safely -> Writing concise and readable transformations -> Small problems still demand careful edge-case thinking Language: Java -> Runtime: 1 ms (Beats 98.72%) ⚡ -> Memory: 47.08 MB Consistency > Complexity. Keep moving forward. 💻🔥 #LeetCode #Java #Arrays #ModularArithmetic #ProblemSolving #100DaysOfCode

🚀 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

🚀 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

🚀 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

🚀 Cracked LeetCode 189 – Rotate Array using an optimized in-place approach. Instead of brute force (O(n × k)), I implemented the reverse algorithm technique to achieve: ✅ Time Complexity: O(n) ✅ Space Complexity: O(1) ✅ 0 ms runtime (Beats 100%) Focused on improving optimization thinking and writing clean, efficient Java code. #LeetCode #Java #DataStructures #Algorithms #ProblemSolving #TechGrowth

Day 10 of Daily DSA 🚀 Solved LeetCode 190: Reverse Bits ✅ 🔍 Approach: Worked with the 32-bit representation of the given integer: • Converted the number into a fixed 32-bit binary string • Used a two-pointer technique to reverse the bits • Converted the reversed binary back to an unsigned integer This approach keeps the logic simple while ensuring correctness for signed integers. ⏱ Complexity: • Time: O(32) = O(1) (constant time) • Space: O(32) = O(1) 📊 LeetCode Stats: • Runtime: 9 ms • Memory: 43.46 MB A good reminder that clarity matters more than micro-optimizations when learning bit manipulation. #DSA #LeetCode #Java #BitManipulation #ProblemSolving #DailyCoding #Consistency

Day 15 of Daily DSA 🚀 Solved LeetCode 540: Single Element in a Sorted Array ✅ Approach: Used Binary Search on indices instead of values. Since elements appear in pairs, the unique element breaks the pairing pattern. By forcing mid to be even and comparing nums[mid] with nums[mid + 1], we can safely discard half of the array each time. ⏱ Complexity: • Time: O(log n) — binary search • Space: O(1) — constant extra space 📊 LeetCode Stats: • Runtime: 0 ms (Beats 100%) ⚡ • Memory: 52.96 MB (Beats 57.60%) A neat example of how index properties make binary search even more powerful. #DSA #LeetCode #Java #BinarySearch #ProblemSolving #DailyCoding #Consistency

Day 58/100 – LeetCode Challenge ✅ Problem: Longest Balanced Substring II Difficulty: Medium Language: Java Approach: Multiple Pattern Detection + Prefix Difference Mapping Time Complexity: O(n) Space Complexity: O(n) Key Insight: Balanced substrings must have equal counts of characters in three possible patterns: Single character run - consecutive same characters Two characters - equal occurrences of two specific chars Three characters - all three chars appear same number of times Solution Brief: countOne(): Track longest consecutive run of same character. countTwo(): For two chars (like 'a'/'b'), use prefix sum where n1=+1, n2=-1; reset on other chars. countThree(): Track differences (a-b) and (a-c) as state; when same state repeats, balanced substring found. #LeetCode #Day58 #100DaysOfCode #String #HashMap #Java #Algorithm #CodingChallenge #ProblemSolving #LongestBalancedSubstringII #HardProblem #PrefixSum #StateTracking #DSA

🚀 Day 84 of #100DaysOfCode Solved LeetCode Problem #3721 – Longest Balanced Subarray II ✅ This one was a solid jump in complexity from Part I. It required a more advanced approach using segment trees with lazy propagation to efficiently handle range updates and queries while tracking balance conditions. Key Takeaways: -> Segment Tree + Lazy Propagation is powerful for range-based balance problems -> Encoding states smartly simplifies transitions -> Performance optimization matters when brute force isn’t an option -> Advanced data structures unlock solutions to “Part II” style problems Language: Java -> Runtime: 349 ms (Beats 73.53%) ⚡ -> Memory: 72.05 MB (Beats 98.53%) One day, one hard problem, steady progress. 💻🔥 #LeetCode #Java #SegmentTree #LazyPropagation #DataStructures #ProblemSolving #100DaysOfCode