LeetCode Challenge: Precalculating Grid Sums for Efficient Cutting

🚀 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/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

🚀 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/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/gYK_vaVH 💡 My thought process: We create a result string of size n + m - 1 filled with placeholders. This size is chosen because each index in str1 corresponds to a substring of length m in the final string. First, we handle all T positions. Wherever str1[i] equals T, we place str2 directly starting at index i in the result. We do this first because these are fixed requirements and cannot be changed later. If we wait, we might create conflicts that are hard to fix. Next, we fill the remaining positions marked with $. For each of these positions, we try characters from a to z and choose the smallest one that does not break any F condition. This keeps the answer lexicographically smallest. To check validity quickly, we use checkLocal. Instead of checking the entire string each time, it only checks the substrings that could be affected by the current position. The idea is simple: changing one position can only impact substrings that include that position, so checking beyond that is unnecessary. If no character can be placed at any position without violating an F condition, we return an empty string since construction is impossible. Finally, we do a full validation. For every index: If it is T, the substring must match str2.   If it is F, the substring must not match str2. This final check ensures that all constraints are satisfied correctly. 👉 My Solution: https://lnkd.in/guMdgMBj 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 27/50 – LeetCode Challenge 🧩 Problem: Count and Say Today’s problem focused on generating a sequence based on reading and describing the previous term, which is a great exercise in string manipulation and pattern building. 📌 Problem Summary: The “Count and Say” sequence is defined as: Start with "1" Each next term is formed by reading the previous term aloud Example: 1 → "one 1" → 11 11 → "two 1s" → 21 21 → "one 2, one 1" → 1211 1211 → "one 1, one 2, two 1s" → 111221 🔍 Approach Used ✔ Started from the base case "1" ✔ Iterated to build each next term ✔ Counted consecutive digits ✔ Formed the new string using: count + digit ✔ Repeated until reaching the required n ⏱ Time Complexity: O(n × m) 📦 Space Complexity: O(m) 💡 Key Learning ✔ Understanding pattern generation problems ✔ Working with string grouping and counting ✔ Building results step by step ✔ Improving attention to detail in iteration This problem shows how simple rules can create complex patterns. Consistency is the key to mastery 🚀 🔗 Problem Link: https://lnkd.in/g_cMVP5a #50DaysOfLeetCode #LeetCode #DSA #Strings #ProblemSolving #CodingJourney #FutureAIEngineer #Consistency

🚀 Day 34 | 100 Days of Coding Challenge #DrGViswanathanChallenge 📘 Problem Solved: Min Stack (LeetCode) 🔧 Approach Used (Stack with Pair): • Stored elements as (value, current minimum) in the stack • On each push, updated the minimum using previous minimum • Allowed retrieving minimum in O(1) without extra traversal 📌 Key Idea: Track the minimum at every step so it’s always available at the top. ⏳ Complexity: Push: O(1) Pop: O(1) Top: O(1) GetMin: O(1) Space: O(n) 🧠 Key Learning: Augmenting data structures with extra information can optimize operations significantly. 📂 Topics Covered: Stack, Design, Optimization #100DaysOfCode #DSA #CPP #LeetCode #CodingJourney #CompetitiveProgramming

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/eC4nGYRx 💡 My thought process: The code checks if two strings can be made equal by swapping characters only between indices that have the same parity. This means characters at even indices can only move among even positions, while those at odd indices can only move among odd positions. To manage this, the code first goes through the first string and records the frequency of characters separately for even and odd indices using two unordered maps. This captures how many times each character appears in each parity group. Next, it goes through the second string. For each character, it checks if it exists in the corresponding parity map. If the current index is even, it checks the even map; if odd, it checks the odd map. When it finds a match, it decreases the frequency and removes the entry from the map if it reaches zero. If a character isn’t found in the required parity map, the function returns false right away. This approach is necessary because checking overall character frequency isn’t enough due to the swapping restrictions. Since characters can’t move between even and odd indices, both parity groups must have identical frequency distributions in both strings. Using separate maps ensures that this rule is enforced. If all characters from the second string match with the corresponding parity groups from the first string, the function returns true, confirming that the transformation is possible. 👉 My Solution: https://lnkd.in/erhV7_HJ 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 15/100 — Coding Challenge Solved Search in Linked List The task is straightforward: 1) Traverse the linked list 2) Check each node’s value 3) Return true if found, else false 💡 Approach: 1) Start from head 2) Move node by node using next 3) Stop when element is found or list ends 🧠 Time Complexity: O(n) 💾 Space Complexity: O(1) 💡 What I realized: Unlike arrays, linked lists don’t allow direct access. You have to traverse sequentially, which makes understanding traversal very important. Focusing on strengthening fundamentals. #LeetCode #DSA #100DaysOfCode #Cpp #LinkedList #CodingJourney