"Day 5 of #100DaysOfCode: Missing Number and First Repeating Element"

💡 Day 57 of #100DaysOfCode 💡 Today, I solved LeetCode Problem #217 – Contains Duplicate 🧩 Given an integer array, the task is simple yet fundamental: 👉 Determine whether any value appears at least twice in the array. 💻 Language: Java ⚡ Runtime: 13 ms — Beats 87.70% 🚀 📉 Memory: 58.28 MB — Beats 63.02% 🧠 Key Learnings: Reinforced my understanding of HashSet and its O(1) lookup time. Learned how sets help efficiently identify duplicates in large datasets. Strengthened array traversal and data structure optimization skills. 🧩 Approach: 1️⃣ Initialize an empty HashSet. 2️⃣ Iterate through each element in the array. 3️⃣ If an element already exists in the set → return true. 4️⃣ Else, add it to the set. 5️⃣ If loop completes, return false. ✅ Complexity: Time: O(n) Space: O(n) ✨ Takeaway: Even basic problems like these teach the power of hash-based structures — simple yet powerful for building optimized systems and preventing redundant computations. #100DaysOfCode #LeetCode #Java #DataStructures #HashSet #ProblemSolving #CodingChallenge #CleanCode #Programming #SoftwareEngineering #TechLearning #Algorithms

✅Day 54 : Leetcode 3228 - Maximum Number of Operations to Move Ones to the End #60DayOfLeetcodeChallenge 🧩 Problem Statement You are given a binary string s. You can perform the following operation any number of times: Choose an index i such that: i + 1 < s.length s[i] == '1' s[i+1] == '0' Then move this '1' to the right until it reaches the next '1' or the end of the string. Your task is to return the maximum number of such operations that can be performed. ✅ My Approach Instead of simulating every movement (which is slow), I used a counting strategy: 🔹 Key Idea Whenever I see a '1', I increase the count of ones seen so far. Whenever I see a pattern like …10…, this '1' can move past all previous ones, contributing extra operations equal to the number of '1's before it. 🔹 Steps Initialize: ones = 0 → counts how many '1' encountered so far. res = 0 → stores total operations. Traverse the string: If current char is '1' → increment ones. Else if the previous char is '1' (meaning we found 10) → add ones to result. Return res as the maximum operations. This avoids simulation and gives the optimal count directly. ⏱ Time Complexity O(n) — Single scan through the string O(1) space #dsa #leetcode #binarysearch #java #coding #problemsolving #100daysofdsa #interviewpreparation #learningeveryday #codingjourney

🌟 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

🚩 Problem: 283. Move Zeroes 🔥 Day 39 of #100DaysOfLeetCode 🔍 Problem Summary: Given an integer array nums, move all zeroes to the end of the array while maintaining the relative order of the non-zero elements. You must do this in-place (without using extra space). ✅ Two-Pointer Efficient Approach: Use one pointer to track the position to place non-zero elements. Iterate through the array: When a non-zero element is found, place it at the current pointer index and move the pointer forward. After processing all non-zero elements, fill the remaining positions with zeroes. 📊 Complexity: Time Complexity: O(n) Space Complexity: O(1) 🎯 Key Takeaway: This problem reinforces in-place array manipulation and the two-pointer strategy, a critical technique for optimizing time and space in real-world applications. Link:[https://lnkd.in/gwQFp3fA] #100DaysOfLeetCode #Day39 #Problem283 #MoveZeroes #TwoPointers #ArrayManipulation #DSA #Java #Algorithms #CodingChallenge #ProblemSolving #LeetCode #InterviewPrep #TechCareers #CodingCommunity #ArjunInfoSolution #SoftwareEngineering #ZeroToHero #CareerGrowth #CodeNewbie

🚀 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 58 of My LeetCode Journey 🚀 🔹 Problem: Search Insert Position 🔹 Topic: Binary Search (Fundamental DSA Technique) Today’s challenge focused on enhancing one core skill: thinking in terms of boundaries and decisions. The task was simple — in a sorted array, find the index of the target element. If it doesn’t exist, return the index where it logically should be inserted. What made this problem interesting is that instead of searching linearly, we leverage Binary Search, which reduces the operations drastically. ✅ Key Takeaways & Learning: Binary Search is not just used to find something — it can also determine the correct position for insertion. Understanding how low, mid, and high move helps build clear logic flow. The index where the search ends (low) gives the correct insert position, even if the element doesn't exist. Rewriting brute-force logic into binary search trains the brain to think in terms of optimization. ⏱️ Complexity Analysis: MetricValueTimeO(log n)SpaceO(1) 🌱 Personal Reflection: Every binary search problem improves clarity of thought. From checking a value to identifying a boundary or insert point — the mindset shifts from “finding an element” to “solving a pattern.” #100DaysOfCode #LeetCode #DSA #BinarySearch #Java #ProblemSolving #LearningInPublic #TechJourney #ConsistencyIsKey

