LeetCode Discipline: Remove Element in Place

🚀 Day 561 of #750DaysOfCode 🚀 📌 Problem: Minimum Distance to the Target Element Today’s problem was simple yet a great reminder of how powerful basic iteration can be when applied correctly. 🔍 The task was to find the minimum distance between a given start index and any index i such that nums[i] == target. 💡 Key Insight: Instead of overthinking, just iterate through the array and track the minimum value of |i - start| whenever the target is found. Clean, efficient, and effective. 🧠 What I Learned: Sometimes brute force with clarity is the best solution Always look for opportunities to minimize operations with simple logic Writing clean and readable code matters as much as solving the problem ⚡ Approach: Traverse the array Check for target Update minimum distance ⏱️ Complexity: Time: O(n) Space: O(1) 💻 Consistency is key. Small steps every day build strong problem-solving skills over time. #leetcode #dsa #programming #java #coding #developers #softwareengineering #100daysofcode #codingjourney #tech #learning #growth

🚀 Day 27/60 — LeetCode Discipline Problem Solved: Length of Last Word Difficulty: Easy Today’s problem looked simple, yet it emphasized a subtle but important skill — handling edge cases cleanly. The goal was to find the length of the last word in a string, ignoring any trailing spaces. Instead of splitting the string or using extra space, the solution efficiently traverses from the end, skipping unnecessary characters and counting only what truly matters. It’s a reminder that strong coding is not always about complexity — sometimes, it’s about precision and clarity in small details. 💡 Focus Areas: • Practiced string traversal from end • Improved handling of trailing spaces • Reinforced clean and efficient logic • Avoided unnecessary extra space usage • Strengthened edge-case thinking ⚡ Performance Highlight: Achieved 0 ms runtime (100% performance). Small problems, when approached with discipline, sharpen the instincts that solve big ones. #LeetCode #60DaysOfCode #100DaysOfCode #DSA #Strings #Algorithms #ProblemSolving #CodingJourney #SoftwareEngineering #Java #Developers #Consistency #TechJourney #LearnToCode

Day 38 of My DSA Journey Today I solved LeetCode 152 – Maximum Product Subarray on LeetCode. 📌 Problem Given an integer array nums, find the contiguous subarray that has the largest product, and return that product. Example: Input: [2,3,-2,4] Output: 6 → subarray [2,3] 🧠 Approach – Dynamic Programming (Tracking Max & Min) This problem is tricky because of negative numbers. Key Idea: • At each index, we maintain: maxProduct → maximum product ending at current index minProduct → minimum product ending at current index 👉 Why min? Because a negative number can turn a small (negative) product into a large positive one. Steps I followed: • Initialize maxProduct, minProduct, and ans with the first element • Traverse the array from left to right • If the current number is negative → swap max and min • Update: maxProduct = max(current, current × maxProduct) minProduct = min(current, current × minProduct) • Update the final answer using maxProduct ⏱ Time Complexity: O(n) — Single pass through the array 📦 Space Complexity: O(1) — No extra space used 💡 Key Learnings ✔ Handling negative numbers in product problems ✔ Using dynamic programming with state tracking ✔ Understanding why tracking both max & min is important This is one of those problems that looks simple but tests deep understanding of edge cases 🚀 Consistency continues — leveling up every day 💪 #100DaysOfCode #DSA #DynamicProgramming #Arrays #LeetCode #Java #ProblemSolving #CodingJourney #DeveloperJourney #Programming #SoftwareEngineering

Today, I tackled a classic dynamic programming problem — Minimum Difficulty of a Job Schedule. This challenge really tested my understanding of: - Breaking problems into subproblems (partitioning jobs across days) - Optimizing brute force recursion using memoization - Thinking carefully about state definition (index, days left) Initially, I made a common mistake by trying to accumulate results greedily instead of exploring all valid partitions. Once I corrected that and properly defined the recurrence: 👉 current_day_max + solve(remaining_jobs, remaining_days) things started to click. Key takeaways include: - Always validate your recurrence before optimizing - DP is all about choosing the right state and transitions - Small indexing mistakes (i+1 vs ind+1) can completely break logic This was a valuable exercise in debugging and refining my dynamic programming approach. #LeetCode #DataStructures #Algorithms #DynamicProgramming #ProblemSolving

Day 69 on LeetCode Search in Rotated Array (Brute Force Insight) 🔍✅ Today’s problem initially felt tricky, but turned out to be one of the simplest when approached directly. 🔹 Approach Used in My Solution Instead of overcomplicating with rotation logic, I used a straightforward linear search. Key idea: • Traverse the array from start to end • Compare each element with the target • Return the index once found, otherwise -1 Sometimes, the simplest approach is the most reliable. 🔹 Initial Thought Process (Overthinking Phase 😅) • Considered rotating the array and then searching • Thought about adjusting indices using formulas like:   (index - rotation + n) % n • But realized that all of this is unnecessary for basic search ⚡ Complexity: • Time Complexity: O(n) • Space Complexity: O(1) 💡 Key Takeaways: • Not every problem needs an optimized approach — clarity > complexity • Avoid overengineering when a simple solution works perfectly • Always validate if a direct approach solves the problem efficiently enough 🔥 Great reminder: Sometimes the “easy way” is the smart way. #LeetCode #DSA #Algorithms #DataStructures #Arrays #LinearSearch #ProblemSolving #Coding #Programming #Cpp #STL #SoftwareEngineering #ComputerScience #CodingPractice #DeveloperLife #TechJourney #CodingDaily #100DaysOfCode #BuildInPublic #AlgorithmPractice #CodingSkills #Developers #TechCommunity #SoftwareDeveloper #EngineeringJourney

