Rotated Sorted Array Search Problem Solution

🚀 Day 11 – DSA Daily Series Today’s Problem: Find Minimum in Rotated Sorted Array (LeetCode 153) Today I solved an interesting problem that involves finding the minimum element in a rotated sorted array. 🧠 Problem You are given a sorted array that has been rotated several times. The task is to find the minimum element in the array. Example: Input: nums = [3,4,5,1,2] Output: 1 💡 Approach I solved this problem using Binary Search. Key idea: • In a rotated sorted array, the minimum element lies at the rotation point • Compare the middle element with the rightmost element • If nums[mid] > nums[high], the minimum lies in the right half • Otherwise, it lies in the left half including mid By narrowing the search space, we efficiently locate the minimum element. ⏱ Complexity Time Complexity: O(log n) Space Complexity: O(1) 🔎 Key Learning Binary Search can be extended beyond simple searching and used to identify structural properties of arrays like rotation points. Continuing the DSA Daily Series — solving and learning one problem at a time. 🚀 #DSA #LeetCode #Python #Algorithms #BinarySearch #CodingJourney #ProblemSolving

🚀 Optimized Solution: Search in Rotated Sorted Array II Solved the problem "Search in Rotated Sorted Array II" using an optimized Binary Search approach! 🔍 Key Idea: Instead of using linear search, I applied a modified binary search to handle: ✔️ Rotated sorted array ✔️ Presence of duplicates ⚙️ Approach: Used binary search to divide the array Identified sorted halves even with duplicates Carefully handled edge cases where duplicates make it hard to decide the sorted side 📈 Complexity: ⏱️ Time Complexity: O(log n) (Worst case O(n) due to duplicates) 💾 Space Complexity: O(1) ✅ Result: ✔️ All test cases passed ⚡ Efficient and scalable solution This problem really helped me strengthen my understanding of edge cases in binary search and real-world problem-solving 💡 #leetcode #binarysearch #algorithms #coding #python #datastructures #softwaredeveloper #problemSolving

🚀 𝗗𝗮𝘆 𝟮𝟮/𝟯𝟬 — 𝗗𝗦𝗔 𝗖𝗵𝗮𝗹𝗹𝗲𝗻𝗴𝗲 Day 22 and today’s focus moved into graph problems. Graphs introduce a different way of thinking — instead of linear structures, it’s all about exploring connections between nodes. Traversal techniques like DFS and BFS start to play a big role here. Today was about understanding how to navigate through connected components and explore all possible paths systematically. 🔎 𝗗𝗮𝘆 𝟮𝟮 𝗙𝗼𝗰𝘂𝘀 • Understanding graph traversal concepts • Practicing DFS/BFS thinking • Solved: ✅ Number of Islands ✅ Flood Fill ✅ Find if Path Exists in Graph Graph problems can look intimidating at first, but once the traversal pattern becomes clear, the logic starts falling into place. Still learning. Still improving. On to Day 23 #DSA #Python #LeetCode #Consistency #SoftwareEngineering #ProblemSolving

✅ Day 78 of 100 Days LeetCode Challenge Problem: 🔹 #3110 – Score of a String 🔗 https://lnkd.in/g9jgxcdK Learning Journey: 🔹 Today’s problem involved calculating the score of a string based on the absolute differences between ASCII values of adjacent characters. 🔹 I iterated through the string starting from index 1 and computed the difference between the current and previous character using ord(). 🔹 For each pair, I added the absolute difference to a running total. 🔹 This approach efficiently accumulates the score in a single pass. Concepts Used: 🔹 String Traversal 🔹 ASCII Value Conversion (ord()) 🔹 Absolute Difference Calculation Key Insight: 🔹 The problem reduces to comparing adjacent elements, making it a straightforward linear scan. 🔹 Using ord() allows direct access to ASCII values, simplifying the computation. Complexity: 🔹 Time: O(n) 🔹 Space: O(1) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

📌 Problem: Increasing Triplet Subsequence 💡 Approach: Traverse the array while maintaining two variables: first and second, representing the smallest and second smallest values found so far. If the current number is smaller than first, update first Else if it’s smaller than second, update second If a number is greater than both, we’ve found an increasing triplet This ensures an optimal single-pass solution. ⚙️ Key Insight: Track only two values instead of checking all triplets Greedy + optimization approach reduces complexity significantly ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) 📚 What I learned: Greedy strategy for subsequence problems Reducing brute-force O(n³) to optimal linear solution #LeetCode #DSA #Algorithms #Coding #ProblemSolving #Python #Greedy #InterviewPreparation #CodingJourney

