3Sum Closest Problem Solution in Java

Leetcode Practice - 8. String to Integer (atoi) The problem is solved using JAVA. Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer. The algorithm for myAtoi(string s) is as follows: ✔ Whitespace: Ignore any leading whitespace (" "). ✔ Signedness: Determine the sign by checking if the next character is '-' or '+', assuming positivity if neither present. ✔ Conversion: Read the integer by skipping leading zeros until a non-digit character is encountered or the end of the string is reached. If no digits were read, then the result is 0. ✔ Rounding: If the integer is out of the 32-bit signed integer range [-231, 231 - 1], then round the integer to remain in the range. Specifically, integers less than -231 should be rounded to -231, and integers greater than 231 - 1 should be rounded to 231 - 1. Return the integer as the final result.   Example 1: Input: s = "42" Output: 42 Explanation: The underlined characters are what is read in and the caret is the current reader position. Step 1: "42" (no characters read because there is no leading whitespace) ^ Step 2: "42" (no characters read because there is neither a '-' nor '+') ^ Step 3: "42" ("42" is read in) #LeetCode #Java #StringHandling #CodingPractice #ProblemSolving #DSA #DeveloperJourney #TechLearning

🚀 Day 52/100 Today’s problem was based on String Manipulation — reversing the first k characters of a string. 🧠 What I learned: - How to efficiently manipulate strings - Using "StringBuilder" for reversal in Java - Writing clean and optimized code 💡 Approach: Reversed the first k characters and appended the remaining string. ⚡ Key Insight: Even simple problems help strengthen fundamentals, which are crucial for solving complex problems later. 👨💻 Code (Java): class Solution { public String reversePrefix(String s, int k) { String first = new StringBuilder(s.substring(0, k)).reverse().toString(); String rest = s.substring(k); return first + rest; } } 📈 Consistency is the key — showing up every day matters more than perfection. #Day52 #CodingChallenge #Java #DSA #LearningJourney #Consistency #100DaysOfCode

Leetcode Practice - 15. 3Sum The problem is solved using JAVA. Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0. Notice that the solution set must not contain duplicate triplets.   Example 1: Input: nums = [-1,0,1,2,-1,-4] Output: [[-1,-1,2],[-1,0,1]] Explanation: nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0. nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0. nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0. The distinct triplets are [-1,0,1] and [-1,-1,2]. Notice that the order of the output and the order of the triplets does not matter. Example 2: Input: nums = [0,1,1] Output: [] Explanation: The only possible triplet does not sum up to 0. Example 3: Input: nums = [0,0,0] Output: [[0,0,0]] Explanation: The only possible triplet sums up to 0.   Constraints: 3 <= nums.length <= 3000 -105 <= nums[i] <= 105 #LeetCode #Java #CodingPractice #ProblemSolving #DSA #Array #DeveloperJourney #TechLearning

🔢 Divisible Sum Pairs Problem | Java Solution 💻 Today I solved an interesting problem based on arrays and modular arithmetic! 📌 Problem: Given an array and a number k, count pairs (i, j) such that: 👉 i < j 👉 (arr[i] + arr[j]) % k == 0 ⚡ Approach: Instead of brute force O(n²), I used an optimized O(n) approach using remainder frequency. ✔️ Key Idea: Store frequency of (element % k) For each element, find its complement remainder Count valid pairs efficiently 💡 This improves performance significantly for large inputs! 🧠 Concepts Used: Arrays Modulo Arithmetic Hashing Technique 👨💻 Language: Java #Java #CodingPractice #DataStructures #Algorithms 

📅 Day 10 – Strings in Java 📚 Topic: String Manipulation (Hashing) Today I solved the problem of checking whether two strings are anagrams using an efficient frequency-count approach. This helped me understand how hashing can be used to compare character distributions instead of sorting. ✔ Key Learnings: • Applied hashing using a frequency array • Improved understanding of character frequency comparison • Learned how to optimize from sorting (O(n log n)) to linear time (O(n)) Building problem-solving skills step by step and focusing on writing efficient code 🚀 #DSA #Strings #Java #ProblemSolving #LearningJourney

Solved LeetCode 17 – Letter Combinations of a Phone Number using backtracking in Java. Approach: Mapped each digit (2–9) to its corresponding characters using a simple array for O(1) access. Then used backtracking to build combinations digit by digit. For every digit: Pick each possible character Append → explore next digit → backtrack Key idea: Treat it like a tree of choices, where each level represents a digit and branches represent possible letters. Key learnings: Backtracking = build → explore → undo StringBuilder helps avoid unnecessary string creation Problems like this are about systematic exploration of choices Time Complexity: O(4^n * n) Space Complexity: O(n) recursion stack + output Consistent DSA practice is strengthening pattern recognition day by day. #Java #DSA #Backtracking #LeetCode #CodingInterview #SoftwareEngineering

This Java snippet trips up even experienced developers 👇 Integer a = 127; Integer b = 127; Integer c = 128; Integer d = 128; System.out.println(a == b); System.out.println(c == d); One prints true. One prints false. At first glance, both comparisons look identical. Same type. Same assignment. Same operator. So what’s going on? If you know the reason, you’ve got a deeper grasp of how Java actually works under the hood. Drop your explanation below — no Googling 👇 #Java #Programming #SoftwareEngineering #DevCommunity #CodeChallenge

#Day86 of #100DaysOfCode Started learning String manipulation in Java. Covered: * Basic string operations * Using built-in methods like length() and charAt() Practiced problems like: * Reversing a string * Checking palindrome strings * Counting vowels Focused on understanding how to process and manipulate text data. #Java #Strings #100DaysOfCode

🚀 DSA Journey — LeetCode Practice 📌 Problem: Binary Tree Inorder Traversal (LeetCode 94) 💻 Language: Java 🔹 Approach: To solve this problem, I used Depth-First Search (DFS) with Recursion to perform inorder traversal of the binary tree. • Recursively traverse the left subtree • Visit the root node and store its value • Recursively traverse the right subtree • Follow Inorder pattern: Left → Root → Right This approach works efficiently because recursion naturally follows the inorder traversal structure. ⏱ Time Complexity: O(n) 🧩 Space Complexity: O(n) (including recursion stack) 📖 Key Learning: This problem strengthened my understanding of DFS, recursion, and tree traversal patterns. It also reinforced the inorder pattern (Left → Root → Right), which is a fundamental concept used in many tree-based problems 💡 #DSA #Java #LeetCode #ProblemSolving #CodingJourney #LearningInPublic #BinaryTree #DFS #Recursion #TreeTraversal

🚀 DSA Journey — LeetCode Practice 📌 Problem: Binary Tree Preorder Traversal (LeetCode 144) 💻 Language: Java 🔹 Approach: To solve this problem, I used Depth-First Search (DFS) with Recursion to perform preorder traversal of the binary tree. • Visit the root node first • Store the root value in the result list • Recursively traverse the left subtree • Recursively traverse the right subtree • Follow Preorder pattern: Root → Left → Right This approach works efficiently because recursion naturally follows the preorder traversal structure. ⏱ Time Complexity: O(n) 🧩 Space Complexity: O(n) (including recursion stack) 📖 Key Learning: This problem strengthened my understanding of DFS, recursion, and tree traversal patterns. It also reinforced the preorder pattern (Root → Left → Right), which is a fundamental concept used in many tree-based problems 💡 #DSA #Java #LeetCode #ProblemSolving #CodingJourney #LearningInPublic #BinaryTree #DFS #Recursion #TreeTraversal