Mastering Binary Search for Coding Challenges

🔹 Day 67 of #100DaysOfLeetCodeChallenge 🔹 Problem: Combination Sum Focus: Backtracking + Recursion 💡 The Challenge: Find all unique combinations of numbers that sum to a target. The twist? You can reuse the same number unlimited times! This makes it more interesting than standard subset problems. 🧠 My Approach: Used backtracking with two key decisions at each step: Pick: Include current element and explore further (stay at same index for reuse) Skip: Move to next element without including current one Base case: When we've explored all elements, check if target reached 0 Optimization: Only pick if element ≤ remaining target 📊 Complexity Analysis: ⏳ Time: O(2^t) where t = target/min(candidates) — worst case recursion tree 💾 Space: O(t) — maximum recursion depth 📌 Example: Input: candidates = [2,3,6,7], target = 7 Output: [[2,2,3],[7]] Explanation: - 2+2+3 = 7 (reused 2 twice!) - 7 = 7 (single element) 🎯 Key Takeaway: The beauty of this problem lies in handling unlimited reuse. By staying at the same index when picking an element, we elegantly allow repetition without creating duplicates. This pick/skip pattern is fundamental to combinatorial backtracking! Real-world analogy: Like making change for a dollar — you can use the same coin denomination multiple times! 💰 Day 67/100 complete. Mastering backtracking, one problem at a time! 💪 #LeetCode #Backtracking #DynamicProgramming #DSA #Algorithms #CodingInterview #100DaysOfCode #SoftwareEngineering #ProblemSolving #TechCareer

💡 Day 2 of #30DaysOfDSA When ‘Searching’ Gets Smarter — Binary Search Explained Simply When I first heard about Binary Search, it sounded fancy — but the concept is surprisingly simple. You just keep halving your search space until you find what you’re looking for. Imagine you’re looking for a name in a phone book (yes, those old ones 😄). You don’t start from page 1 and check each entry — you open the book near the middle, decide which half might contain the name, and repeat. That’s binary search in action. In code, this logic turns into one of the most efficient algorithms we use every day — from searching elements in sorted data to how databases and even search engines optimize lookups. It’s also the foundation of many advanced algorithms and problem-solving techniques we encounter later — especially in competitive programming and system-level optimization. The key takeaway? Binary search isn’t just about searching — it’s about thinking in halves and making decisions with precision. #DSA #CodingJourney #LearnInPublic #SoftwareEngineering #Java #Programming #BinarySearch #TechSeries

Day 24: #100DaysOfCodingChallenge 🔹📊 Today’s focus was on binary search on the answer and understanding stable sorting — perfect for optimizing solutions and maintaining element order! 🔵 LeetCode/GFG: Binary Search on Answer (Basic Example) 🚀 Difficulty: Medium ⏳ Time Complexity: O(log(max_range))** 💾 Space Complexity: O(1)** ⏱ Time Taken: 0.2 sec 🔍 Approach: Applied binary search not on array indices, but on the range of possible answers. Checked feasibility at each mid-value and narrowed down the search space — a key technique for optimization problems! 🟢 LeetCode/GFG: Stable Sort Concept (Practical) 🟢 Difficulty: Easy ⏳ Time Complexity: O(n log n)** 💾 Space Complexity: O(n)** ⏱ Time Taken: 0.2 sec 🔍 Approach: Used a sorting algorithm (like Merge Sort or stable sort in STL) that preserves the relative order of equal elements. Understanding stability is crucial for problems where order matters in secondary attributes. 💡 Reflection: Binary search on the answer transforms brute-force range checks into efficient log-time searches. Stable sorting ensures correctness in multi-criteria sorting problems — both are foundational for advanced algorithmic thinking! Thanks to Nandan Kumar Mishra Sir, Abhishek Kumar Sir, and #KRMangalamUniversity for their guidance and continuous encouragement! #100DaysOfCodingChallenge #DSA #BinarySearchOnAnswer #StableSort #Array #Sorting #CPlusPlus #ProblemSolving #CodingJourney

Day 24: #100DaysOfCodingChallenge 🔹📊 Today’s focus was on binary search on the answer and understanding stable sorting — perfect for optimizing solutions and maintaining element order! 🔵 LeetCode/GFG: Binary Search on Answer (Basic Example) 🚀 Difficulty: Medium ⏳ Time Complexity: O(log(max_range))** 💾 Space Complexity: O(1)** ⏱ Time Taken: 0.2 sec 🔍 Approach: Applied binary search not on array indices, but on the range of possible answers. Checked feasibility at each mid-value and narrowed down the search space — a key technique for optimization problems! 🟢 LeetCode/GFG: Stable Sort Concept (Practical) 🟢 Difficulty: Easy ⏳ Time Complexity: O(n log n)** 💾 Space Complexity: O(n)** ⏱ Time Taken: 0.2 sec 🔍 Approach: Used a sorting algorithm (like Merge Sort or stable sort in STL) that preserves the relative order of equal elements. Understanding stability is crucial for problems where order matters in secondary attributes. 💡 Reflection: Binary search on the answer transforms brute-force range checks into efficient log-time searches. Stable sorting ensures correctness in multi-criteria sorting problems — both are foundational for advanced algorithmic thinking! Thanks to Nandan Kumar Mishra Sir, Abhishek Kumar Sir, and #KRMangalamUniversity for their guidance and continuous encouragement! #100DaysOfCodingChallenge #DSA #BinarySearchOnAnswer #StableSort #Array #Sorting #CPlusPlus #ProblemSolving #CodingJourney

🌟 Day 24 of #100DaysOfCode. Today I come with one challenging problem which is related to the topic Binary Search. 🌟 ✔️ What I solved: *️⃣ Problem of the Day: Binary Search ✴️ Approach & Key Insights: 🔹 Binary Search divides the array into halves to efficiently search for a target in O(log n) time. 🔹 The sorted property is crucial — binary search only works on sorted arrays. ◾ Steps: 🔹 Initialize two pointers, left at the start and right at the end of the array. While left <= right: 🔹 Compute the middle index mid = left + (right - left) // 2. 🔹 If nums[mid] is the target, return mid. 🔹If nums[mid] < target, search right half (left = mid + 1). 🔹If nums [mid] > target, search left half (right = mid - 1). 🔹If you exit the loop, target is not present. Return -1. ✴️ Complexity: 🔹 Time Complexity: O(log⁡n) 🔹 Space Complexity: O(1) ✴️ Key Takeaways: 🔹 Use binary search for efficient lookups in sorted arrays. Always check for overflow when computing mid: mid = left + (right - left) / 2. 🔹 The standard approach is iterative for minimal space, but recursive solutions work as well. 🚀For more information, Click here ⬇️ https://lnkd.in/gaQTpVtk 🪄 I am very thankful to my Mentor Mr.Rajesh Gupta Sir and KR Mangalam University for giving me the support throughout this week. #DSA #100DaysOfCode #KRMU #Competitive #Coding

Title: From "What" to "How": Implementing My First Algorithm Hey LinkedIn fam! 👋 Today, my journey was all about understanding what an array is: a contiguous block of memory for storing elements of the same type. I took the logical next step: learning how to do something useful with it. My focus was on my very first searching algorithm: Linear Search. The task is simple, just like finding a specific dish on a menu: "Find the index of an element in a given array." The logic is exactly what you'd do in real life: Start at the beginning (index 0). Check each element, one by one. If you find what you're looking for (the key), you immediately stop and return its index. But the real "aha!" coding moment for me was handling the "Not Found" case. You can't just return 0, because 0 is a valid index! The elegant solution is to return a sentinel value—a number that's outside the possible range of indices. In this case: -1. Then, in the main code, you just check: if (index == -1) { ... print "Not Found" } It's so cool to see the theory (the diagram) and the practical, clean code come together. It feels great to connect these fundamental building blocks. One step at a time! What was the first algorithm you remember learning after arrays? #Java #Algorithms #LinearSearch #DataStructures #CodingJourney #LearnInPublic #SoftwareDevelopment #ProgrammingFundamentals

