How to Concatenate an Array in Java for LeetCode 1929

#Day_28 Today’s problem was quite an interesting one — “Reach a Target Number” using minimal moves. 💡 Problem Summary: Starting from 0, on each move i, you can go either left or right by i steps. The goal is to find the minimum number of moves required to reach a given target. At first glance, it looked like a simple math problem, but the trick was to notice the parity condition — once the cumulative sum goes beyond the target, the difference (sum - target) must be even to allow flipping directions and still land exactly on target. Here’s the optimized logic I implemented in Java Key Takeaway: Sometimes, problems that look complex are just about observing patterns in numbers rather than brute force. A touch of math can simplify the entire logic! #100DaysOfCode #LeetCode #CodingChallenge #Java #ProblemSolving #DSA #LearnEveryday #CodingJourney

📌 Day 32/100 – Count and Say (LeetCode 38) 🔹 Problem: Given an integer n, return the nth term of the Count and Say sequence, where each term is generated by describing the digits of the previous term. 🔹 Approach: Used recursion to generate the previous sequence (n-1) and built the current one by counting consecutive identical digits. Utilized StringBuilder for efficient string formation during traversal. 🔹 Key Learning: Improved recursion and pattern recognition skills. Practiced string manipulation and efficient concatenation. Understood how to construct sequences based on descriptive logic. 🔹 Complexity: Time: O(m) per level (m = length of previous term) Space: O(n) due to recursion depth #100DaysOfCode #LeetCode #Java #Recursion #Strings #ProblemSolving #DSA #CodingPatterns #Motivation #descipline

Solved: LeetCode Problem 922 – Sort Array By Parity II 📌 Difficulty Level: Easy 💻 Language Used: Java 🧠 Topics: Arrays, Index Manipulation, Two-Pointer 🧩 Problem Overview: Rearrange the array so that elements at even indices are even numbers and elements at odd indices are odd numbers. 🎯 Key Learnings: ✅ Practiced separate pointer movement for even/odd indices ✅ Learned how to skip indices in steps of 2 for efficiency ✅ Strengthened control flow in condition-based pointer increment ✅ Designed an O(n) solution without extra space 🛠 Skills Practiced: Index Management Array Rearrangement Two-Pointer Optimization In-place Logic Building ⚡️ Problems like this sharpen thinking about index constraints and help in mastering array partitioning & arrangement patterns. #LeetCode #Java #RotateArray #ProblemSolving #Arrays #LogicBuilding #100DaysOfCode #CodeNewbie #TechCareers #WomenInTech #LearningInPublic #BuildInPublic #DeveloperJourney #CleanCode #DSA #AlgorithmPractice #CodingInterviewPrep #SoftwareDevelopment #AfrozCodes #DailyCoding #CrackTheCodingInterview #TechForAll

🔢 Today I worked on a classic matrix problem in Java — Diagonal Sum. The goal was simple: Calculate the sum of both the primary and secondary diagonals of a square matrix. Initially, I used a nested loop approach with O(n²) complexity. But then I optimized it to a clean O(n) solution by directly accessing diagonal indexes. While optimizing, I made an interesting mistake: ❌ I wrote if (i != matrix[i][matrix.length - 1 - i]) Here I accidentally compared an index with an element value. ✔ Correct approach: if (i != matrix.length - 1 - i) This ensures the center element in odd-sized matrices isn’t counted twice. 🧠 Key Learning: Indexes and values may look similar, but mixing them can break logic silently. Optimization is not just about speed — it’s about accuracy. #Java #leetcode #Coding #DSA #ProblemSolving #LearningInPublic #SoftwareEngineering

📌 Day 13/100 - Valid Anagram (LeetCode 242) 🔹 Problem: Given two strings s and t, determine whether t is an anagram of s. An anagram is formed by rearranging the letters of one word to create another word — meaning both strings must have the same characters with the same frequency. 🔹 Approach: First, check if both strings are of equal length — if not, they can’t be anagrams. Convert both strings into character arrays. Sort both arrays and compare them using Arrays.equals(). If both sorted arrays are identical, the strings are anagrams. 🔹 Key Learning: Sorting can simplify comparison-based string problems. Always check base conditions (like length) to avoid unnecessary computation. This approach has a time complexity of O(n log n) due to sorting, which is efficient enough for this problem. Every solved challenge is another letter added to the dictionary of progress! 🚀 #100DaysOfCode #LeetCode #Java #ProblemSolving #DSA #CodingJourney #Consistency

🚀 LeetCode Day 3: Reverse Integer Today’s challenge was all about reversing the digits of an integer while handling tricky edge cases — especially negative numbers and integer overflow. 🧠 💡 Problem: Given a signed 32-bit integer, reverse its digits. If the reversed integer overflows, return 0. 🧩 Key Takeaways: Extract digits using modulo and division Manage negative numbers gracefully Check for overflow before building the reversed number 💻 Language: Java ✅ Topic: Math / Integer Manipulation Every day is a new step toward mastering problem-solving and improving coding logic. #LeetCode #100DaysOfCode #Java #CodingChallenge #ProblemSolving #LeetCodeJourney

🚀#PostLog35 🧩 Problem: Frequency of the Most Frequent Element (LeetCode #1838) 🔹 Topic: Sliding Window 🔹 Language: Java Challenge was about maximizing the frequency of an element after performing at most k increment operations. The key idea was to use a sliding window combined with sorting: Sort the array to align potential targets. Expand the window while the total operations needed ≤ k. If it exceeds k, shrink the window from the left. Track the maximum frequency achievable. 💡 This approach efficiently balances elements using prefix sums and window logic — no need for nested loops or complex math tricks. ✅ Result: Accepted (Runtime: 33 ms) Another problem done — one step closer to mastering array and window patterns! #LeetCode #Java #CodingJourney #ProblemSolving #DSA

📌 Day 16/100 - Reverse String (LeetCode 344) 🔹 Problem: Reverse a given string in-place — meaning you must modify the original array of characters without using extra space. 🔹 Approach: Used the two-pointer technique — one starting at the beginning and one at the end of the array. Swap characters at both pointers, then move them closer until they meet. Efficient, clean, and runs in linear time without additional memory allocation. 🔹 Key Learnings: In-place algorithms optimize space complexity significantly. The two-pointer pattern is a versatile tool for many array and string problems. Understanding mutable vs immutable structures in Java is crucial for memory efficiency. Sometimes, the simplest logic beats the most complex one. 🧠 “True efficiency lies in simplicity, not complexity.” #100DaysOfCode #LeetCode #Java #ProblemSolving #DSA #CodingJourney #TwoPointers

🚀 Day 398 of #500DaysOfCode 🔹 Problem: 914. X of a Kind in a Deck of Cards 🔹 Difficulty: Easy 🔹 Language Used: Java ☕ 🧩 Problem Summary: Given a deck of cards where each card has an integer, the goal is to check if we can divide the deck into groups such that: Each group has exactly X cards, where X > 1, and All cards in a group have the same integer. If such a partition is possible, return true — otherwise, return false. 💡 Key Idea: The solution relies on finding the greatest common divisor (GCD) of all card frequencies. If the GCD of all counts is greater than 1, we can form valid groups; otherwise, we can’t. ⚙️ Approach: 1️⃣ Count the frequency of each card using a HashMap. 2️⃣ Compute the GCD of all frequency values. 3️⃣ If GCD > 1, return true; else, false. 🧠 Concepts Used: HashMap GCD (Euclidean Algorithm) Frequency Counting ✅ Example: Input: [1,2,3,4,4,3,2,1] → Output: true Input: [1,1,1,2,2,2,3,3] → Output: false 📘 Lesson Learned: Even simple-looking problems can hide elegant mathematical patterns. Understanding GCD turned out to be the key! 💪 #Day398 #Java #LeetCode #ProblemSolving #CodingChallenge #LearnEveryday #Programming #Developer #100DaysOfCode #500DaysOfCode

🌟 Day 86 of #100DaysOfCodingChallenge 🌟 🚀 Problem: 561. Array Partition 💻 Language: Java Today’s challenge was about maximizing the sum of the minimum values in n pairs formed from an array of 2n integers. 🧩 Approach: 1️⃣ Sort the array in ascending order. 2️⃣ Take every alternate element (starting from index 0). 3️⃣ Add them up — this gives the maximum possible sum of mins in all pairs. 🧠 Key Insight: When the array is sorted, pairing adjacent elements ensures the smallest numbers are always matched optimally, leading to the maximum possible result. 📊 Example: Input → [1,4,3,2] Sorted → [1,2,3,4] Pairs → (1,2), (3,4) Output → 4 ✅ ✨ Learning: Sometimes the simplest approach — sorting and selecting — can yield the most optimal result! #Day86 #LeetCode #Java #100DaysOfCode #CodingChallenge #ProblemSolving #ArrayPartition #DSA #WomenInTech #SathyabamaUniversity