Optimizing Code for Performance with Algorithm Efficiency

🚀 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 61 on LeetCode — Contiguous Array (Find Max Length of Equal 0s and 1s) ⚖️✅ This problem is a classic application of prefix sum + hashmap technique. 🔹 Idea Behind the Solution The key trick is to convert the problem into a prefix sum problem: • Treat 0 as -1 and 1 as +1 • Maintain a running sum (sum) • If the same sum appears again at two indices, it means the subarray between them has equal number of 0s and 1s 🔹 How the HashMap Helps • Store the first occurrence of each prefix sum • If a prefix sum repeats at index i and was previously seen at index j, then:   i - j gives a balanced subarray length 🔹 Initialization Trick • seen{{0, -1}} ensures that subarrays starting from index 0 are correctly handled ⚡ Complexity: • Time: O(n) • Space: O(n) 💡 Key Takeaways: • Converting binary problems into prefix sum transformations simplifies logic • Hashmaps are powerful for tracking previous states of cumulative sums • Recognizing patterns where equal values → zero sum subarray is very useful 🔥 This is a very important pattern for array + hashmap + prefix sum problems! #LeetCode #DSA #Algorithms #DataStructures #PrefixSum #HashMap #Arrays #ProblemSolving #Coding #Programming #Cpp #STL #SoftwareEngineering #ComputerScience #CodingPractice #DeveloperLife #TechJourney #CodingDaily #100DaysOfCode #AlgorithmPractice #CodingSkills #Developers #TechCommunity #SoftwareDeveloper #EngineeringJourney

Day 71 on LeetCode Find Peak Element ⛰️✅ Today’s problem introduced a Binary Search approach, but I first solved it using a brute-force traversal to build intuition. 🔹 Approach Used in My Solution (Brute Force) The goal was to find an element that is greater than its neighbors. Key idea: • Traverse the array from index 1 to n-1 • Check if current element is greater than its neighbors • Handle edge cases like last element separately • Return the index once a peak is found This approach is simple and works reliably. 🔹 Optimized Approach (Binary Search Insight) • If nums[mid] < nums[mid+1] → peak lies on the right side • Else → peak lies on the left side (including mid) • Continue narrowing until left == right ⚡ Complexity: • Brute Force: O(n) • Binary Search: O(log n) 💡 Key Takeaways: • Building intuition with brute force helps understand the problem deeply • Learned how to transition from linear scan → binary search optimization • Reinforced that multiple approaches can lead to the same result, but efficiency matters 🔥 Step by step, moving from basic logic to optimized patterns! #LeetCode #DSA #Algorithms #DataStructures #BinarySearch #Arrays #ProblemSolving #Coding #Programming #Cpp #STL #SoftwareEngineering #ComputerScience #CodingPractice #DeveloperLife #TechJourney #CodingDaily #100DaysOfCode #BuildInPublic #AlgorithmPractice #CodingSkills #Developers #TechCommunity #SoftwareDeveloper #EngineeringJourney

Day 62 on LeetCode Subarrays Divisible by K ➗✅ Today’s problem was a strong extension of Prefix Sum + Hash Map, focusing on modulo arithmetic. 🔹 Approach Used in My Solution The goal was to count subarrays whose sum is divisible by k. Key idea in the solution: • Maintain a prefix sum while iterating • Compute remainder = prefix % k • If the same remainder appears again, it means the subarray between them is divisible by k Why? 👉 Because: (prefix[i] - prefix[j]) % k == 0 ➡️ when both have the same remainder 🔹 Important Handling • If remainder is negative → adjust using rem += k • Use a hash map to store frequency of remainders • Add frequency of current remainder to result 🔹 Initialization Trick • remainderCount[0] = 1 handles cases where subarray from index 0 is valid ⚡ Complexity: • Time: O(n) • Space: O(n) 💡 Key Takeaways: • Learned how modulo with prefix sum helps in divisibility problems • Strengthened understanding of remainder frequency tracking • Reinforced identifying patterns like same remainder → valid subarray 🔥 Another powerful pattern unlocked in prefix sum problems! #LeetCode #DSA #Algorithms #DataStructures #PrefixSum #HashMap #Modulo #Arrays #ProblemSolving #Coding #Programming #Cpp #STL #SoftwareEngineering #ComputerScience #CodingPractice #DeveloperLife #TechJourney #CodingDaily #100DaysOfCode #BuildInPublic #AlgorithmPractice #CodingSkills #Developers #TechCommunity #SoftwareDeveloper #EngineeringJourney

