Solved Leetcode Problem 7: Reverse Integer with Overflow Check

Anagrams and Minimum Steps: A Frequency Counting Solution! Just solved the "Minimum Steps to Make Two Strings Anagram" problem! When strings 's' and 't' are the same length, the key is efficiently determining how many characters in 't' need to be replaced to match the required character distribution of 's'. My approach is simple and highly efficient: 1. Count Requirements: Use a HashMap to store the exact frequency of every character in string 's'. 2. Match & Mismatch: Iterate through string 't'. If a character is needed (count > 0), we use it up. If it's not needed, it represents one step (one character in 't' that must be changed). This lands us in optimal O(N) time complexity and O(1) space complexity (since the alphabet size is constant). Frequency counting is a powerhouse technique for any string problem! #Algorithms #DataStructures #StringManipulation #LeetCode #CodingChallenge #Java #100daysofcode

💡 Day 19 of My LeetCode Journey — 3Sum Closest Today’s problem was “3Sum Closest”, a slight twist on the classic 3Sum challenge. 🧩 Problem Statement: Given an integer array nums and a target value, find three integers whose sum is closest to the target. Return the sum of those three numbers. Example: Input: nums = [-1, 2, 1, -4], target = 1 Output: 2 Explanation: (-1 + 2 + 1 = 2) ⚙️ Approach: Sort the array for easy pointer movement. Use a for loop to fix one element at a time. Apply the two-pointer technique to find the closest sum. Compare differences using Math.abs() to track which combination is nearer to the target. 📘 Key Learning: 🔹 Sorting simplifies pointer logic. 🔹 Math.abs() is crucial for measuring closeness. 🔹 Sometimes, you don’t need the exact answer — just the closest one. #30DaysOfCode #LeetCode #Java #CodingJourney #ProblemSolving #DataStructures #Algorithms #TwoPointers #TechLearning

✅ Day 68 of LeetCode Medium/Hard Edition Today’s challenge was “Number of Ways to Form a Target String Given a Dictionary” — a brilliant Dynamic Programming and Combinatorics problem that tests precision in transitions and precomputation logic 🧩⚙️ 📦 Problem: You’re given an array of equal-length strings words and a target string target. You must form target from left to right by picking characters from the columns of words under these rules: 1️⃣ Once you use column k from any word, all columns ≤ k in every word become unusable. 2️⃣ You can use multiple characters from the same word, respecting column progression. Return the number of ways to form target modulo 1e9 + 7. 🔗 Problem Link: https://lnkd.in/gns9CwWa ✅ My Submission: https://lnkd.in/g7bsgZq9 💡 Thought Process: This problem is a clever mix of frequency compression and DP memoization. We precompute a frequency table freq[26][m], where each cell represents how many words contain a given letter at column m. Then, using recursion with memoization: 🎯 At each step (i, j) Either use freq[target[i]][j] if it exists → multiply by ways for the next position (i+1, j+1) Or skip the current column → move to (i, j+1) The recurrence relation: dp[i][j] = freq[target[i]][j] * dp[i+1][j+1] + dp[i][j+1] All computations are done modulo 1e9 + 7. ⚙️ Complexity: ⏱ Time: O(26 * n + t * m) — efficient due to frequency precomputation 💾 Space: O(t * m) — for memoized DP table 💭 Takeaway: This challenge reinforced how precomputation and state optimization can transform a seemingly exponential recursion into an elegant polynomial-time solution 🚀 #LeetCodeMediumHardEdition #100DaysChallenge #DynamicProgramming #Combinatorics #ProblemSolving #Java #CodingChallenge #DSA

🌟 Day 44 – LeetCode Practice Problem: Product of Array Except Self (LeetCode #238) 📌 Concept: Given an integer array, create a new array where each element equals the product of all other elements except itself, without using division. 🧠 My Approach: First pass → compute prefix products (multiply everything before current index) Second pass → compute suffix products (multiply everything after current index) Combine both to get the result for each index efficiently No extra multiplication array used — optimized and clean logic ⚡ This avoids division and runs in linear time — exactly what the problem demands ✅ 📈 Result: ✅ Accepted ⚡ Runtime: 2 ms (Beats ~87%) 📦 Memory: Efficient usage 💡 Key Learning: This problem reinforces the power of prefix & suffix computation — a common trick in array manipulation + interview favorite. Great way to improve problem-solving without relying on brute force or division. --- 🚀 Growing stronger every day in DSA! #LeetCode #DSA #Java #ProblemSolving #PrefixSuffix #CodingJourney #ArrayProblems #LearningMindset

#100DaysOfCode – Day 74 Count and Say Problem: Given an integer n, return the n-th term of the “Count and Say” sequence a fascinating pattern where each term describes the previous one. Example: Input: n = 4 Output: "1211" My Approach: Used recursion to generate the previous term. Applied run-length encoding logic counted consecutive digits and built the next term using a StringBuilder. Optimized for clean, readable iteration with O(N²) complexity (due to string building). Understanding recursive string construction deepens how we visualize “generation-based” sequences it’s not just about coding, it’s about seeing patterns grow. #100DaysOfCode #Java #LeetCode #ProblemSolving #Recursion #StringManipulation #CodingJourney #TechWithPurpose #takeUforward

