"Day 58: Solved Search Insert Position with Binary Search"

🚀 Day 51 of My LeetCode Journey 🚀 💡 Problem: Binary Search 🧠 Concept: Binary Search is one of the most fundamental algorithms — it efficiently searches a sorted array by repeatedly dividing the search space in half. If the target value is equal to the middle element, return the index. If the target is smaller, search the left half; if larger, search the right half. ⚙️ Complexity: ⏱ Time: O(log n) 💾 Space: O(1) ✨ Learning: Binary search is not just an algorithm — it’s a mindset to “divide and conquer.” Mastering it helps in many advanced problems like searching in rotated arrays, finding peaks, and solving range queries efficiently. #100DaysOfCode #LeetCode #CodingJourney #Java #DSA #BinarySearch #ProblemSolving

🚀 Day 23 of #50DaysLeetCodeChallenge Problem: Binary Search (LeetCode 704) — a timeless classic that reminded me how elegance in programming often lies in simplicity. 🎯 🧠 Thought Process: When dealing with sorted data, you don’t need to search everything — just guide your logic efficiently. Binary Search embodies this idea perfectly: divide, compare, and conquer. It’s all about making smarter moves, not more moves. ⚙️ Approach: 🔹 Initialized two pointers — low and high — to define the search boundaries. 🔸 Calculated the midpoint each time and compared it with the target. 🔹 Adjusted the range intelligently, narrowing down until the target was found (or confirmed missing). 🔸 Returned the index if found, else -1. 📊 Efficiency: ✅ Time Complexity: O(log n) ✅ Space Complexity: O(1) ⏱️ Runtime: 0 ms — Beats 100% of Java submissions! ⚡ 💭 Key Takeaway: Binary Search teaches that speed isn’t always about doing more work, but about doing the right work. It’s a clear reminder that logic and precision go hand in hand when optimizing performance. 💫 Big thanks to Trainer Shishir chaurasiya and PrepInsta for constantly fueling my journey toward sharper logic and smarter problem-solving! 🚀 #LeetCode #Day23 #50DaysOfCode #ProblemSolving #Java #AlgorithmDesign #BinarySearch #CodingChallenge #PrepInsta #DSA #TechLearning #LogicBuilding #DeveloperJourney #ProgrammingLife #CodingLife

🚀 Day 8 of #45DaysOfLeetCode Challenge 😎 📌Today's problem: Palindrome Number (LeetCode #9) 💡 🔹 Concept: Check whether a given integer reads the same backward and forward — without converting it to a string! 🔹 Logic Used: Instead of reversing the entire number, I reversed only half of it to improve efficiency. This avoids unnecessary computation and eliminates integer overflow risks. 🔹 Key Learnings: ✅ Optimized logic using mathematical manipulation ✅ Improved understanding of number reversal techniques ✅ Importance of edge case handling (negative numbers, trailing zeros) ⚙️ Result: 💻 Runtime: 4 ms (Beats 100%) 💾 Memory: 44.84 MB 😎Small optimizations make a big difference in performance! 💪 📌Problem: https://lnkd.in/ef6AC2j6 🔥Each day is a step closer to writing cleaner and more optimized code. ✨ Let’s keep pushing forward and refining our problem-solving skills! 💻🔥 #LeetCode #Day8 #100DaysOfCode #ProblemSolving #Java #CodingChallenge #PalindromeNumber #DataStructures #Algorithms #LearningEveryday

#Day5 of #100DaysOfCode Challenge! 🔥 Today’s focus is on strengthening array fundamentals through two challenging problems: Missing Number and First Repeating Element. Problem 1: LeetCode 268 – Missing Number Objective: Identify the missing number in an array containing n distinct numbers ranging from 0 to n. Approach: Utilized the sum formula, n*(n+1)/2, to calculate the expected sum. By subtracting the actual sum of the array elements, the missing number was readily identified. Time Complexity: O(n) | Space Complexity: O(1) This solution is concise, elegant, and exemplifies a clean approach to solving interview problems. Problem 2: First Repeating Element (GFG) Objective: Determine the first repeating element with the smallest index (1-based). Approach: Traversed the array from right to left, utilizing a HashSet to store elements. Whenever a number was encountered that was already present in the HashSet, the index was updated to ensure the smallest possible value. Time Complexity: O(n) | Space Complexity: O(n) Key Takeaway: Recognizing when a mathematical shortcut, such as the sum formula, is applicable versus when a data structure, like a HashSet, is necessary, cultivates strong problem-solving intuition. Balancing simplicity and optimization is crucial for achieving significant growth in problem-solving skills. Thank you Rajesh Gupta for your assistance in guiding me through this process. #100DaysOfCode #LeetCode #DSA #ProblemSolving #Arrays #CodingChallenge #LearningInPublic #Java #Programming