🚩 Problem: 169. Majority Element 🔥 Day 43 of #100DaysOfLeetCode 🔍 Problem Summary: Given an array nums, find the element that appears more than ⌊ n/2 ⌋ times. You may assume that the majority element always exists in the array. ✅ Approach 1: Using HashMap Count the frequency of each element using a HashMap and return the one that occurs most frequently. ✅ Approach 2 (Optimized): Boyer–Moore Voting Algorithm Maintain a candidate and count. Traverse the array: If count == 0, set candidate = num. If num == candidate, increment count, else decrement. Return the candidate. 📊 Complexity: Time Complexity: O(n) Space Complexity: O(1) ✨ Key Takeaway: The Boyer–Moore Voting Algorithm is a brilliant example of how a deep understanding of problem constraints can lead to an O(1) space solution — an essential pattern in algorithmic thinking. Link:[https://lnkd.in/gsc4XgbK] #100DaysOfLeetCode #Day43 #Problem169 #MajorityElement #BoyerMoore #Algorithms #DSA #Java #CodingChallenge #ProblemSolving #LeetCode #InterviewPreparation #DataStructures #TechCareers #ArjunInfoSolution #CodingCommunity #CodeNewbie #SoftwareEngineering #ZeroToHero #LearnToCode

📌 Day 153 of Coding - Maximum Frequency of an Element After Operations I (LeetCode - Medium) 🎯 Goal: Given an integer array nums, an integer range k, and a number of operations, determine the maximum possible frequency of any number after performing up to numOperations modifications - where each operation can change a number by at most k. 💡Approach & Debugging: Found the maximum value in the array and created a frequency prefix array. For each possible target value i, calculated the total numbers within the [i - k, i + k] range. The frequency of i could be increased using available operations on nearby elements. Kept track of the best achievable frequency. ✔️ Time Complexity: O(N + M) - where M is the value range up to max(nums) + k. ✔️ Space Complexity: O(M) 🧠Key Takeaways: Prefix sums + frequency arrays are powerful for range-based computations. Always optimize loops around range-based frequency calculations. #Day153 #LeetCode #Arrays #PrefixSum #Frequency #Java #ProblemSolving #DSA #CodingChallenge #Algorithms #100DaysOfCode #InterviewPrep #SoftwareEngineering #DataStructures #CodeNewbie #ProgrammersLife

Day 6 of #100DaysOfCode Today’s exploration of two classic array problems, Two Sum and Equilibrium Point, enhanced my comprehension of hashing, prefix sums, and optimization techniques. Problem 1: Two Sum (LeetCode #1) Objective: Identify two numbers that sum to a specified target and return their indices. Approach: Utilizing a one-pass HashMap, I stored each element’s index and promptly verified if the complement (target minus the current element) already existed. This approach ensured both efficiency and elegance. Time Complexity: O(n) Space Complexity: O(n) Problem 2: Equilibrium Point (GFG) Objective: Determine the index where the sum of elements on the left equals the sum on the right. Approach: By calculating the total sum, I maintained a running leftSum while dynamically updating the rightSum in a single pass. Time Complexity: O(n) Space Complexity: O(1) Key Takeaways: These problems demonstrated how clear mathematical logic combined with appropriate data structures can yield powerful yet straightforward solutions. Each line of code imparts a novel perspective on efficient problem-solving. Thank you Rajesh Gupta all for your guidance and support! 🙌 hashtag #100DaysOfCode #LeetCode #DSA #ProblemSolving #Arrays #CodingChallenge #LearningInPublic #Java #Programming

🧩 Day 78 of #100DaysOfCode Challenge 🧩 📘 LeetCode 56 — Merge Intervals Topic: Arrays & Sorting Today’s problem was about merging overlapping intervals — a real test of logical thinking and array manipulation skills! 💡 Problem Summary: Given multiple intervals, the goal is to merge all overlapping ones and return a list of non-overlapping intervals that cover all ranges in the input. ⚙️ Approach (Brute + Optimized): 1️⃣ Sort all intervals based on their start value. 2️⃣ Compare the current interval with the previous one: If they overlap, merge them by updating the end time. If not, add the previous interval to the result. 3️⃣ Continue until all intervals are processed. 🧠 Key Learning: Sorting simplifies overlapping detection. Interval-based problems help strengthen array and logic handling. Real-world application: scheduling problems and timeline merging! ✨ Each problem is another step toward mastering patterns in DSA. #Day78 #LeetCode #100DaysOfCode #Java #DSA #ProblemSolving #WomenInTech #CodingJourney #Arrays #Sorting