LeetCode Challenge: Two-Pointer Approach for Submatrix Reversal

🚀 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

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gqTik3da 💡 My thought process: First, save all indices of each number in a map. The key is the number, and the value is a sorted list of its positions in the array. Then, for every index i, it computes the reverse of nums[i]. While reversing, it skips leading zeros. This means that a number like 120 becomes 21 instead of 021. After getting the reversed number, check the map for all indices where this reversed value exists. Using upper_bound, find the first index that is strictly greater than i. If such an index exists, calculate the distance between the two indices and update the minimum distance. Finally, if no such pair is found, it returns -1. Otherwise, it returns the minimum distance found. 👉 My Solution: https://lnkd.in/gpNu3WGh 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/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/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

🚀 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

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/g9SGp2Qa 💡 My thought process: First, create a mapping from each value to a sorted list of its indices using a map<int, vector<int>>. This allows for easy lookup of all positions where a value occurs. For each query index idx, it retrieves the corresponding value num. If that value appears only once, the result is -1 because no valid pair exists. In case of multiple occurrences, we use binary search on the index list of num: * upper_bound finds the next occurrence that is strictly greater than idx. * lower_bound locates the current position and checks the previous occurrence. It calculates distances in both directions: * Forward distance: either direct (next - idx) or circular wrap (n - (idx - first)). * Backward distance: either direct (idx - prev) or circular wrap (n - (last - idx)). The minimum of these distances is stored as the answer. 👉 My Solution: https://lnkd.in/gjmKVa3y 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 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->26 🚀 LeetCode 1002 — Find Common Characters | Frequency Optimization 🔥 Solved this problem using a frequency array approach and achieved 100% runtime (0 ms) ⚡ 🔍 The goal is to find characters that appear in all strings, including duplicates. 💡 Approach: Use a frequency array of size 26 Initialize with a large value (INT_MAX) For each word: Count character frequency Update global frequency using minimum values 🧠 Key Insight: We take the minimum frequency because a character must exist in every string. ⚡ Complexity: Time → O(n × k) Space → O(1) This problem helped me understand how frequency comparison can efficiently find common elements 🚀 #DSA #LeetCode #Cpp #Coding #ProblemSolving #Arrays #LearningJourney

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gfx2E6jb 💡 My thought process: The rotate90 function rotates the matrix 90 degrees clockwise in place. It starts by transposing the matrix, swapping elements across the diagonal. This changes rows into columns. After that, it reverses each row to achieve the final rotated version. The check function compares two matrices element by element. It goes through all the positions and returns false immediately if it finds any mismatch. If all elements match, it returns true, showing that both matrices are the same. The findRotation function tests all possible rotations of the matrix: 0, 90, 180, and 270 degrees. For each rotation, it checks if the current matrix matches the target matrix. If a match is found at any point, it returns true. If not, it rotates the matrix by 90 degrees and continues. If no rotation matches, it returns false. Overall, the method is efficient. Each rotation and comparison takes quadratic time based on the size of the matrix, and it only performs a constant number of rotations. 👉 My Solution: https://lnkd.in/gcj4S5iQ 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 11 of 100 Days LeetCode Challenge Problem: Complement of Base 10 Integer Today’s problem is a clean example of bit manipulation fundamentals 🔥 💡 Key Insight: To find the complement: Convert the number to binary Flip all bits (0 → 1, 1 → 0) Convert it back to decimal 👉 But here’s the trick: We only flip significant bits (ignore leading zeros) 🔍 Optimized Approach: Create a bitmask with all bits set to 1 (same length as n) Then: complement = n XOR mask 👉 Example: n = 5 → (101) mask = 111 Result = 101 ⊕ 111 = 010 → 2 🔥 What I Learned Today: XOR is powerful for bit flipping operations Bitmasking simplifies binary problems Always consider binary representation constraints 📈 Challenge Progress: Day 11/100 ✅ Bit by bit improving! LeetCode, Bit Manipulation, XOR, Binary Representation, Bitmasking, DSA Practice, Coding Challenge, Problem Solving, Algorithms #100DaysOfCode #LeetCode #DSA #CodingChallenge #BitManipulation #Binary #XOR #ProblemSolving #TechJourney #ProgrammerLife #SoftwareDeveloper #CodingLife #LearnToCode #Developers #Consistency #GrowthMindset #InterviewPrep