Java Solution: Intersection of Two Arrays II with Sorting and Two-Pointer Technique

🚀 Day 31/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 69. Sqrt(x) Used the Binary Search approach to find the integer square root without using built-in power functions. The search space is reduced by checking mid * mid against x. ⏱️ Time Complexity: O(log n) 📦 Space Complexity: O(1) Strengthening problem-solving skills with binary search patterns on answer space. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

LeetCode Problem || Check if Binary String Has at Most One Segment of Ones(1784)🚀 we need to check: The string should have only one continuous block of '1's. After a '0' appears, '1' should never appear again. ✨ Insight: Instead of manually counting segments using loops, we can simply check if "01" exists in the string. 📌 Time Complexity: O(n) Practicing problems like these helps improve pattern recognition and problem-solving efficiency. #LeetCode #DSA #Java #CodingPractice #ProblemSolving

🚀 Day 48/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 1534. Count Good Triplets Used a brute-force triple nested loop to check all possible triplets and validate the given conditions using absolute differences. ⏱️ Time Complexity: O(n³) 📦 Space Complexity: O(1) Strengthening understanding of nested loop patterns and condition-based filtering. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 Day 47/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 448. Find All Numbers Disappeared in an Array Used an in-place marking technique by treating indices as a hash map and marking visited numbers as negative to identify missing elements. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) (excluding output list) Strengthening understanding of array manipulation and in-place hashing tricks. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 Day 13 of #LeetCode Challenge 🔢 Problem: 1758 – Minimum Changes To Make Alternating Binary String Today’s problem was about converting a binary string into an alternating pattern with minimum changes. 💡 Key Insight: An alternating string can only start with either ‘0’ or ‘1’. So instead of trying many possibilities, I compared both patterns and counted mismatches — then returned the minimum. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) ✨ This problem improved my thinking in: Pattern observation Optimization approach Writing clean and efficient Java code Consistency > Motivation 💪 #Day13 #Java #DSA #LeetCode #CodingJourney #FutureFullStackDeveloper

🚀 Day 41/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 3427. Sum of Variable Length Subarrays Used a nested loop approach to calculate subarray sums for each index by dynamically determining the starting point using max(0, i - nums[i]). ⏱️ Time Complexity: O(n²) 📦 Space Complexity: O(1) Strengthening understanding of subarray problems and index-based range calculations. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 Day 24/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 2859. Sum of Values at Indices With K Set Bits Used bit manipulation to count the number of set bits in each index and summed the values whose index contains exactly k set bits. ⏱️ Time Complexity: O(n log n) 📦 Space Complexity: O(1) Strengthening understanding of bitwise operations and binary representation through consistent practice. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 Day 46/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 2315. Count Asterisks Used a boolean flag approach to track whether the current position is inside a pair of '|' and count only the valid '*' outside those sections. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) Strengthening understanding of state-based string traversal and conditional counting. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

Day 6/100 – LeetCode Challenge 🚀 Problem: #189 Rotate Array   Difficulty: Medium   Language: Java   Approach: Array Reversal Technique   Time Complexity: O(n)   Space Complexity: O(1) 🔍 Key Insight: Instead of shifting elements one by one, the array can be rotated efficiently using a three-step reversal strategy. Steps: 1️⃣ Reverse the entire array   2️⃣ Reverse the first k elements   3️⃣ Reverse the remaining elements  This achieves the required rotation in-place with constant extra space. 🧠 Solution Brief: Calculated k % n to handle cases where k is greater than array length.   Reversed the entire array first.   Then reversed the first k elements and the remaining n-k elements.   This sequence correctly rotates the array to the right. 📌 What I Learned: Understanding patterns like array reversal can simplify problems that initially seem complex.   Optimizing from brute force shifting to an in-place O(n) solution improves efficiency. #LeetCode #Day6 #100DaysOfCode #Java #DSA #Arrays #RotateArray #ProblemSolving #CodingJourney

𝗬𝗼𝘂 𝗵𝗮𝘃𝗲 𝗯𝗲𝗲𝗻 𝘂𝘀𝗶𝗻𝗴 𝗕𝗶𝗻𝗮𝗿𝘆 𝗦𝗲𝗮𝗿𝗰𝗵 𝘄𝗿𝗼𝗻𝗴. It is not just for sorted arrays. The real definition — "Eliminate HALF the search space with each decision." That one shift in thinking unlocks 20+ LeetCode problems. Swipe to see the template, live code, and 7 problems you can now solve with one pattern. 👇 Save this. 🔖 #DSA #BinarySearch #Java #LeetCode #CodingInterview #DebugWithPurpose