Converting String to Integer with LeetCode Challenge

💡 LeetCode #58 — Length of Last Word Today I solved LeetCode Problem 58: Length of Last Word 🧠 Problem Summary: Given a string s consisting of words and spaces, return the length of the last word in the string. A word is defined as a sequence of non-space characters. Key Idea: We can traverse the string from end to start — Skip trailing spaces. Count the number of characters until the next space (which marks the last word). This avoids unnecessary splitting and saves extra space. #LeetCode #Java #DSA #Strings #ProblemSolving #CodingChallenge

Day 46 of #100DaysOfLeetCode Challenge 🚀Today’s Problem: Remove Duplicate Letters (LeetCode - Medium) Language: Java In this problem, the goal is to remove duplicate letters from a string such that every letter appears once and the result is the smallest lexicographical order possible. Concepts Covered: Stack for maintaining order Greedy approach for smallest lexicographical sequence Tracking last occurrence index of each character Boolean array to keep track of visited characters Key Takeaways: Understanding when to remove elements from the stack is crucial for optimal ordering. The combination of stack + greedy ensures both uniqueness and lexicographical minimality. #100DaysOfCode #LeetCode #Java #ProblemSolving #CodingChallenge #DSA #DevelopersJourney

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

👇 🚀 Stream API Coding – 1: Check if a String is Palindrome Exploring the power of Java 8 Stream API with a simple yet elegant example — checking whether a string is a palindrome using functional style. 💡 String s = "racecar"; boolean isPalindrome = IntStream.range(0, s.length() / 2) .allMatch(i -> s.charAt(i) == s.charAt(s.length() - i - 1)); System.out.println(isPalindrome ? "String is Palindrome" : "String is not palindrome"); ✨ Key Learnings: ✅ Use IntStream.range() for index-based iteration ✅ allMatch() ensures all comparisons pass ✅ Clean and concise approach — no loops, no extra variables 📚 Output: 👉 String is Palindrome This is just the beginning — more Stream API challenges coming soon! 🔥 #Java #StreamAPI #CodingChallenge #JavaDeveloper #FunctionalProgramming #CleanCode #CodingSeries #InterviewPreparation

🚀 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

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

💻 LeetCode 50 Days Challenge — Day 8: Median of Two Sorted Arrays Day 8 of my #LeetCode50DaysChallenge ✅ Today’s problem was about finding the median of two sorted arrays — Median of Two Sorted Arrays ✨ 🧩 Problem: Given two sorted arrays nums1 and nums2, return the median of the two sorted arrays. The challenge was to merge them efficiently and then determine the middle value(s). 💡 Approach: I used Java 8 Streams to merge both arrays in a single line using IntStream.concat(), followed by Arrays.sort() to sort the combined array. Finally, I calculated the median by checking if the array length is even or odd. ⏱️ Time Complexity: O((m + n) log (m + n)) (due to sorting) 📊 Example: Input: nums1 = [1, 2], nums2 = [3, 4] Output: 2.5 Each day, I’m getting more comfortable with cleaner and modern Java techniques like Streams — small steps toward writing concise and efficient code 🚀 #LeetCode #CodingChallenge #Day8 #ProblemSolving #Java #SoftwareDevelopment #Consistency #50DaysOfCode

🚀 Day 33 of My Daily Coding Challenge 🚀 Today I implemented and revised Cyclic Sort (Java) — one of the most efficient sorting techniques for arrays containing numbers from 1 to N. 💡 Key takeaway: Instead of comparing adjacent elements, Cyclic Sort keeps placing each number at its correct index — making it super fast for such problems. 🔗 Solution: [https://lnkd.in/epUjThMt] Building consistency, one algorithm at a time 💻💪 #DailyCodingChallenge #Java #DSA #CyclicSort #SortingAlgorithms #100DaysOfCode #LearningEveryday

🚀 Day 56 of my #100DaysOfLeetCode Challenge 🚀 Today's problem: Matrix Diagonal Sum Language: Java In this challenge, I practiced calculating the sum of both primary and secondary diagonals of a square matrix. The goal was to improve my understanding of nested loops and conditional logic for index-based problems. Key Takeaways: Strengthened logic for 2D array traversal Improved debugging and edge case handling Reinforced clean, readable code habits Here’s the core idea from my solution: If i == j (primary diagonal) or i + j == n - 1 (secondary diagonal), add that element to the sum. #LeetCode #Java #CodingChallenge #100DaysOfCode #ProblemSolving #SoftwareEngineering