"Mastering Binary Search: A Fundamental Algorithm"

🚀 Day 58 of My LeetCode Journey 🚀 🔹 Problem: Search Insert Position 🔹 Topic: Binary Search (Fundamental DSA Technique) Today’s challenge focused on enhancing one core skill: thinking in terms of boundaries and decisions. The task was simple — in a sorted array, find the index of the target element. If it doesn’t exist, return the index where it logically should be inserted. What made this problem interesting is that instead of searching linearly, we leverage Binary Search, which reduces the operations drastically. ✅ Key Takeaways & Learning: Binary Search is not just used to find something — it can also determine the correct position for insertion. Understanding how low, mid, and high move helps build clear logic flow. The index where the search ends (low) gives the correct insert position, even if the element doesn't exist. Rewriting brute-force logic into binary search trains the brain to think in terms of optimization. ⏱️ Complexity Analysis: MetricValueTimeO(log n)SpaceO(1) 🌱 Personal Reflection: Every binary search problem improves clarity of thought. From checking a value to identifying a boundary or insert point — the mindset shifts from “finding an element” to “solving a pattern.” #100DaysOfCode #LeetCode #DSA #BinarySearch #Java #ProblemSolving #LearningInPublic #TechJourney #ConsistencyIsKey

🚀 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 48 of #50DaysOfLeetCodeChallenge Problem : Daily Temperatures 💡Approach: Used a monotonic stack to store indices of temperatures. For each day, I checked if the current temperature is higher than the one at the top of the stack. If yes, that means we’ve found a warmer day — so I popped the index and calculated the number of days waited. This gave an efficient O(n) solution instead of checking every pair. 🔥Key Takeaways: Learned how stacks can simplify problems involving “next greater element.” Improved understanding of monotonic structures and how they reduce unnecessary comparisons. Continuing my journey of learning, growing, and coding every day! #LeetCode #50DaysOfCode #Java #DataStructures #Stack #ProblemSolving #CodeToLearn #CharuCodes

🚀 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

🚩 Problem: 46. Permutations 🔥 Day 49 of #100DaysOfLeetCode 🔍 Problem Summary: Given an array nums of distinct integers, return all possible permutations in any order. 🧠 Intuition: A permutation is a rearrangement of elements. We can use backtracking to explore all possible arrangements: Build partial permutations recursively. Add unused elements one by one. Backtrack after exploring each path. ✅ Approach (Backtracking): Use a temporary list to store the current permutation. If its size equals nums.length, add it to the result. For each element, if not already in the temp list, choose it and recurse. Backtrack by removing the last chosen element. ⚙️ Performance: ⏱️ Runtime: 1 ms 🚀 💪 Beats: 99.95% of Java solutions 💾 Memory: 44.5 MB ⚡ (Beats 91.87% of users) 📊 Complexity: Time Complexity: O(n × n!) Space Complexity: O(n) (recursion stack + temporary list) ✨ Key Takeaway: This problem builds a solid foundation in backtracking, teaching how to systematically generate all possible outcomes through recursive exploration and reversal. Link:[https://lnkd.in/gncncqK8] #100DaysOfLeetCode #Day49 #Problem46 #Permutations #Backtracking #Recursion #DSA #Algorithms #Java #LeetCode #ProblemSolving #CodingChallenge #InterviewPreparation #CrackingTheCodingInterview #CodingCommunity #SoftwareEngineering #DataStructures #DeveloperJourney #ArjunInfoSolution #Programming #CodeNewbie #LearnToCode #TechCareers #CareerGrowth #ComputerScience #CodeEveryday #ZeroToHero #CodingIsFun #CodeLife #JavaDeveloper #GameDeveloper #Unity #AI #MachineLearnin

🔥 LeetCode Day 27 Challenge Problem Attempted: Search in Rotated Sorted Array II 💡 Approach & Progress: Today, I explored how to search for a target element in a rotated sorted array that contains duplicates. This version is trickier than the original because duplicates can make it hard to determine which half of the array is sorted. I learned that the key is to handle edge cases carefully — especially when the elements at low, mid, and high are equal. In such cases, it becomes impossible to decide which side is sorted, so the best approach is to shrink the search space by moving both pointers inward. After managing duplicates, I applied the standard binary search logic by identifying which half of the array is sorted and checking if the target lies within that range. 🧠 Key Learnings: Understood how duplicates affect binary search in rotated arrays. Learned to identify and manage ambiguous cases where sorting order is unclear. Improved logical reasoning for range-based conditions in binary search. Strengthened confidence in solving complex search variations efficiently. #LeetCode #Day27 #BinarySearch #DSA #ProblemSolving #CodingChallenge #Java #Algorithm #100DaysOfCode #LeetCodeJourney #CodeEveryday #TechLearning #ProgrammerLife

🚀 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

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 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

NeetCode 30 | LeetCode #33 | Search in Rotated Sorted Array Implemented modified binary search logic: Determine which side is sorted Adjust search boundaries based on the target’s position Time: O(log n) | Space: O(1) #LeetCode #NeetCode #Java #Algorithms #DataStructures #Optimization #Coding