How to Use Templates for Efficient Coding in MQL5

🚀 Day 83 | Trees & Structural Matching Today’s problem focused on verifying if one binary tree is a subtree of another — a great blend of recursion, structure comparison, and logic. 🧩 Problem Solved: 572. Subtree of Another Tree • Approach: Used recursion — for each node in the main tree, checked if the subtree starting at that node matched the given subtree using a helper comparison function. • Insight: Tree problems teach the art of recursion — breaking complex hierarchies into repeatable subproblems. ✨ Key Takeaway: Recursion builds clarity — by trusting smaller calls to do their part, big problems become manageable. 📚 Topics: Binary Tree · Recursion · DFS · Comparison Logic 💻 Platform: LeetCode #DSA #LeetCode #ProblemSolving #DailyCoding #BinaryTree #Recursion #DFS #Consistency

🚀 Day 60 of #100DaysOfLeetCodeHard — LeetCode 3725: Count Ways to Choose Coprime Integers from Rows (Hard) My Submission:https://lnkd.in/gFY6Yg_f Today’s problem was a Dynamic Programming + GCD challenge — a great exercise in combining mathematical reasoning with state-based recursion. The task was to count the number of ways to select exactly one integer from each row of a given matrix such that the GCD of all chosen integers equals 1, with results taken modulo 109+710^9 + 7109+7. 💡 Approach: I used recursive DP with memoization where each state represents: ind: current row index gcdNum: current running GCD value At each step, we iterate through all numbers in the current row and recursively compute the number of valid combinations leading to a final GCD of 1. A 2D DP array dp[ind][gcdNum] stores intermediate results to avoid recomputation. 📘 Key Concepts Used: Dynamic Programming over GCD states Modulo arithmetic for large result handling Efficient bottom-up recursion with pruning ⏱️ Time Complexity: O(n × m × log(maxValue)) 💾 Space Complexity: O(n × maxValue) This problem beautifully highlights how mathematical insight (GCD) and DP optimization come together to handle complex combinatorial counting problems. 💪 #LeetCode #DynamicProgramming #Math #GCD #ProblemSolving #C++ #100DaysOfCode #LearningEveryday #CodingChallenge

After solving 500+ LeetCode questions, I concluded: · Linked List · If you can draw the solution with paper and pen, you can code it. · Trees · If you have a command of recursion, you can solve it. The only things you need are a base case and a recurrence relation. · Sliding Window · When you get the feeling of comparing two strings at each step, go with it. You only need a hashmap and a comparison function. · Binary Search · If you know anything is sorted, or you get the feeling of eliminating half of the input set in a divide-and-conquer manner, go ahead with it. · Stack · When you need to track elements for a later comparison or reverse the order of processing, use this data structure. #leetcode #dsa #datastructure #algorithms

🚀 Problem Solving Update Today’s problem-solving session was all about optimizing thought processes and tackling both classic and hard-level challenges 💪 🟩 GFG POTD – Get Minimum Squares Approach: Used Dynamic Programming to compute the minimum number of perfect squares that sum up to a given integer n. Built a DP array where dp[i] represents the minimum number of squares required to form i. For every i, iterated over all possible squares j*j ≤ i and updated dp[i] = min(dp[i], dp[i - j*j] + 1). Final answer is dp[n]. A neat and efficient bottom-up approach to a fundamental DP problem! ⚙️ 🟥 LeetCode Daily Challenge – Find X-Sum of All K-Long Subarrays II (Hard) Approach: Implemented an advanced sliding window combined with frequency tracking using two multisets — one maintaining the top x most significant (frequency × value) elements, and another storing the rest. Used custom comparators to dynamically manage transitions between the sets. Maintained a running sum of the top elements (sumTop) to efficiently update answers as the window slides. Smart rebalancing ensured that insertions/removals kept the sets valid for every window. This one was quite intense 🔥 — required careful balancing of logic, data structures, and optimization to achieve efficiency. #ProblemSolving #LeetCode #GeeksforGeeks

Ever wondered how a tiny ++ can turn a perfectly good program into a mystery? I’ve been digging into some of the most notorious ways the increment operator can cause undefined behavior (UB). Below are seven classic (and definitely not‑to‑be‑copied) examples that illustrate why you should always respect sequence points and operator precedence. 1. i+++++i – Multiple increments without an intervening sequence point. The compiler can evaluate the left‑hand and right‑hand sides in any order, leading to unpredictable results. 2. i = i++ – The assignment and the post‑increment are unsequenced; the final value of i is undefined. 3. i = ++i + i – Mixing pre‑increment with other reads of the same variable creates a race condition in the abstract machine. 4. i = i = i + 1 – Chained assignments with side effects are a recipe for UB. 5. *i = i**++* – The ** is not a valid operator; even if it were, the combination of dereference and increment without a sequence point would be illegal. 6. std::cout << i+++++i; – The stream insertion operator evaluates its operands in an unspecified order, so the increments may happen before or after the output. 7. i = ++i + ++i; – Two pre‑increments on the same variable in a single expression are unsequenced, giving different results on different compilers. > These snippets are pure examples of bad practice. Do NOT use them in production code. Why does UB matter?Undefined behavior isn’t just a theoretical concern—it can cause crashes, silent data corruption, or security vulnerabilities that are incredibly hard to track down. Modern compilers often exploit UB for optimization, so what looks “harmless” today may break spectacularly tomorrow. I’d love to hear from youHave you ever stumbled upon a tricky UB case in C/C++? How did you debug it, and what lesson did you take away? Share your stories in the comments—let’s help each other write safer, more predictable code! #Cpp #UndefinedBehavior #CodeQuality #SoftwareEngineering #LearningFromMistakes

