Day 14 of #50DaysOfDSA: LeetCode 131. Minimum One Bit Operations to Make Integers Zero (Hard)

📌 Day 3/100 - Remove Duplicates from Sorted Array (LeetCode 26) 🔹 Problem: Given a sorted array, remove the duplicates in-place so that each element appears only once and return the new length. You must modify the array without using extra space for another array. 🔹 Approach: I used a simple counting-based approach: Iterate through the array using a single loop. If the current element is the same as the next, skip it. Otherwise, place it at the current count index and increment count. Finally, return count as the number of unique elements. 🔹 Key Learning: Practiced in-place array modification efficiently without extra space. Improved understanding of loop-based filtering logic. Realized that sometimes the simplest linear approach works best! Consistency compounds — each problem adds a new layer of confidence! 🚀#100DaysOfCode #LeetCode #Java #ProblemSolving #Array #DSA #TwoPointers

🧩 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

🚀 Day 7️⃣7️⃣ of #100DaysOfCode Solved LeetCode Problem #1437 — Check If All 1’s Are at Least K Places Away 🔍✨ 🔧 Concept: Given a binary array, ensure that every 1 is separated by at least K zeros. The solution simply tracks the previous index of 1 and checks the distance. 🧠 Key Idea: Scan the array once When a 1 is found, verify: 👉 currentIndex - previousIndex > k If any pair violates this, return false ⚡ Efficient: Time Complexity: O(n) Space Complexity: O(1) Another clean and efficient logical problem solved. #LeetCode #Java #Arrays #ProblemSolving #DSA #CodingJourney #100DaysOfCode #DeveloperLife #Motivation

#100DaysOfCode – Day 89 Linked Lists & Big Number Addition 🔹 Problem: Add Two Numbers You’re given two non-empty linked lists representing numbers. Each node stores a single digit, and the digits are stored in reverse order. 🔹 My Approach: Used a dummy node to simplify list construction. Traversed both lists simultaneously while managing the carry. Created new nodes for each digit of the sum using modular arithmetic. Continued until both lists and carry were fully processed. Time Complexity: O(max(N, M)) Space Complexity: O(max(N, M)) Even in linked lists, clean logic + careful pointer handling can simplify complex problems. This challenge reinforced how breaking big tasks into smaller, consistent steps leads to efficient solutions. #takeUforward #100DaysOfCode #LinkedList #Java #ProblemSolving #LeetCode #DSA #CodingJourney #CodeNewbie #ChitkaraUniversity

🗓 Day 6 / 100 – #100DaysOfLeetCode 📘 Problem: 3228. Maximum Number of Operations to Move Ones to the End Difficulty: Medium 💡 Problem Summary: Given a binary string s, you can repeatedly choose an index i where s[i] == '1' and s[i+1] == '0', and move that '1' to the right until it reaches the end of the string or hits another '1'. The goal is to find the maximum number of such operations possible. 🧠 My Approach: Instead of simulating the moves (which would be inefficient), I used a counting strategy: Keep a running count of the number of '1's seen so far (cnt). Whenever a '0' appears after one or more '1's, we can perform cnt operations involving those ones. Sum these up for the final result. 📊 Complexity Analysis: Time Complexity: O(n) Space Complexity: O(1) #LeetCode #Java #ProblemSolving #CodingChallenge #100DaysOfCode #DSA #LearningEveryday

📅 Day 52 of #160DaysofDSA 🔹 Count Pairs with Sum Less Than Target Today’s problem was an elegant use of the Two Pointer Technique — a powerful approach for solving array problems efficiently ⚡ 🧩 Problem Statement: Given an array arr[] and an integer target, find the number of pairs whose sum is strictly less than the target. 💡 Approach (Two Pointer Technique): 1️⃣ Sort the array. 2️⃣ Use two pointers: left = 0, right = n - 1. 3️⃣ If arr[left] + arr[right] < target: ➤ All pairs between left and right will also have smaller sums. ➤ Add (right - left) to count and move left++. 4️⃣ Else, move right--. ⏱ Time Complexity: O(N log N) 💾 Space Complexity: O(1) #DSA #Java #Coding #TwoPointers #ProblemSolving #Sum #Count #Target #GFG #Day52

