Final Value of Variable After Operations - LeetCode Challenge

🚀LeetCode Daily Challenge 🧩 Problem: Make Array Elements Equal to Zero 🔍 Problem Intuition: You’re given an integer array where you can make selections at positions containing 0. Each selection tries to balance the sum of elements to the left and right of that zero. Your task is to count how many such valid selections can make both sides equal or nearly equal (difference of 0 or 1) — leading the entire array toward equilibrium (eventually all zeros). 💡Example Insight: Imagine scanning through each 0 in the array — If the sum of elements on both sides is the same → that position contributes 2 valid ways. If the sums differ by 1 → it contributes 1 valid way. You repeat this for all zeros and count the total valid selections. 🧠 Approach & Strategy: ✔ Traverse through the array and check positions where the element is 0. ✔ For each zero, compute: sum1 = sum of elements to the left (including that zero). sum2 = sum of elements to the right (including that zero). ✔ If |sum1 - sum2| == 0, add 2 to the count. ✔ If |sum1 - sum2| == 1, add 1 to the count. ✔ Return the total count of valid selections. ⚙️ Complexity Analysis: Time Complexity: O(n²) → For each zero, we calculate left and right sums. Space Complexity: O(1) → Only integer variables used. 🔗 Problem: https://lnkd.in/dTKS6N-F 💻 Solution: https://lnkd.in/djxzBeMF #LeetCode #DSA #Cpp #ProblemSolving #CodingChallenge #LeetCodeDailyChallenge

🚀LeetCode Daily Challenge 🧩 Problem: Check If All 1's Are at Least Length K Places Away Problem Intuition: You are given a binary array nums and an integer k. Your task is to check whether every pair of consecutive 1s in the array are at least k zeros apart. In other words, for any two indices i and j where nums[i] = nums[j] = 1, we must ensure: j - i - 1 ≥ k This guarantees enough spacing between two ’1’s. If even one pair violates this, the answer is false. Example Insight: Let’s consider: nums = [1, 0, 0, 1, 0, 1], k = 2 Positions of 1s: 0, 3, 5 Distance: Between index 0 & 3 → 3 - 0 - 1 = 2 ✔️ Between index 3 & 5 → 5 - 3 - 1 = 1 ❌ (violates the condition) Thus, the array is not valid. Approach & Strategy: ✔ First, find the index of the first occurrence of 1. ✔ Then iterate through the array: • Whenever you encounter a new 1, check the distance from the previous 1. • If the gap is less than k, return false. • Otherwise, update the last index of 1 and continue. This ensures a clean and efficient linear scan. 📈 Complexity Analysis: Time Complexity: O(n) — single pass through the array Space Complexity: O(1) — no extra space used 🔗 Problem: https://lnkd.in/dP5T3BAH  💻 Solution: https://lnkd.in/dVYvDktq #LeetCode #DSA #Cpp #Arrays #ProblemSolving #CodingChallenge #LeetCodeDailyChallenge

🚀LeetCode Daily Challenge 🧩 Problem: Smallest Number With All Set Bits Problem Intuition: Given an integer n, your task is to find the smallest number that has the same number of bits as n when represented in binary, but with all bits set to 1. In simple terms — if n requires k bits to represent, you must find the smallest integer whose binary form has exactly k ones (111...1). 💡Example Insight: Let’s say n = 6 → Binary: 110 → requires 3 bits. The smallest number with all 3 bits set to 1 is 111 → which is 7 in decimal. Hence, the answer is 7. 🧠 Approach & Strategy: ✔ First, determine the number of bits needed to represent n: res = floor(log2(n)) + 1. ✔ Then, iterate from n to an upper limit and check for the first number having exactly res set bits. ✔ The check is performed using the built-in function __builtin_popcount(x) which counts the number of 1’s in the binary representation. ✔ Return the first number satisfying this condition. ⚙️ Complexity Analysis: Time Complexity: O(k) → Iterates until the matching number is found (within small bounds). Space Complexity: O(1) → Uses only integer variables. 🔗Problem: https://lnkd.in/dwR7Gp8i 💻 Solution: https://lnkd.in/dJU6Wrra #LeetCode #DSA #Cpp #ProblemSolving #CodingChallenge #LeetCodeDailyChallenge

🚀LeetCode Daily Challenge 🧩 Problem: Count Operations to Obtain Zero Problem Intuition: You are given two non-negative integers, num1 and num2. In one operation, you subtract the smaller number from the larger one. Your goal is to count how many operations are required until either of them becomes zero. Essentially, this problem simulates a repeated subtraction process, similar to how the Euclidean algorithm works for finding the GCD — but instead of stopping at the GCD, we stop when one number becomes 0 and count every subtraction step. Example Insight: Let’s take an example: num1 = 5, num2 = 2 Step 1: 5 > 2 → num1 = 5 - 2 = 3 Step 2: 3 > 2 → num1 = 3 - 2 = 1 Step 3: 2 > 1 → num2 = 2 - 1 = 1 Step 4: 1 == 1 → final subtraction makes one of them zero. Total operations = 4 Approach & Strategy: ✔ Initialize a counter ans = 0. ✔ Keep subtracting the smaller number from the larger one until they are equal. ✔ Increment the counter after each operation. ✔ Once both become equal, one final operation makes the result zero. Key Idea: This iterative subtraction continues until one of the numbers becomes zero, and each subtraction step contributes to the total operation count. 📈 Complexity Analysis: Time Complexity: O(max(num1, num2)) — since we reduce the larger number step by step. Space Complexity: O(1) — constant extra space. 🔗 Problem: https://lnkd.in/dzn6_K9U 💻 Solution: https://lnkd.in/deg7Ku7e #LeetCode #DSA #Cpp #ProblemSolving #CodingChallenge #LeetCodeDailyChallenge

