Java Binary Search Insert Position Optimized

🚀 Day 21 of #128DaysOfCode 🔍 Key Learnings: Efficient searching in O(log n) Using low & high to narrow the range Finding correct position even if target is absent 🧠 Approach: Compare mid with target → move left/right → return index or insert position Consistency is key 🔥 #DSA #Java #CodingJourney #PlacementPreparation

Did you know that List.of() and Set.of() create immutable collections in Java? What does that mean? You cannot add, remove, or modify elements after creation. Key Points: ❌ No modification allowed ❌ No null values allowed ✅ Safer (no accidental changes) ✅ Cleaner than Arrays.asList() When should you use it? - Constants - Read-only data - Defensive programming (protect your APIs) Common mistake: Treating it like a normal list → runtime exception #java #interview #immutable #certification

𝐃𝐚𝐲 𝟗𝟏/𝟑𝟔𝟓 🚀 📌 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞 𝐏𝐎𝐓𝐃: 𝐌𝐚𝐱𝐢𝐦𝐮𝐦 𝐃𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝐁𝐞𝐭𝐰𝐞𝐞𝐧 𝐚 𝐏𝐚𝐢𝐫 𝐨𝐟 𝐕𝐚𝐥𝐮𝐞𝐬 🔎 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: • Use two pointers (i, j)   • If A[i] <= B[j], update answer and move j   • Else move i  🔍 𝐀𝐥𝐠𝐨𝐫𝐢𝐭𝐡𝐦: • Two Pointer (Greedy) ⏱ 𝐓𝐢𝐦𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: • O(n + m) 🧠 𝐒𝐩𝐚𝐜𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: • O(1) 📈 𝐊𝐞𝐲 𝐓𝐚𝐤𝐞𝐚𝐰𝐚𝐲: • Expanding one pointer while fixing the other helps maximize distance efficiently  https://lnkd.in/dRYupYYp #LeetCode #365DaysOfCode #DSA #Java #TwoPointer #ProblemSolving

Day 65 — LeetCode Progress (Java) Problem: Find All Numbers Disappeared in an Array Required: Given an array of size n containing numbers in the range [1, n], return all the numbers that are missing from the array. Idea: Compare the expected range [1…n] with the actual elements to identify missing values. Approach: Initialize a set containing all numbers from 1 to n. Traverse the array: Remove each element from the set The remaining elements in the set are the missing numbers. Time Complexity: O(n) Space Complexity: O(n) #LeetCode #DSA #Java #HashSet #Arrays #Algorithms #CodingJourney #100DaysOfCode

Day 78 – Next Greater Node in Linked List Solved a problem to find the next greater value for each node in a linked list by combining linked list traversal with a monotonic stack approach. Key Learnings: Converted a linked list into an array to enable index-based processing Applied the monotonic stack technique to efficiently find next greater elements #DSA #Java #LinkedList #Stack #ProblemSolving #CodingPractice

#Day95 of #100DaysOfCode Extended the previous number-to-words program to handle multiple digits. The program takes a number as input and prints each digit in its word form. Focused on improving logic by combining loops, string conversion, and conditional handling. #Java #MiniProject #100DaysOfCode

Day 61 — LeetCode Progress (Java) Problem: First Unique Character in a String Required: Given a string s, find the first non-repeating character and return its index. If none exists, return -1. Idea: Track the occurrence of each character and identify the one that appears only once with the smallest index. Approach: Use a HashMap to store character information. Traverse the string: If a character appears for the first time, store its index If it appears again, mark it as invalid (e.g., update with a special value) Traverse the map (or string again): Find the character with a valid index and minimum position Return that index, or -1 if no unique character exists. Time Complexity: O(n) Space Complexity: O(1) #LeetCode #DSA #Java #HashMap #Strings #Algorithms #CodingJourney #100DaysOfCode

Most people solve this problem by checking every possible pair. It works… but it doesn’t scale. The key insight is simple: The container is always limited by the shorter line. Once you understand that, you don’t need to check everything. You can eliminate unnecessary pairs and reach an optimal solution in O(n). This is where the Two Pointer approach becomes powerful. Small shift in thinking → big improvement in efficiency. Swipe to see how it works → #DSA #CodingInterview #Java #TwoPointers #ProblemSolving #SoftwareEngineering #LeetCode

🚀 𝐃𝐚𝐲 94/100  ✅ 𝐏𝐫𝐨𝐛𝐥𝐞𝐦: 𝐑𝐞𝐩𝐞𝐚𝐭𝐞𝐝 𝐒𝐮𝐛𝐬𝐭𝐫𝐢𝐧𝐠 𝐏𝐚𝐭𝐭𝐞𝐫𝐧 Today’s problem was about checking whether a string can be formed by repeating one of its substrings multiple times. 💡 𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠𝐬: Try all possible substring lengths up to n/2 Only consider lengths that divide the string completely Build and compare repeated string with original Time Complexity: O(n²) in worst case  𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Iterate over possible substring sizes If length divides string → repeat substring Compare with original string If match found → return true #Day94 #100DaysOfCode #LeetCode #Java #DSA #CodingJourney #ProblemSolving

Day 66 — LeetCode Progress (Java) Problem: Find the Difference of Two Arrays Required: Given two integer arrays, return: Elements present in nums1 but not in nums2 Elements present in nums2 but not in nums1 Idea: Use sets to remove duplicates and quickly check membership. Approach: Convert both arrays into sets Iterate over nums1 set: Add elements not present in nums2 set to result1 Iterate over nums2 set: Add elements not present in nums1 set to result2 Return both lists Time Complexity: O(n + m) Space Complexity: O(n + m) #LeetCode #DSA #Java #HashSet #Arrays #ProblemSolving #CodingJourney