🚀 Day 23/30 – 30DaysOfDSAChallenge 🚀 🧠 Topic: Subsets II (Handling Duplicates) 📘 Concepts Covered: Recursion | Backtracking | Duplicate Handling Today I solved the Subsets II problem — a tricky variant of the power set problem where the array can contain duplicate elements. The challenge was to ensure that no duplicate subsets appear in the final output. 🔹 Approach – Recursive Backtracking with Skip Logic 💡 Logic: Sort the array first to group duplicate elements together. While generating subsets recursively, skip consecutive duplicates using a simple condition. This ensures unique subsets without extra space for duplicate checks. ⚙️ Time Complexity: O(2ⁿ) ⚙️ Space Complexity: O(2ⁿ * n) ✅ Result: Accepted ✅ | Runtime: 0 ms 🟢 | Beats: 100% 🚀 ✨ Key Takeaway: Understanding how to prune recursion trees helps eliminate redundancy and boosts performance — a crucial concept in advanced DSA problems. #Day23 #30DaysOfDSAChallenge #LeetCode #Backtracking #Recursion #SubsetsII #ProblemSolving #CPlusPlus #CodingChallenge #DSA #CodeEveryday #DeveloperJourney #Algorithms

🚀 Day 34 of My DSA Journey — Exploring Recursion & Backtracking 🧩 Problem 1: Generate Parentheses LeetCode #22 | Medium 🔍 Introduction: Given n pairs of parentheses, the task is to generate all possible combinations of well-formed parentheses. This problem is a classic example of backtracking, where we make decisions step-by-step and discard invalid ones along the way. 💡 Approach: We use recursion to build valid strings by keeping track of the number of open and close brackets used so far: Add '(' if open count < n. Add ')' if close count < open. Continue until both counts equal n. 🕒 Time Complexity: O(4^n / √n) (Catalan number complexity) 💾 Space Complexity: O(n) (recursion stack + temporary string) This approach ensures only valid combinations are explored, avoiding unnecessary computations. 🧩 Problem 2: Print All Subsequences / Power Set Concept: Recursion & Subset Generation 🔍 Introduction: Given an array or string, we need to generate all possible subsequences (or subsets). Each element has two choices — include or exclude — leading to a total of 2^n possible combinations. This problem strengthens the understanding of recursive decision-making and forms the base for subset-related problems. 💡 Approach: At each recursion step, choose to include or skip the current element. Continue until the end of the array/string is reached. Print or store the formed subset. 🕒 Time Complexity: O(2^n) 💾 Space Complexity: O(n) (recursion depth) 🎯 Key Takeaway: Both problems revolve around exploring all possibilities — but with different constraints: Generate Parentheses focuses on valid structured output. Power Set explores all possible outcomes. Understanding these recursive foundations is crucial before tackling advanced backtracking problems like Combination Sum, Subsets II, or Palindrome Partitioning. #Day34 #100DaysOfDSA #Backtracking #Recursion #LeetCode22 #CodingJourney #ProblemSolving #LearnWithMe #DSA

🔹 Day 70 of #100DaysOfLeetCodeChallenge 🔹 🚀 Problem: Combination Sum III 🔑 Topic: Backtracking 🧠 Approach: We need to find all combinations of k distinct numbers (1–9) that sum to n. Here’s how I solved it 👇 Use backtracking to explore all combinations starting from 1 to 9. Add the number and move forward recursively, reducing both k (count) and n (target). Stop when k == 0 and n == 0 → valid combination found! Backtrack by removing the last number and continue exploring. ⏳ Time Complexity: O(2⁹) ≈ O(512) 💾 Space Complexity: O(k) (recursion + temp list) 📌 Example: Input: k = 3, n = 9 Output: [[1,2,6],[1,3,5],[2,3,4]] ✅ 🎯 Takeaway: This problem teaches the power of controlled recursion — when both depth and sum conditions guide the search efficiently. 💡 #LeetCode #DSA #Backtracking #ProblemSolving #CodingChallenge #100DaysOfLeetCodeChallenge 🚀

🔍 Day 66 of #100DaysOfCode 🔍 🔹 Problem: Binary Search – LeetCode ✨ Approach: Implemented an iterative binary search to efficiently locate the target element within a sorted array. By halving the search range each time — adjusting low and high around the mid-point — the algorithm achieves blazing-fast lookups! ⚡ 📊 Complexity Analysis: Time Complexity: O(log n) — array size halves with each iteration Space Complexity: O(1) — constant extra space ✅ Runtime: 0 ms (Beats 100.00%) ✅ Memory: 46.02 MB 🔑 Key Insight: Binary Search is proof that efficiency isn’t about doing more — it’s about eliminating what’s unnecessary. 🚀 #LeetCode #100DaysOfCode #ProblemSolving #DSA #AlgorithmDesign #BinarySearch #LogicBuilding #Efficiency #ProgrammingChallenge #CodeJourney #CodingDaily