LeetCode Challenge: Top-Down Dynamic Programming Solution

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/giAcdv2U 💡 My thought process: This solution uses 3D dynamic programming to find the maximum coins that can be collected from the top-left to the bottom-right of a grid, allowing up to 2 neutralizations for negative cells. The state dp[i][j][k] stores the maximum coins reachable at cell (i, j) using k neutralizations. For each cell, transitions are taken from the left and top cells. If the current cell is negative, two options are considered: either neutralize it (if k > 0) and carry forward the previous value without adding the negative, or do not neutralize and add the cell value. For non-negative cells, the value is simply added to the best of the previous states. The starting cell is initialized separately based on whether it is neutralized or not. The final answer is the maximum value among dp[m-1][n-1][0], dp[m-1][n-1][1], and dp[m-1][n-1][2]. 👉 My Solution: https://lnkd.in/gZWS3Kns 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

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gBvd5iWy 💡 My thought process: For this question, I am precalculating the total sum of the grid and row-wise and column-wise sums. Store the precalculated values in a vector and then simply do a cumulative sum over the vector. If the cumulative sum equals the total sum - cumulative sum, this means that a single cut is possible to split the array. The calculation part is done in another method, and I simply call the method twice for row-wise check and then column-wise. 👉 My Solution: https://lnkd.in/gNYsk6iz 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 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

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gbtaBXYW 💡 My thought process: The algorithm first calculates a normalized shift value using k % n. Shifting a row by its total length, n, returns the original configuration. Therefore, any shift k is the same as k modulo n. Instead of rotating memory physically or creating a temporary matrix, the code uses a virtual mapping strategy. It goes through each cell (i, j) and finds the index of the element that would be in that position after the shift. For even rows (left shift), to find the original value that moves into position j after a left shift, the code looks at the index (j - shift). It manages negative results by wrapping around with an addition of n. For odd rows (right shift), it checks the index (j + shift) % n. The function has a fail-fast mechanism; it returns false right away if it finds the first difference between a cell and its shifted counterpart. 👉 My Solution: https://lnkd.in/g5myRYRj 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 32/50 – LeetCode Challenge 🧩 Problem #416 – Partition Equal Subset Sum Today’s problem focused on determining whether an array can be divided into two subsets with equal sum, which is a classic Dynamic Programming (Subset Sum) problem. 📌 Problem Summary: Given an array of integers, check if it can be partitioned into two subsets such that their sums are equal. 🔍 Approach Used ✔ First calculated the total sum of the array ✔ If the sum is odd → not possible to divide equally ✔ Converted the problem into: Can we find a subset with sum = total / 2 ? ✔ Used a set (DP approach) to store all possible sums ✔ Iteratively built new sums by adding each number ✔ Checked if the target sum exists ⏱ Time Complexity: O(n × target) 📦 Space Complexity: O(target) 💡 Key Learning ✔ Converting problems into Subset Sum pattern ✔ Using Dynamic Programming with sets ✔ Thinking in terms of possibility instead of brute force ✔ Recognizing important DP patterns This problem helped me understand how to simplify complex problems into familiar patterns. Consistency builds strong problem-solving intuition 🚀 🔗 Problem Link: https://lnkd.in/gCRhSyaz #50DaysOfLeetCode #LeetCode #DSA #DynamicProgramming #SubsetSum #ProblemSolving #CodingJourney #FutureAIEngineer #Consistency

Now DP is new the stage 😁 Solved the classic problem Climbing Stairs using Dynamic Programming. The problem is simple: Given n steps, you can climb either 1 or 2 steps at a time. Find the total number of distinct ways to reach the top. At first, I thought of solving it using recursion, but it leads to repeated calculations and becomes inefficient. Then I moved to memoization (top-down DP) to store already computed results and avoid recomputation. Finally, I implemented a bottom-up approach, which builds the solution iteratively. Key Idea: To reach step i: • You can come from (i-1) • Or from (i-2) So, ways[i] = ways[i-1] + ways[i-2] This makes the problem similar to the Fibonacci sequence. Time Complexity: O(N) Space Complexity: O(N) What I liked about this problem is how it clearly shows the transition from a simple recursive solution to an optimized dynamic programming approach. It’s a great example of understanding overlapping subproblems and optimizing step by step. #DSA #DynamicProgramming #Algorithms #Coding #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

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 36 of My DSA Journey Today I solved LeetCode 150 – Evaluate Reverse Polish Notation (RPN) on LeetCode. 📌 Problem Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are: +, -, *, / Each operand may be an integer. Example: Input: ["2","1","+","3","*"] Output: 9 Explanation: (2 + 1) * 3 = 9 🧠 Approach – Stack This problem is a perfect application of the Stack data structure. Steps I followed: • Traverse the tokens array. • If the element is a number → push it onto the stack. • If it is an operator: Pop the top two elements (a and b) Perform the operation (a operator b) Push the result back into the stack • At the end, the stack contains the final result. ⏱ Time Complexity: O(n) — We process each token once 📦 Space Complexity: O(n) — Stack to store operands 💡 Key Learnings ✔ Understanding Reverse Polish Notation (Postfix Expression) ✔ Applying stack for expression evaluation ✔ Handling operator precedence implicitly using stack This problem connects DSA with compiler/interpreter concepts, which makes it even more interesting 🚀 Consistency continues — improving every day 💪 #100DaysOfCode #DSA #Stack #LeetCode #Java #ProblemSolving #CodingJourney #DeveloperJourney #Programming #SoftwareEngineering

Copy vs Move semantics in C++ — small change, big performance impact. At first glance, both can look like a simple assignment. But under the hood, they behave very differently. Copy → duplicates data → allocates new memory → keeps the source unchanged Move → transfers ownership of resources → avoids unnecessary allocations → leaves the source in a valid but unspecified state 💡 This difference becomes especially important with resource-heavy types such as std::string, std::vector, or user-defined classes managing dynamic memory. Another key point: std::move does not move anything by itself. It only casts an object to an rvalue, enabling move semantics if the type supports them. I put together this visual to make the difference easier to understand at a glance. Curious how others approach this in real code. #cpp #cplusplus #moderncpp #performance #softwareengineering #programming #coding