How to count 1 bits in binary with LeetCode

✅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

✅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 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 84 of #100DaysDSAChallenge 📌 Topic: Array — Squares of a Sorted Array ✅ Problem Solved on #LeetCode: 977. Squares of a Sorted Array (🟢 Easy) 💡 Key Learnings: • Strengthened understanding of the two-pointer technique for sorted array manipulation. • Learned to efficiently handle both negative and positive numbers when squaring values. • Optimized time complexity from O(n²) to O(n) by eliminating nested loops. • Enhanced focus on writing clean, readable, and efficient Java code. 🚀 Consistency Builds Clarity: Every problem you solve sharpens your logical thinking and builds confidence. Keep learning, improving, and moving forward — one problem at a time 💪 🏷️ #Day84 #100DaysChallenge #100DaysDSAChallenge #LeetCode #Java #CodingChallenge #ProblemSolving #DataStructures #Algorithms #Arrays #Programming #LearningJourney #10kCoders #10000Coders #Consistency #LogicBuilding

✅Day 59 of #100DaysOfLeetCode 📌Problem: Given a fixed-length integer array arr, duplicate each occurrence of zero, shifting the remaining elements to the right. 🟢 Difficulty: Easy 📍Topic: Array 🎯 Goal: To modify the input array in place by duplicating every zero and shifting other elements accordingly, without returning anything. 🧠key idea: Approach 1: Create a new result array to hold the modified elements. Iterate through the original array; if the element is a zero, add it twice to the result array. Otherwise, add the element once. Finally, copy the contents of the result array back into the original array. #100DaysOfCode #LeetCode #CodingChallenge #Java #DataStructures #Algorithms #ProblemSolving #Developer #SoftwareEngineering #CodeNewbie #Tech #Programming #ArrayManipulation #DuplicateZeros #Day59

🚀 Problem 2: Reverse Integer 💡 Leetcode Problem No: 7 Today, I solved an interesting Leetcode-style problem that involves reversing an integer — while handling tricky cases like integer overflow 🔄 ✨ Challenge: Given an integer x, reverse its digits and return the reversed number. If reversing x causes the value to go outside the signed 32-bit integer range, return 0. 💻 Key Learnings: 🔹 Used Math.abs(x) to handle negatives elegantly 🔹 Applied overflow check using: if (rev > (Integer.MAX_VALUE - d) / 10) 🔹 Mastered modulus (%) and division (/) logic for digit extraction 🔹 Ensured both positive and negative numbers are correctly reversed 💙 Tip: Always consider edge cases and overflow conditions when working with integer manipulation problems. #Java #Coding #LeetCode #ProblemSolving #JavaDeveloper #ProgrammingChallenge #LeetcodeCoding

NeetCode 29 | LeetCode #153 | Find Minimum in Rotated Sorted Array Used binary search to find the rotation pivot. Compared nums[mid] with nums[right] to shrink search space efficiently. Time complexity: O(log n) | Space: O(1) #LeetCode #NeetCode #Java #Algorithms #DataStructures #Optimization #Coding

🌟 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

✅Day 80 of #100DaysOfLeetCode 1.📌Problem: Set Mismatch You are given an array that represents a set of integers from 1 1 to n n, but due to an error, one number is duplicated and another is missing. The task is to find the duplicated number and the missing number and return them as an array. 2.🟢 Difficulty: Easy 3.📍Topic: Hashing,Array 4.🎯 Goal: Find the number that occurs twice and the number that is missing, then return them in an array. 5.🧠 key idea: Approach 1: Use a HashMap to count occurrences of each number in the input array. Identify the duplicated number by finding the one which appears twice. Compute the total sum expected (n(n+1)/2 n(n+1)/2) and compare with the actual sum (excluding the duplicate). The missing number equals the difference between the expected total and the corrected actual sum. Return the duplicate and missing numbers as the result array. #LeetCode #100DaysOfCode #CodingChallenge #Java #Programming #InterviewPrep #Algorithms #DataStructures #Hashing #ProblemSolving #Array #SoftwareEngineering #TechCareers #LearnToCode #CodeNewbie #DailyCode #ChallengeYourself

Day 53 of My DSA Challenge Problem: Find the pair in an array whose sum is closest to and less than or equal to a given target. Approach: Instead of checking every possible pair (O(N²)), I optimized the solution using the Two-Pointer Technique. Sort the array. Initialize two pointers — one at the start and one at the end. Move the pointers based on the sum compared to the target: If the sum is less than or equal to the target, record it and move the left pointer forward. If the sum exceeds the target, move the right pointer backward. This ensures that every pair is checked efficiently, and the closest valid sum is captured. Complexity: Time: O(N log N) (due to sorting) Space: O(1) #Day53 #DSAChallenge #TwoPointers #Sorting #Optimization #ProblemSolving #DSA #Java #CodingChallenge #Algorithms #DataStructures #100DaysOfCode #GeeksforGeeks #LeetCode #ProgrammingJourney #CodingCommunity