Hi DataSceintist, Today I programmed a 95% CI for the labda for my modern factoranalysis. Simply by a combination of bootstrapping and procrustes analysis. Mostly you will use a pd.dataframe as input. Do not use de d.sample method of Pandas , but the random.choice method of Numpy. Which mean you have to alter a dataframe d into a 2D array by d.values So, you first calculate true labda L. Secondly you sample the 2D array with replacement using np.random.choice and then compute again and again labda matrices. Repeat this cycle 100 times and specify the 2.5 and 97.5 percentiles. Store the output in dataframe with [True_value[2,5 , 97.5]] cell -values Ready! Happy coding

🔥 Day 222 - Daily DSA Challenge! 🔥 Problem: 🌳 Flatten Binary Tree to Linked List Given the root of a binary tree, flatten it into a linked list in-place, following the preorder traversal order (root → left → right). 💡 Key Insights: 🔹 The goal is to modify the tree without using extra space. 🔹 For each node, if a left subtree exists: 1️⃣ Find the rightmost node of the left subtree. 2️⃣ Connect that node’s right pointer to the current node’s right subtree. 3️⃣ Move the left subtree to the right and set left = null. 🔹 Repeat this process while traversing through the tree. ⚡ Optimized Plan: ✅ Traverse using a pointer (curr). ✅ If curr.left exists, attach it in place of curr.right and re-link the subtrees. ✅ Continue the process until the end of the tree. ✅ Time Complexity: O(n) — each node visited once. ✅ Space Complexity: O(1) — in-place transformation. 💬 Challenge for you: 1️⃣ Can you flatten the tree using recursion instead of iteration? 2️⃣ How would you modify this to produce a postorder or inorder flattened list? #DSA #BinaryTree #LinkedList #LeetCode #ProblemSolving #CodingChallenge #KeepCoding #100DaysOfCode

Day 65 of My DSA Challenge Problem: Find all unique permutations of a given string. Example: Input: "AAB" Output: ["AAB", "ABA", "BAA"] Approach Used: → This is a backtracking problem where we recursively build all possible strings by choosing one unused character at a time. → To avoid duplicate permutations (like for "AAB"), I used a HashSet to ensure only unique strings are added to the final result. → A boolean vis[] array keeps track of which characters are already included in the current recursive path. Key Concept: Backtracking explores all possibilities, then “undoes” each choice to explore others. Using a HashSet avoids expensive O(n) “contains” checks on an ArrayList. Time Complexity: O(n!) Space Complexity: O(n!) (for storing permutations) Pattern: Backtracking + HashSet for Uniqueness #Day65 #DSAChallenge #Backtracking #StringPermutation #Recursion #JavaCoding #ProblemSolving #DataStructures #Algorithms #LearnDSA #GeeksforGeeks #CodeOptimization #CodingJourney #TechGrowth #160DaysOfCode #DSAWithJava #PermutationProblem

🌟𝐒𝐭𝐫𝐞𝐚𝐤 𝐃𝐚𝐲 647 💻 𝐍𝐨𝐯𝐞𝐦𝐛𝐞𝐫 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐢𝐧𝐠 𝐂𝐡𝐚𝐥𝐥𝐞𝐧𝐠𝐞 - 𝐃𝐚𝐲 08 🧠 Problem of the Day: 1611. Minimum One Bit Operations to Make Integers Zero (https://lnkd.in/gKBqgaYV) 🔍 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: - This problem computes the minimum number of operations to reduce an integer n to 0, where each operation flips one bit following specific rules. - The insight is that this transformation follows a Gray code pattern — every step flips one bit from the previous number. - The algorithm works by successively XORing n with its right-shifted self (mask >>= 1; n ^= mask), which effectively converts n from binary to Gray code representation (or reverses it). - Once all bits are processed, the resulting n equals the minimum number of required operations. 📚 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐋𝐢𝐧𝐤: https://lnkd.in/gk5ydAJF 📈 𝐃𝐞𝐭𝐚𝐢𝐥𝐞𝐝 𝐃𝐫𝐲-𝐑𝐮𝐧: https://lnkd.in/gh7iii7C #leetcode #leetcodechallenge #dsa #leetcodestreak #potd #JobSeeker #DataStructures #Algorithms

🧩 LeetCode Challenge – Day 88 ✅ Focused on a binary string problem today — simple structure, elegant counting logic. 🔗 LeetCode 1513 – Number of Substrings With Only 1s The challenge is to count all substrings made entirely of 1s. Instead of checking each substring individually, the key is recognizing that a continuous block of length k contributes k × (k + 1) / 2 substrings. This transforms a brute-force approach into a clean, mathematical solution. 💡 Key Takeaways: • Pattern recognition turns linear scanning into efficient counting. • Math-based formulas reduce complexity and eliminate unnecessary loops. • Binary string problems highlight how structure can be leveraged for quick solutions. #Day88 #LeetCodeChallenge #100DaysOfCode #DSA #CodingJourney #ProblemSolving #Strings #Math #Counting #Optimization