How to duplicate zeros in a fixed-length array in place

#Day13 2 – Add Two Numbers Solved this classic linked list problem by implementing an efficient algorithm to add two numbers represented as reverse-digit linked lists. 💡 Approach: Used a dummy node to simplify list construction, iterated through both lists while handling different lengths, and maintained a carry value for proper digit summation. The solution elegantly handles cases where lists have different sizes and final carry propagation. #LeetCode #Java #LinkedList #DataStructures #Algorithms #ProblemSolving #Coding #Tech #SoftwareEngineering #DailyCoding #Programming

✅Day 58 of #100DaysOfLeetCode 1.📌Problem: 2011. Final Value of Variable After Performing Operations 2.✅ Difficulty: Easy 3.📍Topic: Array 4.🎯 Goal: Given an array of strings operations containing a list of operations, return the final value of x after performing all the operations. 5.🧠key idea: Approach 1: Initialize a counter variable to zero. Iterate through the input array of operations. For each operation string, check if it represents an increment ("++X" or "X++"). If it does, increase the counter by one. Otherwise, it's a decrement, so decrease the counter by one. After checking all the operations, return the final value of the counter. #100DaysOfLeetCode #CodingChallenge #Java #ProblemSolving #SoftwareDevelopment #Developer #Programming #CodeNewbie #Tech #DataStructures #Algorithms #Array #LeetCodeChallenge #SoftwareEngineer #DailyChallenge

✅Day 78 of #100DaysOfLeetCode 1.📌Problem: Find the Difference Given two strings s and t, where t is generated by shuffling s and adding one extra letter at a random position, return the letter that was added. 2.🟢 Difficulty: Easy 3.📍Topic: Strings 4.🎯 Goal: Find the extra character that was added to string t after shuffling and insertion. 5.🧠Key idea: Approach 1: Use the XOR operation to efficiently find the difference. Iterate over both strings, XOR all characters together, and the result will be the added character. This works because XOR-ing identical characters cancels them out, leaving only the unique/added character. #LeetCode #100DaysChallenge #Programming #Coding #CodeNewbie #Java #Tech #SoftwareEngineering #ProblemSolving #Developer #DailyChallenge #HashTable #Algorithms #CodingInterview #LearnToCode

✅Day 74 of #100DaysOfLeetCode 1.📌Problem: Reverse Vowels of a String 2.🟩 Difficulty: Easy 3.📍Topic: Two Pointers 4.🎯 Goal: Given a string, reverse only the vowels in the string and return the modified string. Vowels include 'a', 'e', 'i', 'o', 'u' in both lower and upper cases. 5.🧠key idea: Approach 1: Use two pointers, one starting from the beginning and one from the end. Move towards each other, swap the vowels found at each pointer, and continue until all vowels are reversed. #LeetCode #CodingChallenge #100DaysOfCode #Algorithms #TechCareers #CodeNewbie #WomenWhoCode #Java #Programming #DailyCoding #Developer #InterviewPrep #TwoPointers #DataStructures #LearnToCode #ProblemSolving

✅Day 57 of #100DaysOfLeetCode 1.📌Problem: 350. Intersection of Two Arrays II 2.🟢 Difficulty: Easy 3.📍Topic: Arrays 4.🎯 Goal: Given two integer arrays, nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays, and you may return the result in any order. 5.🧠key idea: Approach 1: Sort both arrays and use a two-pointer technique. Initialize two pointers, one for each array, starting at the beginning. If the elements at the current pointers are the same, add the element to the result and advance both pointers. If the element in the first array is smaller, advance its pointer. If the element in the second array is smaller, advance its pointer. Continue until one of the pointers reaches the end of its array. #100DaysOfLeetCode #LeetCode #CodingChallenge #Java #DataStructures #Algorithms #SoftwareDevelopment #Programming #CodeNewbie #Tech #Developer #SoftwareEngineer #ProblemSolving #TwoPointers #Array

✅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

✅Day 79 of #100DaysOfLeetCode 1.📌Problem: Can Make Arithmetic Progression From Sequence 2.🟢 Difficulty: Easy 3.📍Topic: Arrays, sorting 4.🎯 Goal: Given an array of numbers, return true if the array can be rearranged to form an arithmetic progression; otherwise, return false. 5.🧠 Key idea: Approach 1: Sort the array to bring numbers in order. Calculate the common difference between the first two elements. Iterate through the array and check if the difference between consecutive elements matches the common difference. If any element does not satisfy the condition, return false; otherwise, return true. #LeetCode #100DaysChallenge #Programming #Coding #Java #Algorithms #CodingPractice #Developer #Tech #CodeNewbie #LearnToCode #ProblemSolving #DataStructures #Arithmetics #InterviewPreparation #CodingSkills #DailyCoding #LinkedInChallenge

✅Day 62 of #100DaysOfLeetCode 📌Problem: You are given a sorted array consisting of only integers where every element appears exactly twice, except for one element which appears exactly once. 🟠 Difficulty: Medium 📍Topic: Array, Binary Search 🎯 Goal: Return the single element that appears only once in O(log n) time and O(1) space. 🧠key idea: Approach 1: The problem can be efficiently solved using a modified binary search. The core logic relies on the properties of the sorted array. If we are at an even index, its duplicate pair should be at the next (odd) index. If we are at an odd index, its pair should be at the previous (even) index. By examining the middle element and its neighbor, we can determine if the single, non-duplicated element lies in the left or right half of the array, allowing us to discard one half in each step and achieve a logarithmic time complexity. #100DaysOfCode #LeetCode #CodingChallenge #Java #DataStructures #Algorithms #BinarySearch #ProblemSolving #SoftwareDeveloper #TechJobs #DeveloperLife #CodeNewbie #Programming #LearnToCode #JobSeekers

✅Day 83 of #100DaysOfLeetCode 📌Problem: Given a binary array and an integer k, return true if all 1's are at least k places away from each other, otherwise return false. 🟢 Difficulty: Easy 📍Topic: Array 🎯 Goal: Check if all 1's in the array are at least k positions apart. 🧠 Key idea: Approach 1: Iterate through the array, track the position of the previous 1, and for each new 1, calculate the distance from the previous 1. If the distance is less than k at any point, return false. Otherwise, return true after the loop ends. #LeetCode #CodingChallenge #100DaysOfCode #DataStructures #Algorithms #Array #Java #CodingInterview #Programmers #DailyCoding #CodeNewbie #TechCareers #ProblemSolving #CodeChallenge

NeetCode 16 | LeetCode #42 | Trapping Rain Water Applied two-pointer technique with max-left and max-right tracking to compute water trapped between bars. Optimized for linear time and constant space. #NeetCode #LeetCode #Java #Optimization #Algorithms #DataStructures #Coding