Sateesh Sonkamble’s Post

🚀 Mastering Efficiency: Solving the Maximum Subarray Problem | LeetCode Just cleared the Maximum Subarray challenge! This problem is a perfect example of how a simple shift in logic—moving from brute force to a greedy approach—can drastically optimize performance. 💡 The Problem: Given an integer array nums, find the subarray with the largest sum and return its sum. ⚡ My Approach (Kadane's Algorithm): Instead of calculating every possible subarray sum, I used a greedy strategy to decide at each step whether to keep the current running sum or "start fresh" from the current element. 👉 The Logic: Initialize: Start with max_sum as the first element and a curr_sum of 0. The "Fresh Start" Rule: As I iterate, if curr_sum becomes negative, it’s a burden. I reset it to 0 because any subarray starting with a negative sum will only decrease the potential total. Accumulate: Add the current number to curr_sum. Update Global Max: Compare curr_sum with max_sum and store the higher value. 🔥 Complexity Analysis: ⏱ Time Complexity: $O(n)$ – A single, clean pass through the array. 📦 Space Complexity: $O(1)$ – Constant space; no extra data structures needed. 🏆 The Result: ✔️ Accepted: Passed all 210 test cases. ✔️ Performance: 28 ms runtime, beating 79.77% of Python3 submissions! 📌 Key Learning: Kadane’s Algorithm is a masterclass in dynamic programming/greedy logic. It teaches you to discard "baggage" (negative sums) and focus only on what contributes to the optimal goal. 💻 Tech Stack: #Python | #Algorithms | #DataStructures #leetcode #dsa #coding #programming #100DaysOfCode #softwareengineering #kadanesalgorithm #optimization #techcommunity

Day 68 on LeetCode Guess Number Higher or Lower 🎯✅ Today’s problem reinforced the power of Binary Search on an answer space. 🔹 Approach Used in My Solution The goal was to identify a hidden number using the provided guess() API. Key idea in the solution: • Apply binary search between 1 and n • Pick mid and call guess(mid) • Based on the response:   – 0 → correct number found   – -1 → guessed number is too high → move left   – 1 → guessed number is too low → move right • Continue narrowing the search space until the number is found This is a perfect example of searching efficiently using feedback. ⚡ Complexity: • Time Complexity: O(log n) • Space Complexity: O(1) 💡 Key Takeaways: • Strengthened understanding of binary search with external APIs • Learned how to adjust search space based on feedback • Reinforced the concept of searching on answer space 🔥 Another solid step in mastering binary search patterns! #LeetCode #DSA #Algorithms #DataStructures #BinarySearch #DivideAndConquer #ProblemSolving #Coding #Programming #Cpp #STL #SoftwareEngineering #ComputerScience #CodingPractice #DeveloperLife #TechJourney #CodingDaily #100DaysOfCode #BuildInPublic #AlgorithmPractice #CodingSkills #Developers #TechCommunity #SoftwareDeveloper #EngineeringJourney

🚀 Cracked the Spiral Matrix problem on LeetCode — and here’s the mindset behind it 👇 Most people jump straight into coding this problem. I didn’t. Instead, I approached it like a boundary management problem 🧠 🔍 My thought process: • Treat the matrix like a shrinking box • Maintain 4 pointers: top, bottom, left, right • Traverse layer by layer — not element by element • After each traversal, shrink the boundaries inward This helped me: ✅ Avoid unnecessary conditions ✅ Prevent duplicate traversals ✅ Keep the logic clean and scalable The real learning wasn’t just solving it — it was realizing how visualizing the structure simplifies the code. 💡 Sometimes the difference between confusion and clarity is just how you frame the problem. 🔥 Consistency + clarity > brute force coding #LeetCode #DSA #CodingJourney #ProblemSolving #Python #WomenInTech #TechLearning #SoftwareEngineering #100DaysOfCode #StudentDeveloper #CodingMindset #LearnInPublic

Today I learned that not all loops are the same and it actually blew my mind.I'm not going to lie. I thought a loop was just a loop.Turns out, there's a whole world in there. Here's what I uncovered today: 🔹 for loop → when you already know how many times you need to repeat something. Clean, predictable, in control. 🔹 while loop → when you don't know how many times, you just keep going until a condition is met. A little unpredictable just like real life. 🔹 while True → runs forever until YOU decide to stop it. At first it scared me. Then I realized how powerful that actually is. 🔹 Nested loops → a loop inside a loop. Each one exists for a reason. Choosing the right loop isn't just about making code work it's about making it make sense. That's the part nobody tells you when you start coding. It's not just logic. It's judgment. Still a beginner. Still figuring it out. But days like today remind me why I started. #Python #Loops #LearningToCode #CodingJourney #PythonProgramming #GrowthMindset #TechCommunity

