Design Ordered Stream with Java

🚀 Day 52/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 3174. Clear Digits Used a stack-like approach with StringBuilder to remove the closest non-digit character whenever a digit is encountered. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(n) Strengthening understanding of string processing and stack-based logic simulation. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 Day 65/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 2535. Difference Between Element Sum and Digit Sum of an Array Used a single loop + digit extraction approach to calculate both element sum and digit sum efficiently. ⏱️ Time Complexity: O(n × d) 📦 Space Complexity: O(1) Strengthening understanding of number manipulation and digit-based operations. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 Day 69/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 2103. Rings and Rods Used a 2D boolean array to track presence of colors (R, G, B) on each rod and counted rods containing all three. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) Strengthening understanding of mapping, indexing, and state tracking techniques. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 Day 57/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 242. Valid Anagram Used a HashMap frequency counting approach (Unicode-safe using code points) to compare character frequencies of both strings. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(n) Strengthening understanding of string frequency analysis and Unicode handling. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency 

🚀 Day 48/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 1534. Count Good Triplets Used a brute-force triple nested loop to check all possible triplets and validate the given conditions using absolute differences. ⏱️ Time Complexity: O(n³) 📦 Space Complexity: O(1) Strengthening understanding of nested loop patterns and condition-based filtering. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 Day 47/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 448. Find All Numbers Disappeared in an Array Used an in-place marking technique by treating indices as a hash map and marking visited numbers as negative to identify missing elements. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) (excluding output list) Strengthening understanding of array manipulation and in-place hashing tricks. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 Day 48 of My #LeetCode Journey Today’s problem: 2615. Sum of Distances 💡 Key Idea: Instead of calculating distances between equal elements using brute force (O(n²)), I used: HashMap to group indices of same values Prefix Sum to efficiently compute distances This reduced the complexity to O(n) 🔥 🧠 What I Learned: How prefix sums can optimize distance calculations Efficient handling of repeated elements Writing clean and optimized code using Java ⚡ Approach: Store indices of each number Use prefix sums to calculate left & right distances Combine both to get final answer 📈 Time Complexity: O(n) 📦 Space Complexity: O(n) Consistency is key. Small progress every day leads to big results 💪 #Day48 #Java #FullStackDeveloper #CodingJourney #100DaysOfCode #DSA #LeetCode

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

🚀 Day 23 of #75DaysOfCode Solved: Equal Row and Column Pairs 🔹 Problem: Given an n x n matrix, count the number of pairs (ri, cj) such that row ri and column cj are exactly the same. 🔹 Approach: • Stored all rows in a HashMap with their frequency • Converted each column into a comparable string format • Matched columns with stored rows to count equal pairs 🔹 Key Learning: Using StringBuilder instead of String concatenation significantly improves performance due to immutability of Strings in Java. 🔹 Time Complexity: O(n²) 🔹 Takeaway: Smart use of HashMap + efficient string handling can simplify and optimize matrix-based problems. #Day22 #75DaysOfCode #Java #DSA #LeetCode #CodingJourney #ProblemSolving