🚀 Day 65 of My LeetCode Journey 🚀 🔹 Problem: Find the Duplicate Number (Binary Search Approach) Today, I realized that binary search is not always about searching in a sorted array… Sometimes, we apply binary search on the range of values, not the array itself ✅ 🧠 Key Insight: Choose a mid value. Count how many values in the array are <= mid. If the count is more than mid, the duplicate is in the left half. Otherwise, it's in the right half. We keep narrowing the range until start and end meet — that number is the duplicate. ⚙️ Time Complexity: O(n log n) (no array modification, no extra space) 🧠 Learning: Before jumping to brute-force or extra space solutions, try to reason about the properties of the input. Sometimes the pattern is hidden in the constraints, not in the array. 🔸 Continuous learning. 🔸 Continuous improvement. #100DaysOfCode #LeetCode #Java #DSA #CodingJourney #ProblemSolving #LearningEveryday #BinarySearch #TechSkills

🚀 Day 08/10 of Linked List Series: Delete a Node by its Index🎯 🧠 Algorithm (Step-by-Step Explanation) 1️⃣ Start from the head of the linked list. 2️⃣ If the index = 0, it means we want to delete the first node — so just move the head to the next node. 3️⃣ Otherwise, use a loop to reach the node just before the target index (that is, index - 1). 4️⃣ Once you reach there, skip the next node by connecting the current node’s next pointer to the node after the one being deleted. 5️⃣ Return the (possibly updated) head of the linked list. 🪄 Simple Understanding: Imagine a chain of people holding hands (each one pointing to the next). If you want to remove the person at position index, just tell the person before them to hold the hand of the one after them — the middle one is now “unlinked” and effectively deleted! 🔥 Keep Learning — Keep Building #CodeWithLakkojuEswaraSai #CodeWithLakkojuEswaraSai_LinkedList #Java #DSA #DataStructures #LinkedList #CodingCommunity #100DaysOfCode #ProgrammerLife #LearnWithMe #CodeEveryday #TechJourney #CodingIsFun #JavaDevelopers #SoftwareEngineering #CodeNewbie

🚀 Day 47 of My LeetCode Journey 🚀 Problem: Sum of Square Numbers Topic: Math / Two Pointers 🧠 Approach: To check if a number c can be expressed as the sum of squares of two integers (a² + b² = c), we use the two-pointer technique: Start a = 0 and b = √c Calculate sum = a² + b² If sum == c → ✅ return true If sum < c → increase a If sum > c → decrease b Repeat until a <= b. ⏱️ Complexity: Time: O(√c) Space: O(1) This problem beautifully combines mathematical logic with an efficient two-pointer approach — clean and elegant! 💡 #100DaysOfCode #LeetCode #Java #CodingChallenge #ProblemSolving #DSA

💡 Day 98 of My DSA Challenge – Longest Subsequence With Limited Sum 🔷 Problem: 2389. Longest Subsequence With Limited Sum 🔷 Goal: For each query, find the maximum number of elements from nums that can form a subsequence with a sum ≤ query value. 🔷 Key Insight: To maximize the subsequence size, we should always pick smaller elements first — this is a greedy choice. Hence, sorting the array ensures we can accumulate the smallest elements until the sum exceeds the query limit. Approach outline: 1️⃣ Sort the array nums. 2️⃣ For each query, add elements one by one until the sum crosses the limit. 3️⃣ Return the count of elements that fit within the limit. This logic can also be optimized further using prefix sums + binary search, but the greedy approach works perfectly within given constraints. 🔷 My Java Approach: Sort the array. Use a helper function to count how many elements fit under the query sum. Store each result in the answer array. 🔷 Complexity: Time → O(n log n + n × m) Space → O(1) Sometimes, simplicity wins. This problem reinforces the power of greedy thinking — starting small, building up gradually, and knowing when to stop. Recognizing such patterns is key to writing clean and intuitive code. 🚀 #100DaysOfCode #Day98 #LeetCode #DSA #Java #ProblemSolving #GreedyAlgorithm #CodingChallenge #Programming #LearnToCode #CodingLife #SoftwareEngineering #Algorithms #DataStructures #TechJourney #CodeEveryday #EngineerMindset #DeveloperJourney #GrowthMindset #CodeNewbie #KeepLearning #ApnaCollege #AlphaBatch #ShraddhaKhapra

🚀 Day 54 of My #100DaysOfLeetCode Challenge 🚀 Today I worked on a problem related to making parentheses valid — a classic stack-based problem that tests your understanding of balanced brackets and edge cases. 💡 Problem: Given a string containing only '(' and ')', determine the minimum number of parentheses that must be added to make the string valid. 🧠 Concepts Used: Stack data structure String traversal Counting unmatched parentheses 🧩 Approach: Traverse through the string character by character. Use a stack to track unmatched '('. When encountering ')', pop from the stack if possible, else increment the counter. Finally, the total unmatched parentheses = stack size + counter. #100DaysOfCode #LeetCode #Java #ProblemSolving #CodingJourney #DSA #WomenInTech #TechLearning #CodeNewbie