Binary Search on Rotated Arrays: Adapting Logarithmic Search to Broken Invariants Standard binary search requires sorted data. When an array is rotated (e.g., [4,5,6,7,0,1,2]), the sorted property breaks globally but persists locally — one half is always properly sorted. The adaptation: determine which half is sorted by comparing endpoints, then check if the target falls within that sorted range. This preserves O(log n) complexity despite the rotation disrupting global order. The Design Lesson: When invariants break, look for partial invariants. Here, global sorting is lost but local sorting remains. This "find the preserved property" approach applies broadly — searching in nearly-sorted data, handling corrupted indices with known structure, or working with time-series data with periodic gaps. The algorithm adapts to what guarantees still hold. Time: O(log n) | Space: O(1) #BinarySearch #AdaptiveAlgorithms #RotatedArrays #InvariantPreservation #Python #AlgorithmDesign #SoftwareEngineering

📌 Problem: 1679. Max Number of K-Sum Pairs 💡 Approach: First, sort the array to efficiently apply the two-pointer technique. Initialize two pointers: one at the start (left) and one at the end (right). If the sum equals k, we found a valid pair → increment count and move both pointers If the sum is greater than k, move the right pointer to reduce the sum If the sum is smaller than k, move the left pointer to increase the sum Continue until both pointers meet. ⚙️ Key Insight: Sorting enables efficient pair finding Two-pointer approach avoids checking all pairs (O(n²)) Greedy selection ensures maximum number of operations ⏱️ Time Complexity: O(n log n) (due to sorting) 📦 Space Complexity: O(1) (ignoring sorting space) 📚 What I learned: Two-pointer technique on sorted arrays Optimizing pair problems from brute force to efficient solutions #LeetCode #DSA #Algorithms #Coding #ProblemSolving #Python #TwoPointers #Greedy #InterviewPreparation #CodingJourney

🚀 Day 22 of Consistency | #75DaysLeetCodeChallenge 🧠 Today’s Problem : Daily Temperatures (#739) 💡 Key Learning: This problem introduces the concept of a monotonic stack, useful for solving “next greater element” type problems efficiently. ⚡ Approach: Use a stack to store indices of temperatures Traverse the array → While current temp > stack top temp → pop & calculate days difference Store result for popped indices Push current index to stack 🧠 Why this works: Monotonic stack keeps decreasing order Helps find next greater element in O(n) Avoids brute force O(n²) 🔥 Result : ✔️ Runtime: 116 ms (Beats 28.07%) 📈 A very important pattern for stack-based optimization problems. A big thanks to Shivam Singh, Nikhil Yadav & Akshat Tiwari for this amazing challenge 🙌 Consistency is compounding. Keep going. 💪 #Day22 #LeetCode #DSA #CodingJourney #100DaysOfCode #Python #Stack #Consistency

Why can’t we just search embeddings with a Python script instead of using a vector database? While working with RAG systems, this question came to mind. A simple approach would be to compare a query embedding with every vector using linear search. But that takes O(n) time — which becomes very slow when you have millions of vectors. Vector databases solve this using Approximate Nearest Neighbor (ANN) algorithms like HNSW and LSH. These structures allow fast similarity search by navigating the vector space efficiently instead of checking every vector. That’s why most modern RAG pipelines rely on vector databases. Interesting how much engineering goes into making retrieval fast. #MachineLearning #GenAI #RAG #VectorDB

🚀 Day 25 of Consistency | #75DaysLeetCodeChallenge 🧠 Today’s Problem: Search a 2D Matrix (#74) 💡 Key Learning: This problem teaches how to apply binary search on a 2D matrix by treating it as a flattened sorted array. It’s all about visualizing the structure differently. ⚡ Approach: First apply binary search on rows to find the potential row Check range using first and last element of each row Once row is identified → apply binary search on that row Return true if target found, else false 🧠 Why this works: Matrix follows sorted properties row-wise and across rows Binary search efficiently reduces search space Achieves optimal time complexity → O(log(m × n)) 🔥 Result: ✔️ Runtime: 0 ms (Beats 100%) 📈 This is a classic problem combining matrix + binary search, frequently asked in interviews. A big thanks to Shivam Singh, Nikhil Yadav & Akshat Tiwari for this amazing challenge 🙌 Consistency is compounding. Keep going. 💪 #Day25 #LeetCode #DSA #CodingJourney #100DaysOfCode #Python #BinarySearch #Consistency