🚀 Day 10 of 100 Days LeetCode Challenge Problem: Find All Possible Stable Binary Arrays II Today’s problem is an extension of Day 9—but with optimized Dynamic Programming ⚡ 💡 Key Insight: Same constraints: Fixed number of 0s and 1s No more than limit consecutive identical elements 👉 Which means: We must carefully control streak length And efficiently count all valid combinations 🔍 Approach (Optimized DP): Use DP + Prefix Sum Optimization State includes: Count of 0s used Count of 1s used Ending with 0 or 1 💡 Optimization: Instead of recalculating ranges repeatedly, use prefix sums This reduces time complexity significantly 👉 Apply modulo (10⁹ + 7) for large answers 🔥 What I Learned Today: Same problem can have multiple levels of optimization Prefix sum is powerful in reducing DP transitions Moving from brute → DP → optimized DP is real growth 📈 📈 Challenge Progress: Day 10/100 ✅ Double digits achieved! LeetCode, Dynamic Programming, Prefix Sum, Optimization, Combinatorics, Binary Arrays, DSA Practice, Coding Challenge, Problem Solving #100DaysOfCode #LeetCode #DSA #CodingChallenge #DynamicProgramming #PrefixSum #Optimization #ProblemSolving #TechJourney #ProgrammerLife #SoftwareDeveloper #CodingLife #LearnToCode #Developers #Consistency #GrowthMindset #InterviewPrep

🚀 Day 29/60 — LeetCode Discipline Problem Solved: Longest Common Prefix Difficulty: Easy Today’s problem focused on identifying the longest common starting pattern across multiple strings — a classic exercise in string comparison and iteration. The approach involved checking characters position by position across all strings until a mismatch is found. This reinforces how simple iterative logic can elegantly solve problems involving multiple inputs. It’s a reminder that sometimes, clarity in approach matters more than complexity in code. 💡 Focus Areas: • Strengthened string traversal techniques • Practiced comparing multiple inputs efficiently • Improved understanding of edge cases (empty strings, mismatch handling) • Reinforced early stopping conditions for optimization • Focused on clean and readable logic ⚡ Performance Highlight: Achieved efficient runtime with minimal overhead. Finding common ground across multiple inputs is not just a coding skill — it’s a pattern recognition mindset. #LeetCode #60DaysOfCode #100DaysOfCode #DSA #Strings #Algorithms #ProblemSolving #CodingJourney #SoftwareEngineering #Java #Developers #Consistency #TechGrowth #LearnToCode

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gK993p-S 💡 My thought process: The main function, minAbsDiff, goes through all possible top-left positions of the k × k submatrices. For each position (i, j), it calls the helper function solve to find the result for that submatrix and stores it in the answer matrix, which has a size of (m − k + 1) × (n − k + 1). The solve function collects all elements from the k × k submatrix starting at (i, j) and puts them into a set. This automatically sorts the elements and removes duplicates. If the set has only one unique element, the function returns 0 because all values are the same. If there are multiple unique elements, the function goes through the sorted set and calculates the absolute difference between each pair of consecutive elements. Since the set is sorted, the minimum absolute difference will be between adjacent elements. The smallest difference is tracked and returned. The overall time complexity is O((m − k + 1) × (n − k + 1) × k² log(k²)) because each submatrix requires inserting k² elements into a set and iterating through it. The space complexity is O(k²) for storing elements of each submatrix. 👉 My Solution: https://lnkd.in/gj6juuUU If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

🚀 Day 6 of 100 Days LeetCode Challenge Problem: Check if Binary String Has at Most One Segment of Ones Today’s problem is all about pattern observation in strings—simple, but easy to overthink. 💡 Key Insight: The string should contain only one continuous block of '1's. 👉 That means: Once a 0 appears after a 1, There should be no more '1's later 🔍 Simplest Trick: Just check if the pattern "01" appears more than once OR even better → check if "10" appears followed by another "1" 💡 Cleaner approach: Traverse the string Count transitions from 1 → 0 If you ever see 1 again after that → ❌ Invalid 🔥 What I Learned Today: Many problems are just pattern validation Clean logic beats complex conditions Always try to reduce the problem to a simple rule 📈 Challenge Progress: Day 6/100 ✅ Consistency building strong! LeetCode, Strings, Pattern Recognition, Greedy, DSA Practice, Coding Challenge, Problem Solving, Algorithm Thinking, Programming #100DaysOfCode #LeetCode #DSA #CodingChallenge #Strings #ProblemSolving #TechJourney #ProgrammerLife #SoftwareDeveloper #CodingLife #LearnToCode #Developers #Consistency #GrowthMindset #InterviewPrep

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gVK8xX3p 💡 My thought process: The solution uses a top-down dynamic programming approach with memoization. Each cell stores a pair of values: the maximum product and the minimum product achievable from that cell to the destination. The minimum value is necessary because multiplying two negative numbers can lead to a larger positive result later. The recursive function explores both possible moves—right and down—from the current cell and retrieves their stored results. For each direction, it considers both the maximum and minimum products returned and multiplies them by the current cell’s value. All combinations are evaluated, and the overall maximum and minimum are selected and stored in the DP table for that cell. Memoization ensures that each cell is computed only once, reducing the time complexity to O(m × n). The base case occurs at the bottom-right cell, where both maximum and minimum values equal the cell’s value. Invalid positions return sentinel values to indicate they should not be considered. Finally, the function checks the result from the starting cell. If the maximum product is negative, it returns -1. Otherwise, it returns the result modulo 1e9 + 7. 👉 My Solution: https://lnkd.in/gcKfTVNq If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari