Solved Median of Two Sorted Arrays with Java 8 Streams

Today I worked on a LeetCode problem that helped me understand integer overflow in Java more clearly. The challenge was to divide two integers and handle a special case where the result goes beyond the limit of a 32-bit integer. Example tricky case: -2147483648 / -1 The real answer should be 2147483648, but Java can’t store this because the maximum value of a 32-bit integer is 2147483647. So without handling it, the value overflows and becomes negative again! What I learned?? Integer values have limits in programming Going beyond that range causes overflow Always consider edge cases when dealing with math operations in code Solving problems like this builds confidence and improves my problem-solving mindset. One small step forward every day! #Java #LeetCode #CodingPractice #ProblemSolving #DeveloperJourney

💻 Day 45 of #LeetCode Journey 🚀 ✅ Problem: Count and Say 📘 Language: Java 🔹 Status: Accepted (30/30 test cases passed) 🔹 Runtime: 4 ms | Beats 55.38% 🔹 Memory: 41.86 MB 🧠 Concept: This problem is about generating the n-th term in the “count and say” sequence. Each term is built by describing the previous term — count the number of digits and say them in order. 🧩 Approach: Start with "1". For each iteration, use a StringBuilder to construct the next sequence. Track consecutive digits using a counter. Append the count and digit when the sequence changes. 💡 Efficient string manipulation and iteration give optimal performance. 🔥 Every solved problem builds confidence. One step closer to mastering patterns in strings! #Day45 #LeetCode #Java #CodingChallenge #ProblemSolving #CountAndSay #50DaysOfCode

📌 Day 11/100 – Longest Common Prefix (LeetCode 14) 🔹 Problem: Given an array of strings, find the longest common prefix shared among them. If no common prefix exists, return an empty string. 🔹 Approach: I first sorted the array of strings in alphabetical order. The idea is that only the first and last strings in the sorted order will differ the most — so the common prefix between them is also the prefix for the entire array. I then compared characters one by one between these two strings to extract the prefix efficiently. 🔹 Key Learning: Sorting simplifies comparison by aligning similar prefixes together. Smart observation can reduce unnecessary iterations. Sometimes, the most optimal logic lies in choosing the right perspective. Step by step, improving problem-solving clarity and confidence 💪 #100DaysOfCode #LeetCode #Java #ProblemSolving #DSA #CodingJourney #LearnByDoing #CodingChallenge

Day 89 of #100DaysOfCode Solved Find Numbers with Even Number of Digits in Java 🔢 Approach The goal of this problem was to count how many integers in a given array have an even number of digits. I implemented two methods: findNumbers(int[] nums): This is the main function that iterates through every number in the input array nums. isEvenOrOdd(int n): This helper function takes an integer and determines if its digit count is even or odd. Inside the helper function, I used a while loop to count the digits: I initialized a count to 0. The loop continues as long as $n > 0$. In each iteration, I increment count and then perform integer division by 10 (n = n / 10) to remove the least significant digit. Finally, I return true if the total count of digits is even (count \% 2 == 0). This efficient, digit-by-digit checking approach resulted in a strong performance, beating 98.90% of other submissions. #Java #100DaysOfCode #LeetCode #CodingChallenge #Algorithms #Array #ProblemSolving #Optimization

🚀 Day 29 of 100 Days of LeetCode 📘 Problem: Combination Sum 💻 Language: Java ✅ Status: Accepted — Runtime ⚡ 2 ms (Beats 86%) Today’s problem was all about backtracking — exploring all possible combinations to reach a target sum. It’s a perfect blend of recursion, decision-making, and pruning unnecessary paths 🧠 ✨ Key Learnings: Backtracking is like exploring a maze — try, backtrack, and try again until you find the exit 🌀 Each recursive call represents a “choice point” — include or skip the element Clean recursion with controlled base cases leads to clarity and precision This problem reminded me how persistence works in both code and life: 💬 “Sometimes, the path to the solution is not straightforward — you just need to keep exploring.” #Day29 #100DaysOfCode #LeetCode #Java #Backtracking #Recursion #DSA #ProblemSolving #CodingJourney #SoftwareDevelopment #KeepLearning #CodeEveryday

🧠 Daily LeetCode Grind — Java Edition Today’s challenge: ✅ Container With Most Water (#11 - Medium) 📌 Goal: Given an integer array height[], find two lines that, together with the x-axis, form a container that holds the most water. 📌 Approach: 🔹 Use the Two-Pointer Technique to optimize the search for the maximum area. 🔹 Start with two pointers at both ends of the array and calculate the area at each step. 🔹 Move the pointer pointing to the smaller height inward to maximize potential area. 🔹 Continue until both pointers meet. 🧩 Test Case: Input: height = [1,8,6,2,5,4,8,3,7] Output: 49 💡 Key Takeaways: 🔹 Learned to balance area optimization using width × height. 🔹 Reinforced the power of the two-pointer approach for reducing time complexity. 🔹 Strengthened problem-solving speed through mathematical reasoning. 💻 Language: Java 🧠 Complexity: O(n) — single pass using two pointers. #LeetCode #Java #CodingPractice #ProblemSolving #TwoPointer #Algorithm #DeveloperLife #CybernautEdTech #AcceptedSolution

✨ Day 74 of #100DaysOfCode Today, I solved LeetCode 22. Generate Parentheses using Backtracking in Java 🧠💻 📝 Problem Summary: Given n pairs of parentheses, generate all combinations of well-formed parentheses. ✅ Example: Input: n = 3 Output: ["((()))","(()())","(())()","()(())","()()()"] 💡 Key Idea: Use backtracking to build all valid strings. Keep track of the number of opening and closing brackets. Add '(' if open < n. Add ')' if close < open. When the string reaches length 2 * n, it's a valid combination. 🚀 Learning: Backtracking is a powerful technique for exploring all valid combinations. Keeping proper state (open, close) helps prune invalid paths early. #100DaysOfCode #LeetCode #DSA #Java #ProblemSolving #Backtracking #CodingChallenge #SowmiyaCodes #PlacementPrep