Modified Binary Search for Sorted Arrays with Duplicates

🚀 Day 35 of my #100DaysOfCode Journey Today, I solved the LeetCode problem: Valid Anagram Problem Insight: Given two strings, check if one is an anagram of the other. Approach: • First, check if the strings have the same length; if not, return false • Convert both strings to character arrays • Sort both arrays • Compare the sorted arrays — if equal, the strings are anagrams Time Complexity: • O(n log n) — due to sorting the arrays Space Complexity: • O(n) — for the character arrays Key Learnings: • Sorting is a simple and effective way to compare character compositions • Edge cases like different lengths should be handled first • Breaking the problem into small steps makes it easy to reason about Takeaway: Sometimes, sorting can reduce a seemingly complex problem into a simple comparison. #DSA #Java #LeetCode #100DaysOfCode #CodingJourney #ProblemSolving #Strings

Day 25/100: Finding the "Gap" 🎯 Today's challenge: Search Insert Position. We all know Binary Search finds an element in O(log n), but what if the element isn't there? I learned that by the end of the search, the `left` pointer doesn't just give up—it points exactly to where that missing number *should* be inserted to keep the array sorted. It’s a powerful way to handle dynamic data without breaking the order. Quarter of the way through the challenge! 🚀 #100DaysOfCode #Java #DSA #BinarySearch #ProblemSolving #Unit3 #Day25 #LearnInPublic

🚀 #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

Day 68/100 Completed ✅ 🚀 Solved LeetCode – Split Array Largest Sum (Java) ⚡ Applied an efficient Binary Search on Answer approach to split the array into k subarrays such that the largest subarray sum is minimized. Instead of trying all possible splits, optimized the solution by searching within the range of maximum element and total sum, and validating using a greedy strategy. 🧠 Key Learnings: Identifying binary search applicability in partition-based problems Defining search space using max element (lower bound) and total sum (upper bound) Using greedy logic to check feasibility of splitting into k subarrays Minimizing the maximum subarray sum through efficient validation 💯 This problem strengthened my understanding of partition problems and reinforced how binary search can be used for optimization rather than direct searching. 🔗 Profile: https://lnkd.in/gaJmKdrA #leetcode #datastructures #algorithms #java #binarysearch #problemSolving #100DaysOfCode

🔥 Day 58 of #100DaysOfCode Solved – Check Balanced String 🔍 What I did: Traversed the string and calculated the sum of digits at even and odd indices separately, then compared both sums to determine if the string is balanced. 💡 Key Learning: Practiced string traversal and indexing Improved understanding of even vs odd index logic Learned how to convert char to integer (c - '0') 🎯 Takeaway: Even simple index-based problems can strengthen your fundamentals and attention to detail. #Day58 #LeetCode #Java #ProblemSolving #CodingJourney #100DaysOfCode #DSA

Day 70 - Next Greater Node Understanding how to find the next larger value for each node using stack optimization. Approach: • Convert linked list → array for easy indexing • Use a stack to track indices • Compare current value with stack top • Update results when a greater value is found Key Insight: Monotonic stack helps avoid unnecessary comparisons Time Complexity: O(n) Space Complexity: O(n) #Day70 #LeetCode #Java #CodingPractice #TechJourney #DSA #LinkedList

#Day98 of #100DaysOfCode Learned Binary Search, an efficient searching algorithm for sorted arrays. Implemented logic to find an element by repeatedly dividing the search space. Focused on understanding how optimized searching works compared to linear search. #Java #DSA #BinarySearch #100DaysOfCode

Day 90 of #100DaysOfCode – Unique Binary Search Trees II Today’s problem was all about generating all structurally unique BSTs for values from 1 to n. At first glance, it feels tricky because it's not just counting trees — we actually need to build every possible tree structure. 💡 Key Insight: Pick each number i as the root Recursively build: Left subtree from [1 ... i-1] Right subtree from [i+1 ... n] Combine every left & right subtree pair 🌳 This is a classic Recursion + Backtracking problem and is closely related to Catalan Numbers. 📌 Example: For n = 3, we get 5 unique BSTs ⚡ What I learned today: How recursion can generate combinations of structures Importance of base case (start > end → null) Combining subproblems effectively 💻 Time Complexity: Approximately O(4ⁿ / √n) Consistency is key — 90 days strong and still going 💪 Next target: 💯 #DSA #Java #LeetCode #Recursion #BinaryTree #CodingJourney #100DaysOfCode

🚀 Day 64/100 Today’s problem: Find all strings that are substrings of another word 🧠 What I learned: - How to compare strings using nested loops - Using ".contains()" to check substrings efficiently - Importance of breaking early to optimize performance - Strengthening problem-solving with brute-force approach 💡 Key Insight: Sometimes simple solutions (O(n²)) are enough when constraints are small. No need to overcomplicate! 🔁 Consistency > Perfection #Day64 #DSA #Java #CodingJourney #Consistency #KeepLearning #100DaysOfCode

Day 77 of #100DaysOfLeetCode 💻✅ Solved #153. Find Minimum in Rotated Sorted Array problem in Java. Approach: • Used Binary Search to find minimum element • Compared mid with right to decide direction • Reduced search space until pointers converged Performance: ✓ Runtime: 0 ms (Beats 100.00% submissions) 🚀 ✓ Memory: 43.62 MB (Beats 78.18% submissions) Key Learning: ✓ Strengthened Binary Search on rotated arrays ✓ Learned how to identify unsorted portion efficiently ✓ Improved optimization over linear search Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinarySearch #Arrays #ProblemSolving #CodingJourney #100DaysOfCode