Binary Search on Rotated Arrays with Java

🚀 #100DaysOfCode | Day 47 🔍 Solved: Find Minimum in Rotated Sorted Array Today I explored another interesting variation of Binary Search. Instead of searching for a target, the goal was to find the minimum element in a rotated sorted array. 💡 Key Insight: By comparing the middle element with the last element, we can determine which half contains the minimum value. Approach: ✔ Used Binary Search to achieve O(log n) time complexity ✔ Compared mid with end to identify the unsorted portion ✔ Narrowed down the search space efficiently What I Learned: This problem helped me understand how binary search can be applied beyond simple searching—especially in rotated and partially sorted arrays. #Java #DSA #LeetCode #BinarySearch #CodingJourney #ProblemSolving #TechSkills

📘 DSA Journey — Day 28 Today’s focus: Binary Search for minimum in rotated arrays. Problem solved: • Find Minimum in Rotated Sorted Array (LeetCode 153) Concepts used: • Binary Search • Identifying unsorted half • Search space reduction Key takeaway: The goal is to find the minimum element in a rotated sorted array. Using binary search, we compare the mid element with the rightmost element: • If nums[mid] > nums[right] → minimum lies in the right half • Else → minimum lies in the left half (including mid) This works because the rotation creates one unsorted region, and the minimum always lies in that region. By narrowing the search space each time, we achieve O(log n) time complexity. This problem highlights how slight modifications in array structure still allow binary search to work efficiently with the right observations. Continuing to strengthen binary search patterns and consistency in problem solving. #DSA #Java #LeetCode #CodingJourney

📘 DSA Journey — Day 33 Today’s focus: Binary Search for next greater element. Problem solved: • Find Smallest Letter Greater Than Target (LeetCode 744) Concepts used: • Binary Search • Upper bound concept • Circular handling Key takeaway: The goal is to find the smallest character strictly greater than a given target in a sorted array. This is a classic upper bound problem: We use binary search to find the first element greater than the target. During search: • If letters[mid] > target, store it as a possible answer and move left • Else move right An important edge case: If no character is greater than the target, we return the first element (circular behavior). Continuing to strengthen binary search patterns and problem-solving consistency. #DSA #Java #LeetCode #CodingJourney

𝐃𝐚𝐲 𝟓𝟔 – 𝐃𝐒𝐀 𝐉𝐨𝐮𝐫𝐧𝐞𝐲 | 𝐀𝐫𝐫𝐚𝐲𝐬 🚀 Today’s problem focused on finding a peak element using binary search. 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝐒𝐨𝐥𝐯𝐞𝐝 • Find Peak Element 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡 • Used binary search instead of linear scan • Compared the middle element with its next element Logic: • If nums[mid] > nums[mid + 1] → peak lies on the left side (including mid) • Else → peak lies on the right side • Continued until left == right 𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠𝐬 • Binary search can be applied on patterns, not just sorted arrays • A peak always exists due to problem constraints • Comparing adjacent elements helps determine direction • Reducing the search space is the key idea 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲 • Time: O(log n) • Space: O(1) 𝐓𝐚𝐤𝐞𝐚𝐰𝐚𝐲 Binary search is not about sorted arrays — it’s about eliminating half of the search space using logic. 56 days consistent 🚀 On to Day 57. #DSA #Arrays #BinarySearch #LeetCode #Java #ProblemSolving #DailyCoding #LearningInPublic #SoftwareDeveloper

#Day77 of my second #100DaysOfCode Finally wrapped up binary search on 1D arrays, now moving to binary search on answers. DSA • Solved Sqrt(x) (LeetCode 69) — finding the integer square root of a number without using built-in functions – Brute: iterate until square exceeds target → O(n) – Optimal: binary search on answer space → O(log n) • Key idea: instead of searching an index, we search for the answer itself • Learned how to narrow down the range based on mid² comparison Nice shift in thinking — binary search feels much more powerful now. #DSA #BinarySearch #LeetCode #Algorithms #Java #100DaysOfCode #WomenWhoCode #BuildInPublic #LearningInPublic

