LeetCode Challenge: Search in Rotated Sorted Array II

🚀 Day 8/100 — LeetCode Challenge Solved Find Minimum in Rotated Sorted Array At first, it looks tricky because the array is rotated. But the key is to identify which part is sorted. 💡 Approach: 1) If the left half is sorted → the minimum could be at low 2) Otherwise → the minimum lies in the right half 3) Keep narrowing the search space using binary search 👉 Instead of searching for a target, here we’re searching for the minimum element. 🧠 Time Complexity: O(log n) 💾 Space Complexity: O(1) 💡 What I’m realizing: Many problems look different, but they’re just variations of binary search with slight tweaks. Slowly building intuition around it. #LeetCode #DSA #100DaysOfCode #Cpp #BinarySearch #CodingJourney

🚀 Day 26/100 — LeetCode Challenge Today's problem: Search in Rotated Sorted Array II 🧠 Concept: Binary Search with Edge Cases 💡 Key Idea: Duplicates can make it difficult to identify the sorted half. In such cases, shrink the search space from both ends. ⚡ Time Complexity: O(log n) average, O(n) worst case 📂 Solutions Repository https://lnkd.in/gkFh2mPZ A great example of how edge cases can complicate an otherwise straightforward approach. #100DaysOfLeetCode #DSA #LeetCode #CodingChallenge #SoftwareEngineering

🚀 Day 25/100 — LeetCode Challenge Today's problem: Find Minimum in Rotated Sorted Array 🧠 Concept: Binary Search on Rotated Array 💡 Key Idea: The minimum element represents the pivot point of rotation. By comparing mid with the rightmost element, we can efficiently narrow down the search space. ⚡ Time Complexity: O(log n) 📂 Solutions Repository https://lnkd.in/gkFh2mPZ Continuing to explore how binary search can be adapted for different problem variations. #100DaysOfLeetCode #DSA #LeetCode #CodingChallenge #SoftwareEngineering

🚀 Day 29/100 — LeetCode Challenge Today's problem: Find First and Last Position of Element in Sorted Array 🧠 Concept: Binary Search (Boundary Finding) 💡 Key Idea: Use binary search twice to find the first and last occurrence of a target element in a sorted array. ⚡ Time Complexity: O(log n) 📂 Solutions Repository https://lnkd.in/gkFh2mPZ Understanding how to modify binary search to handle duplicates and find boundaries. #100DaysOfLeetCode #DSA #LeetCode #CodingChallenge #SoftwareEngineering

🚀 Day 23/100 – LeetCode Challenge ✅ Problem Solved: Sort an Array Today’s problem was all about implementing sorting from scratch without using any built-in functions. I used the Merge Sort algorithm to achieve the required O(n log n) time complexity. This was a great revision of divide-and-conquer concepts, where the array is recursively divided and then merged in sorted order. 💡 Key Learning: Understanding sorting algorithms is fundamental for interviews Merge Sort guarantees O(n log n) performance Writing algorithms from scratch improves problem-solving depth ⚡ Complexity: Time: O(n log n) Space: O(n) Building strong fundamentals, one day at a time 🚀 #Day23 #100DaysOfCode #LeetCode #DSA #Cpp #CodingJourney #Sorting #MergeSort

Day 29 – LeetCode Journey 🚀 Solved Two Sum II – Input Array Is Sorted using the Two Pointer technique, leveraging the sorted nature of the array for an optimal solution. 🔹 Time Complexity: O(n) 🔹 Runtime: 2 ms (Beats 96.37%) 🔹 Memory Usage: 48.58 MB This problem is a great reminder that recognizing patterns (like sorted arrays) can significantly reduce complexity and improve efficiency. Small optimizations, big impact 📈 Staying consistent and sharpening problem-solving skills every day. #LeetCode #DSA #ProblemSolving #CodingJourney #SoftwareEngineering #Consistency #Learning

🚀 Day 10/100 — LeetCode Challenge Solved 3Sum The brute force approach is simple: Try all triplets → O(n³) But that’s not scalable. 💡 Optimized Approach: 1) First, sort the array 2) Fix one element 3) Use two pointers to find the remaining pair 👉 This reduces the complexity significantly. Also handled duplicates carefully to avoid repeated triplets. 🧠 Time Complexity: O(n²) 💾 Space Complexity: O(1) (ignoring output) 💡 What I learned: Sometimes optimization is not about complex logic, but about: 1) Sorting 2) Using patterns like two pointers 3) Eliminating unnecessary work This problem felt like a step up from previous ones. Staying consistent. #LeetCode #DSA #100DaysOfCode #Cpp #TwoPointers #CodingJourney

🚀 Day 26/100 – LeetCode Challenge ✅ Problem Solved: Valid Perfect Square Today’s problem was about checking whether a number is a perfect square without using any built-in functions like sqrt. I solved it using Binary Search, which helped efficiently determine whether there exists an integer whose square equals the given number. 💡 Key Learning: Binary Search can be applied beyond arrays Handling large numbers requires careful use of data types Avoiding overflow is important in mathematical problems ⚡ Complexity: Time: O(log n) Space: O(1) Learning to apply fundamentals in different ways every day 🚀 #Day26 #100DaysOfCode #LeetCode #DSA #Cpp #CodingJourney #BinarySearch #ProblemSolving

Solved “Letter Combinations of a Phone Number” on LeetCode 💡 What looks simple on the surface is actually a clean exercise in recursion + backtracking. 🔍 Key takeaways: - Mapping digits to characters is straightforward, but generating all combinations efficiently is where the thinking comes in - Backtracking helps explore all possibilities while keeping the solution clean and scalable - Important to control recursion depth and maintain state (temp) correctly 💭 What I focused on: - Building combinations incrementally - Reverting state after each recursive call (classic backtracking pattern) - Keeping the solution readable and modular This problem reinforced how powerful recursion is when combined with proper state management. On to the next one 🚀 #DSA #LeetCode #Backtracking #Recursion #ProblemSolving #CodingJourney

Day 12/100 – Building Consistency in DSA Solved “Sqrt(x)” (LeetCode #69) today using a Binary Search approach. The challenge was to compute the square root of a non-negative integer without using built-in functions, while ensuring the result is rounded down. What stood out: This problem highlights how Binary Search can be applied beyond sorted arrays—specifically in narrowing down an answer space efficiently. Key Insights: Reduced time complexity from O(n) to O(log n) Strengthened understanding of boundary conditions and edge cases Practiced writing precise and overflow-safe logic Takeaway: Problems labeled “Easy” often carry fundamental patterns that are crucial for solving harder problems. Consistency > Intensity. Showing up every day. #Day12 #100DaysOfCode #DSA #LeetCode #BinarySearch #ProblemSolving #SoftwareEngineering