💻 Day 36 of #100DaysOfLeetCode 🚩 Problem: 141. Linked List Cycle 🔍 Problem Summary: Given the head of a linked list, determine if it contains a cycle. A cycle exists if a node’s next pointer points to a previous node, causing an infinite loop. ✅ Approach: Floyd’s Cycle Detection Algorithm (Tortoise & Hare) We use two pointers: Slow pointer moves 1 step at a time. Fast pointer moves 2 steps at a time. 🔄 If both meet at the same node, a cycle exists. Link:[https://lnkd.in/gvrqQ6K6] #100DaysOfLeetCode #Day36 #Problem141 #LinkedListCycle #LinkedList #Java #DSA #DataStructures #Algorithms #CodingChallenge #ProblemSolving #LearnToCode #CodeNewbie #ProgrammerLife #SoftwareDeveloper #TechCareers #InterviewPreparation #CrackingTheCodingInterview #LeetCode #JavaDeveloper #CodingCommunity #WomenInTech #FutureEngineer #PlacementPreparation #MastersInCS #ArjunInfoSolution #TechJourney #AIEnthusiast #SoftwareEngineering #EngineeringStudents #CollegeToCorporate

💡 LeetCode #15 — 3Sum Today I solved LeetCode Problem 15: 3Sum 🔍 Problem Summary: Given an integer array nums, return all unique triplets (a, b, c) such that: a + b + c = 0 The solution must not contain duplicate triplets. Key Idea: Use sorting + two pointers to reduce the time from O(n³) to O(n²). Steps: Sort the array. Fix one number (nums[f]) in each iteration. Use two pointers (i and j) to find pairs whose sum equals -nums[f]. Skip duplicates for both the fixed index and the pointer values. This efficiently finds all unique triplets. #LeetCode #Java #DSA #TwoPointers #ProblemSolving #CodingChallenge #Algorithms

🌟 Day 93 of #100DaysOfCodingChallenge 🌟 🔹 Problem: 442. Find All Duplicates in an Array 🔹 Platform: LeetCode 🔹 Language: Java 🔹 Difficulty Level: Medium 🧩 Problem Statement: Given an integer array nums containing numbers from 1 to n, find all the elements that appear twice in the array. 💡 Intuition: To identify duplicates efficiently, I used two HashSets — one to store visited elements and another to store duplicates when an element is already seen. This method makes the solution concise and easy to understand while maintaining good performance. ✅ Time Complexity: O(n) ✅ Space Complexity: O(n) 🚀 Key Takeaway: Sometimes, clarity and simplicity in code matter more than optimization — this approach is easy to implement and perfectly suitable for most scenarios. #Day93 #LeetCode #100DaysOfCode #Java #ProblemSolving #CodingChallenge #DSA #LearnByCoding #WomenInTech

🚩 Problem: 349. Intersection of Two Arrays 🔥 Day 42 of #100DaysOfLeetCode 🔍 Problem Summary: Given two integer arrays nums1 and nums2, return an array of unique elements that appear in both arrays. The result must not contain duplicates. The order of elements in the output does not matter. ✅ Approach: Using HashSet Store elements of nums1 in a HashSet. Loop through nums2, and if an element exists in HashSet, add to result set. Convert the set into an array. 📊 Complexity: Time Complexity: O(m + n) Space Complexity: O(m + n) 🎯 Key Takeaway: This problem reinforces the concept of using HashSet to remove duplicates and perform efficient membership checks, which is fundamental for solving array and hashing problems optimally. Link:[https://lnkd.in/gz9Y99Ak] #100DaysOfLeetCode #Day42 #Problem349 #IntersectionOfArrays #HashSet #Java #DSA #Algorithms #CodingChallenge #ProblemSolving #LeetCode #InterviewPreparation #DataStructures #TechCareers #ArjunInfoSolution #CodingCommunity #CodeNewbie #ZeroToHero #SoftwareEngineering