When I’m solving problems on LeetCode, I usually hit a wall—not because I don't understand the logic, but because I struggle to recall the exact template or pattern for specific categories. While Binary Search is straightforward, I found myself constantly falling short on Graphs, Backtracking, and Dynamic Programming. It gets even trickier when a problem requires layering multiple patterns to get that optimized solution. To help myself, I started documenting at-a-glance patterns and templates. I realized this could help others who are stuck in the same "grind" as me, so I put everything together into a central hub called AlgoSheet. Check it out here: https://lnkd.in/gCNPzDwj It’s currently a GitHub page where I’ve compiled the most frequent patterns. It's built by a dev, for devs. If you find it useful, feel free to use it for your interview prep. I’m also totally open to suggestions or PRs if you want to help expand the pattern library! #LeetCode #CodingInterview #SoftwareEngineering #DSA #AlgoSheet

Solved one of the most interesting stack-based problems recently: Maximum Subarray Min-Product. At first glance, it looks like a variation of subarray problems, but the real challenge lies in combining multiple concepts efficiently: Monotonic Stack to determine the next smaller element boundaries Prefix Sum to compute subarray sums in constant time BigInt handling in JavaScript to safely manage large intermediate values The key insight was realizing that for every element, we can treat it as the minimum of a subarray and expand left and right until a smaller element blocks it. Using a monotonic increasing stack allows us to compute this in linear time. What made this problem particularly valuable: Reinforced how stack patterns extend beyond classic histogram problems Highlighted the importance of boundary management in array problems Demonstrated practical use of BigInt in algorithmic challenges Time Complexity: O(n) Space Complexity: O(n) Problems like this are a great reminder that mastering patterns, not just problems, is what builds real problem-solving ability. If you're working on Data Structures and Algorithms, this is definitely a problem worth understanding deeply. #DataStructures #Algorithms #DSA #Coding #Programming #SoftwareEngineering  #JavaScript #ProblemSolving #CompetitiveProgramming #LeetCode  #TechLearning #DeveloperJourney #CodeNewbie #LearnToCode  #InterviewPrep #CodingInterview #ComputerScience #100DaysOfCode

Day 2: Leetcode – 704. Binary Search Today’s problem looked straightforward, but it exposed how small mistakes in control flow can completely break the logic. The Key Insight: Binary search isn’t just about comparing values — it’s about maintaining the correct search space and letting the loop run until it’s fully exhausted. Returning too early can silently ruin the entire algorithm. My Mistake: I placed return -1 inside the loop, which caused the function to exit after just one iteration if the target wasn’t found immediately. The code compiled fine but the logic was wrong — a good reminder that passing compilation doesn’t mean correctness. My Approach: Initialize low = 0 and high = n-1 Calculate mid each iteration Compare nums[mid] with target Adjust search space accordingly Only return -1 after the loop ends Takeaway: Binary search is simple in theory but demands discipline in implementation. One misplaced return or bracket can turn a correct idea into a faulty solution. Day 2 Completed #LeetCode #DSA #BinarySearch #Learning #CodingJourney

Anthropic accidentally leaked Claude Code's entire source code. 512,000 lines of TypeScript across 1,900 files, shipped in an npm update. A developer built the entire thing in Python within a few hours with the help of a couple of humans and a few AI agents. The repo became the fastest growing in GitHub history. Half a million lines of production code from one of the best funded AI companies on the planet, rewritten in a few hours. This is proof that writing code is not the bottleneck anymore. Agents do that. The hard part now is thinking clearly, designing the system, and knowing what to build and why. Thinking is software engineering now.