Day 59/75 — Merge Sorted Array Today’s problem was about merging two sorted arrays into one sorted array in-place. Approach: • Start from the end to avoid overwriting elements • Use three pointers: i (nums1), j (nums2), k (merged position) • Place the larger element at the end and move backwards Key logic: while (i >= 0 && j >= 0) { if (nums1[i] > nums2[j]) { nums1[k--] = nums1[i--]; } else { nums1[k--] = nums2[j--]; } } while (j >= 0) { nums1[k--] = nums2[j--]; } Time Complexity: O(m + n) Space Complexity: O(1) A simple yet important problem that reinforces two-pointer technique. 59/75 🚀 #Day59 #DSA #TwoPointers #Arrays #Java #Algorithms #LeetCode

Day 32/50 🚀 — Search Insert Position (Binary Search) Today’s problem reinforced one of the most important concepts in DSA — Binary Search 🔍 🔹 Applied efficient search in a sorted array 🔹 Focused on boundary conditions 🔹 Learned how to return the correct insert position when target isn’t found Key insight: Binary search isn’t just about finding an element — it’s about narrowing down the answer space intelligently. 💡 Returning left at the end ensures we get the exact position where the target should be inserted. Performance highlights: ⚡ Runtime: 0 ms (100%) 📦 Memory: Optimized #Day32 #LeetCode #BinarySearch #DSA #Java #CodingJourney #50DaysOfCode #ProblemSolving

Day 8/30 — DSA Challenge 🚀 Problem: Search a 2D Matrix II Topic: Matrix + Binary Search Pattern Difficulty: Medium Approach: Started from top-right corner of the matrix If current element == target → return true If current element < target → move down If current element > target → move left Mistake / Challenge: Initially tried applying binary search like previous problem (LeetCode 74) Realized matrix is not globally sorted, so that approach fails Fix: Used optimal “staircase search” approach Reduced time complexity efficiently Key Learning: Not all sorted matrices can be treated as 1D arrays Recognize pattern: row-wise + column-wise sorted → use pointer approach Time Taken: 45 minutes Consistency check ✅ See you on Day 9. GitHub Repo: https://lnkd.in/gHW9vKUf #DSA #LeetCode #Java #Matrix #BinarySearch #LearningInPublic

📘 DSA Journey — Day 35 Today’s focus: Binary Search with index patterns. Problem solved: • Single Element in a Sorted Array (LeetCode 540) Concepts used: • Binary Search • Index parity (even/odd pattern) • Search space reduction Key takeaway: The array is sorted and every element appears twice except one. A key observation: Before the single element, pairs start at even indices After the single element, this pattern breaks. Using binary search: • If mid is even and nums[mid] == nums[mid + 1], the single element lies on the right side • Else, it lies on the left side (including mid) By leveraging this pattern, we can find the answer in O(log n) time and O(1) space. Continuing to strengthen binary search intuition and consistency in problem solving. #DSA #Java #LeetCode #CodingJourney

#100DaysOfCode – Day 2 Today I worked on a DSA problem based on arrays: Check if an array is sorted and rotated 🔍 Approach: Instead of finding the exact rotation point, I focused on identifying a pattern: In a sorted and rotated array, the order should break at most once. So, I checked how many times an element is greater than the next element while traversing the array in a circular manner. ✔️ If the count of such breaks is 0 or 1 → valid ❌ If it’s more than 1 → not a sorted rotated array 🧠 Key Takeaway: This problem taught me how pattern observation can simplify logic and avoid unnecessary complexity. Sometimes the best solution is not the most obvious one! 📈 Staying consistent and improving step by step 💪 #100DaysOfCode #DSA #DataStructures #Algorithms #Java #CodingJourney #ProblemSolving #LeetCode #Consistency