🔥 Day 8/100 of My LeetCode Challenge — 0 ms (100%) 🔥 Hook: Most people fail this “easy” problem because they ignore one tiny detail — can you spot it? 💭 Problem: Add 1 to a number represented as an array of digits. Sounds simple, right? But here’s the catch 👇 👉 What if the last digit is 9? 👉 What if ALL digits are 9? Example: [9,9,9] → [1,0,0,0] 💥 🧠 What I learned today: Always think about edge cases Simple problems can hide tricky logic Traversing backwards can simplify carry problems ⚡ My Approach: Start from the end If digit < 9 → add 1 and return If digit == 9 → make it 0 and continue If all become 0 → add 1 at front 💻 Result: ✅ Runtime: 0 ms (100%) ✅ Clean & optimized solution 📈 Consistency Check Day 8 complete. No excuses. No breaks. Small wins daily → Big results later 🚀 #LeetCode #Day8 #CodingJourney #Python #ProblemSolving #100DaysOfCode

Day 10/100 – DSA Challenge Today’s problem: Move Zeroes (LeetCode 283) What I Learned: The goal was to move all zeroes to the end of an array while maintaining the relative order of non-zero elements — and importantly, doing it in-place. Key Idea: Two-Pointer Technique I used a two-pointer approach: One pointer (fast) iterates through the array Another pointer (slow) tracks where the next non-zero element should go Whenever a non-zero element is found, it is swapped with the element at the slow pointer, ensuring all non-zero elements are shifted forward while zeroes naturally move to the end. Why this approach? Maintains order of elements Works in O(n) time complexity Uses O(1) extra space (in-place) Takeaway: This problem reinforced how powerful the two-pointer technique is for array manipulation problems, especially when constraints require in-place operations. Looking forward to tackling more problems and improving consistency! #Day10 #100DaysOfCode #DSA #Python #CodingJourney #LeetCode #ProblemSolving

Code that writes code. 🌌 There is a point in every developer's journey where you stop thinking about Logic and start thinking about Patterns. Advanced Python is about: ~Abstraction: Using Protocols and Generics for structural typing. ~Automation: Using advanced Decorators to inject behavior across entire systems. ~Reliability: Understanding the memory manager so you can prevent leaks before they start. We use these "hidden" features not to make the code more complex, but to make the usage of our code more simple for everyone else. Which part of the "Advanced Mindset" was the hardest for you to learn? For me, it was finally mastering asyncio flow control. #CleanCode #Pythonic #ProgrammingPrinciples #SystemDesign #AdvancedCoding

Every developer has lived this exact moment. Your code is working perfectly. You make one small change just to test. Everything breaks completely. This is not a beginner problem. This is a coding problem. It happens at every level. Here is what I have learned when this happens: Step 1 — Do not panic. Breathe. Step 2 — Undo the change. Get back to what worked. Step 3 — Test one change at a time, never multiple at once. Step 4 — Read the error message carefully. It always tells you something. Step 5 — Google is not cheating. It is a tool. Use it. The best developers are not the ones who never break their code. They are the ones who know how to fix it. Breaking things is how you truly learn to build things. #Python #Coding #DataScience #LearnToCode #BeginnerCoder #Debugging #StudentLife

🚀 Mastering the art of loops! 🔄 Discover how loops help your code execute repetitive tasks efficiently. Essentially, loops are like a magical chant that tells your program to keep doing something until a certain condition is met. For developers, mastering loops is crucial for automating tasks and iterating over data structures with ease. Here's the breakdown: 1️⃣ Initialize a counter variable 2️⃣ Set the condition for the loop 3️⃣ Define the action to perform in each iteration Sample code using a "for" loop in Python: ``` for i in range(5): print("Hello, World!") ``` 🌟 Pro Tip: Use loops to reduce redundancy in your code and boost efficiency. 💡 ⚠️ Common Mistake: Forgetting to update the counter variable in the loop, leading to an infinite loop! 🔄 🌟 What's your favorite use case for loops in your projects? Let's discuss! 💬 #Coding101 #LearnToCode #TechTips #CodeNewbie #PythonProgramming #DeveloperCommunity #LoopLogic #CodeEfficiency #ProDevSkills 🌐 View my full portfolio and more dev resources at tharindunipun.lk

LeetCode Problem 155 "Min Stack": Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. Implement the MinStack class: MinStack() initializes the stack object. void push(int val) pushes the element val onto the stack. void pop() removes the element on the top of the stack. int top() gets the top element of the stack. int getMin() retrieves the minimum element in the stack. You must implement a solution with O(1) time complexity for each function. Approach: we can implement a single stack and store values as tuples (current element, min so far). But wait ... consider edge case when we pop and the stack becomes empty, so to handle this we consider two conditions- one when stack becomes empty, we set min so far to +ve infinity again, second whenever we pop elements, if stack is not empty, we update min so far to the top of the stack's min so far. In this way we can keep track of the minimum element upto a specific stack level. Time complexity of all the operations: O(1) #Python #LeetCode #ProblemSolving #DSA #InterviewPrep #DataStructures #Algorithms #Programming