LeetCode Daily Challenge: Shift Matrix Algorithm Explanation

🚀 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/gZ52pSZR 💡 My thought process: The solution operates in two separate passes: 1. Left-to-Right Pass: For each element at index "i", calculate the distance to all previous occurrences of that value. If a value has appeared "k" times before, the sum of distances is found using this formula: (count * current_index) - (sum of previous_indices). By storing the count and the running sum of indices in a hash map, we can compute this in constant time. 2. Right-to-Left Pass: The same logic is applied in reverse to calculate the distance from the current index to all future occurrences. In this case, the formula changes to: (sum of future_indices) - (count * current_index). By adding the results from both directions, the code captures the total absolute difference for every identical pair without the performance cost of a nested loop. 👉 My Solution: https://lnkd.in/gfEc7Kis If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering

🚀 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

🚀 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

Day 29 – LeetCode Journey 🚀 Solved Two Sum II – Input Array Is Sorted using the Two Pointer technique, leveraging the sorted nature of the array for an optimal solution. 🔹 Time Complexity: O(n) 🔹 Runtime: 2 ms (Beats 96.37%) 🔹 Memory Usage: 48.58 MB This problem is a great reminder that recognizing patterns (like sorted arrays) can significantly reduce complexity and improve efficiency. Small optimizations, big impact 📈 Staying consistent and sharpening problem-solving skills every day. #LeetCode #DSA #ProblemSolving #CodingJourney #SoftwareEngineering #Consistency #Learning

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

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gdjyFXCw First, we note that swaps aren't random; they are only allowed between specific index pairs. If index `a` can swap with `b`, and `b` can swap with `c`, then `a`, `b`, and `c` become connected. We can rearrange values freely among them. This means we need to identify groups of indices where swapping is allowed within the group. To find these groups efficiently, we use Disjoint Set Union (DSU). Each index starts in its own set. For every allowed swap pair `[x, y]`, we combine their sets. After processing all the swaps, indices that share the same parent belong to the same connected component or group. Next, we organize all indices by their DSU parent. Each group shows a set of positions where we can rearrange values without restrictions. For each group, we aim to match values from `source` to `target` as closely as possible: * First, we count how often each value appears in `source` for that group. * Then, we go through the same indices in `target`. For each value:  * If the value exists in our frequency map (meaning we have it in `source`), we use it and decrease its count.  * If it doesn't exist, it means we can't match this value in the group, contributing to the Hamming distance. By doing this for all groups, we maximize matches within each connected component, which minimizes the overall Hamming distance. 👉 My Solution: https://lnkd.in/gfx3iMut 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

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gExeYEjW 💡 My thought process: The distance formula simplifies to 2 × (k − i) when indices are sorted in order (i < j < k). This means the problem is about finding three occurrences of the same value while minimizing the gap between the first and third indices. Instead of saving all indices for each value, the approach only keeps track of the last two occurrences of every number using a hash map. For each element, the map holds a pair {prev2, prev1}, where prev1 is the most recent index and prev2 is the one before that. While going through the array for the current index i: If the element has been seen before, get its last two indices.   If a valid prev2 exists, calculate the distance using 2 × (i − prev2).  This forms a triplet using prev2, prev1, and the current index. Update the answer with the minimum value obtained.   Shift the indices forward by setting prev2 to prev1 and prev1 to i. If the element appears for the first time, set its entry to {-1, i}, indicating that only one occurrence has been seen so far. This method ensures that only consecutive triplets of occurrences are considered. This is enough because any non-consecutive triplet would result in a larger distance. The algorithm runs in O(n) time and O(n) space, while avoiding the extra cost of storing full index lists. 👉 My Solution: https://lnkd.in/gBFCq7QW 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

I spent 30 minutes overthinking a problem that had a 2-minute solution. Problem: Construct Uniform Parity Array I At first, it feels like a construction + constraints problem. You start thinking about cases, patterns, edge conditions… But then I stepped back and looked at the operations: nums2[i] = nums1[i] nums2[i] = nums1[i] - nums1[j] (j ≠ i) Now ask a better question: 👉 What happens to parity under these operations? Same value => parity unchanged Difference of two numbers => can be even or odd depending on choice That’s the key. You can control parity freely. Just to make the entire array all even or all odd And that’s always possible. So the answer would be always TRUE . #FirstPrinciples #ProblemSolving #CodingJourney #LeetCode #DSA #Algorithms #SoftwareEngineering #Developers #CodeNewbie #LearnToCode #ProgrammingLife #TechThinking #LogicalThinking #BuildInPublic #DeveloperMindset #CleanThinking #CodingLife #TechCareers #GrowthMindset #ThinkDifferent