🌳 Day 36/100 ✅ Preorder Traversal of Binary Tree Today I solved the Preorder Traversal problem using recursion. In this traversal method, the order of visiting nodes is simple yet powerful: 👉 Root → Left → Right This problem helped me understand how recursion can naturally handle tree structures by breaking the problem into smaller subproblems. At each recursive call, we visit the root node first, then move to the left subtree, and finally to the right subtree — making the traversal flow smooth and logical. 💡 Key Takeaways: Preorder traversal always starts from the root. Recursion simplifies complex tree navigation into cleaner and readable logic. Visualizing the function call stack helps to truly understand the flow of recursion. Every day, I’m realizing how recursion is not just a coding tool — it’s a way to think step-by-step through structured problems. 🌱 #Day36 #DSA #CodingJourney #BinaryTree #PreorderTraversal #Recursion #ProblemSolving #LearningEveryday #100DaysOfCode #Java

🚀 Day 8 of #45DaysOfLeetCode Challenge 😎 📌Today's problem: Palindrome Number (LeetCode #9) 💡 🔹 Concept: Check whether a given integer reads the same backward and forward — without converting it to a string! 🔹 Logic Used: Instead of reversing the entire number, I reversed only half of it to improve efficiency. This avoids unnecessary computation and eliminates integer overflow risks. 🔹 Key Learnings: ✅ Optimized logic using mathematical manipulation ✅ Improved understanding of number reversal techniques ✅ Importance of edge case handling (negative numbers, trailing zeros) ⚙️ Result: 💻 Runtime: 4 ms (Beats 100%) 💾 Memory: 44.84 MB 😎Small optimizations make a big difference in performance! 💪 📌Problem: https://lnkd.in/ef6AC2j6 🔥Each day is a step closer to writing cleaner and more optimized code. ✨ Let’s keep pushing forward and refining our problem-solving skills! 💻🔥 #LeetCode #Day8 #100DaysOfCode #ProblemSolving #Java #CodingChallenge #PalindromeNumber #DataStructures #Algorithms #LearningEveryday

✅Day 71 of #100DaysOfLeetCode 1.📌Problem: Number of 1 Bits Given a positive integer n, write a function that returns the number of set bits in its binary representation (also known as the Hamming weight). 2.🟢 Difficulty: Easy 3.📍Topic: Bit Manipulation 4.🎯 Goal: Find the number of '1' bits present in the binary representation of the given integer n. 5.🧠 Key idea: Approach 1: Keep dividing n by 2 and count if the remainder is 1. This counts the number of 1's in binary form by repeatedly checking and incrementing a counter. A simple while-loop approach that efficiently solves the problem for any positive integer n.image.jpg #100DaysOfCode #LeetCode #CodingChallenge #BitManipulation #DataStructures #Algorithms #Java #Programming #Tech #CodeNewbie #InterviewPrep #LeetCodeEasy #ProblemSolving #LearnToCode #Developer

🚩 Problem: 41. First Missing Positive 🔥 Day 54 of #100DaysOfLeetCode 🔍 Problem Summary: Given an unsorted integer array nums, return the smallest positive integer that is missing. You must solve it in O(n) time and O(1) extra space. This is one of the most elegant and clever array manipulation problems. 🧠 Intuition: Key observations: The answer must be in the range [1, n+1] We want each number x (1 ≤ x ≤ n) to be placed at index x - 1 This leads to the index marking / cyclic placement trick: Iterate through the array Place each number x into its correct position: At index x - 1 By swapping until the number is either out of range or already in correct place After rearrangement, the first index i where nums[i] != i + 1 gives the answer If all numbers are in correct places → answer is n + 1 Simple idea, extremely powerful.⚙️ Performance: ⏱️ Runtime: 2 ms 🚀 💪 Beats: 99.8% of Java solutions 💾 Memory: 56 MB ⚡ (Beats 95% of users) 📊 Complexity: Time Complexity: O(n) Space Complexity: O(1) ✨ Key Takeaway: This problem teaches the Cyclic Sort pattern — an essential technique for problems involving: Missing numbers Duplicates Correct placement In-place array rearrangement Mastering this trick gives you an edge in advanced array manipulation question Link:[https://lnkd.in/gV8y8dhc] #100DaysOfLeetCode #Day54 #Problem41 #FirstMissingPositive #CyclicSort #IndexMapping #Algorithms #DSA #Java #LeetCode #ProblemSolving #CodingChallenge #InterviewPreparation #CrackingTheCodingInterview #SoftwareEngineering #CodingCommunity #DataStructures #DeveloperJourney #ArjunInfoSolution #TechCareers #CareerGrowth #Programming #CodeNewbie #ComputerScience #LearnToCode #ZeroToHero #CodingIsFun #CodeLife #JavaDeveloper #GameDeveloper #Unity #AI #MachineLearning

#Day36 Of Problem Solving Successfully solved LeetCode #343 – Integer Break 💡 Achieved an Accepted Solution with 0 ms runtime (beats 100% of Java submissions) 🎯 This problem was a great exercise in mathematical optimization and problem decomposition. The key insight was to maximize the product by breaking the integer into 3’s — leveraging both mathematical reasoning and efficient implementation. 📘 Concepts Used: Mathematical analysis of number partitioning Power function and modular logic Time complexity: O(1) Every solved problem is another step toward mastering algorithmic thinking and optimization ⚡ #LeetCode #Java #Coding #ProblemSolving #Mathematics #DSA #DynamicProgramming #CompetitiveProgramming #LearningJourney #SoftwareEngineering #Consistency #Linkedin #HackerRank