🚀 𝐒𝐨𝐥𝐯𝐞𝐝: 𝐌𝐚𝐱𝐢𝐦𝐮𝐦 𝐏𝐫𝐨𝐝𝐮𝐜𝐭 𝐒𝐮𝐛𝐚𝐫𝐫𝐚𝐲 (𝐁𝐫𝐮𝐭𝐞 𝐅𝐨𝐫𝐜𝐞 → 𝐎𝐩𝐭𝐢𝐦𝐢𝐳𝐞𝐝) Today I worked on an interesting problem that highlights how tricky multiplication can be compared to sum problems. 🔍 𝐏𝐫𝐨𝐛𝐥𝐞𝐦: Find the contiguous subarray with the maximum product. 💡 𝐊𝐞𝐲 𝐂𝐡𝐚𝐥𝐥𝐞𝐧𝐠𝐞𝐬: 1. Negative numbers can flip the result 2. Zero breaks the product chain 3.Edge cases make it more complex than it looks 🛠 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡𝐞𝐬 𝐈 𝐢𝐦𝐩𝐥𝐞𝐦𝐞𝐧𝐭𝐞𝐝: 𝟏️.𝐁𝐫𝐮𝐭𝐞 𝐅𝐨𝐫𝐜𝐞 > Check all subarrays > Time Complexity: O(n²) 𝟐️.𝐎𝐩𝐭𝐢𝐦𝐢𝐳𝐞𝐝 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡 (𝐏𝐫𝐞𝐟𝐢𝐱 + 𝐒𝐮𝐟𝐟𝐢𝐱) > Traverse from both directions > Handle negatives and zeros efficiently > Time Complexity: O(n) ✨ 𝐊𝐞𝐲 𝐈𝐧𝐬𝐢𝐠𝐡𝐭: The maximum product can come from either left-to-right or right-to-left traversal due to sign changes caused by negative numbers. 📌 𝐄𝐱𝐚𝐦𝐩𝐥𝐞: Input: [-2, 0, -1, 2] Output: 2 💻 𝐆𝐢𝐭𝐇𝐮𝐛 𝐂𝐨𝐝𝐞: 👉 https://lnkd.in/g2sxhjMZ Clean, readable, and structured code for easy understanding. Would love feedback from the community 🙌 #Python #DSA #Algorithms #Coding #ProblemSolving #Developers #GitHub #Learning

Day 07: Cracking the "Non-Divisible Subset" Logic 🧩 Today was a true test of algorithmic thinking. I tackled a problem that looks like a standard array search but is actually a brilliant exercise in Number Theory and Remainder Math. The Challenge: Given a set of numbers, find the maximum size of a subset where the sum of any two numbers is not divisible by K The Strategy (Remainder Frequency): Instead of checking every possible pair (which would be very slow), I focused on remainders . If two numbers sum to a multiple of K, their remainders (r1+r2)  must sum to K. const s = [19,10,12,10,24,25,22]; const k = 4; function nonDivisibleSubset(k, s) { let freq = new Array(k).fill(0); // count remainders for (let num of s) { freq[num % k]++; } let count = 0; // remainder 0 case if (freq[0] > 0) count++; // check pairs for (let i = 1; i <= Math.floor(k / 2); i++) { if (i === k - i) { // special case when k is even if (freq[i] > 0) count++; } else { count += Math.max(freq[i], freq[k - i]); } } return count; } console.log(nonDivisibleSubset(k,s)) Key Takeaway: When a problem involves divisibility, don't look at the numbers—look at the remainders. It turns a complex pairing problem into a simple counting one! One full week of coding done. The momentum is real! 🚀 #JavaScript #Algorithms #NumberTheory #100DaysOfCode #CodingChallenge #ProblemSolving #SoftwareEngineering