🚀LeetCode Daily Challenge 🧩 Problem: Check if Digits Are Equal in String After Operations I Problem Intuition: You are given a string s consisting of digits. You repeatedly perform the following operation until the string’s length becomes 2: Replace each pair of adjacent digits with their sum modulo 10. Your task is to check whether the last two remaining digits are equal after all operations. Example: Input: s = "3902" Output: true Explanation: Step 1: (3+9)%10=2, (9+0)%10=9, (0+2)%10=2 → "292" Step 2: (2+9)%10=1, (9+2)%10=1 → "11" Since both digits are equal, return true 🧠 Approach & Strategy: ✔ Keep reducing the string by replacing each pair with (s[i] + s[i+1]) % 10. ✔ Continue the process until the string size becomes 2. ✔ Finally, compare the last two digits — if equal, return true; otherwise, false. ✔ This approach simulates the operation directly and efficiently. ⚙️ Complexity Analysis: Time Complexity: O(n²) — each operation reduces the string by one. Space Complexity: O(n) — for constructing new strings in each iteration. 🔗 Problem: https://lnkd.in/d8CHrNeQ 💻 Solution: https://lnkd.in/dkierwNS #LeetCode #DSA #ProblemSolving #Cpp #CodingChallenge #LeetCodeDailyChallenge #StringManipulation 

💡 Code Clarity Tip: The Power of Single-Argument Methods I've been thinking about what truly makes code clean, maintainable, and easy to test. For me, one critical principle stands out: strive for methods/functions that accept at most one argument. The moment a method starts taking two, three, or even more arguments, you introduce several problems: * Increased Cognitive Load: Each argument is a dependency the developer has to track and remember the order of. * Order Dependency: Swapping two arguments of the same type (like two strings or two integers) can silently break the code without a compiler error. * Tougher Testing: You exponentially increase the number of combinations and edge cases you need to test. The Solution? Introduce a Parameter Object. Instead of this: public Order processOrder(String customerId, String productId, int quantity, double discountRate) Refactor to this: public Order processOrder(OrderProcessingDetails details) // 'details' is a dedicated class/struct containing the four parameters. What are the benefits of the Parameter Object? * Self-Describing: The code is immediately clearer. The details object explicitly tells you what context is required. * Encapsulation: You can enforce validation and consistency inside the OrderProcessingDetails object. * Future-Proof: Need to add a shippingAddress? You just modify the parameter object; the method signature remains the same! Simple rule: If you find yourself needing more than one argument, it's a strong signal to introduce a well-named class or struct to encapsulate that data. Your future self (and your teammates) will thank you. What's your take on the Single-Argument Rule? Do you strictly follow it, or do you have a practical limit? #CleanCode #SoftwareEngineering #ProgrammingTips #CodeRefactoring #SoftwareArchitecture

🚀 LeetCode Daily Challenge Complete! Just solved "Final Value of Variable After Performing Operations" - a simple yet elegant problem that teaches the value of finding shortcuts in pattern matching! 💡 Solution Approach: ✅ Iterate through all operation strings ✅ Check middle character (index 1) ✅ If '-': decrement, if '+': increment ✅ Return final value The key insight: Instead of comparing full strings ("--X", "X--", "++X", "X++"), we can observe that ALL operations have the operator in the middle position! By checking just op[1], we instantly know whether to increment or decrement. Pattern recognition: "--X" and "X--" both have '-' at index 1 → decrement "++X" and "X++" both have '+' at index 1 → increment This single-character check is more efficient than string comparisons! Time Complexity: O(n) | Space Complexity: O(1) #LeetCode #StringParsing #CPlusPlus #Algorithms #SoftwareEngineering #ProblemSolving #CleanCode #PatternRecognition

🚀LeetCode Daily Challenge 🧩 Problem: Delete Nodes From Linked List Present in Array Problem Intuition: Given a linked list and an array of integers, you need to remove all nodes from the linked list whose values are present in the given array. In simple terms, delete every node that matches any number in the array and return the updated linked list. 💡Example Insight: Let’s say: nums = [1, 2, 3] Linked List = 1 → 2 → 4 → 3 → 5 After removing nodes with values 1, 2, and 3, the linked list becomes: 4 → 5 🧠 Approach & Strategy: ✔ Store all numbers from the array nums into an unordered_set for quick lookup (O(1) average). ✔ Traverse the linked list, and for each node: Skip it if its value is present in the set. Otherwise, add it to the new linked list. ✔ Finally, return the head of this new linked list. ⚙️ Complexity Analysis: Time Complexity: O(n + m) → n = length of linked list, m = size of array. Space Complexity: O(m) → For storing array elements in a set. 🔗 Problem: https://lnkd.in/dWA2hsDr 💻 Solution: https://lnkd.in/dypBvguK #LeetCode #DSA #Cpp #ProblemSolving #CodingChallenge #LeetCodeDailyChallenge

🔹 Day 67 of #100DaysOfLeetCodeChallenge 🔹 Problem: Combination Sum Focus: Backtracking + Recursion 💡 The Challenge: Find all unique combinations of numbers that sum to a target. The twist? You can reuse the same number unlimited times! This makes it more interesting than standard subset problems. 🧠 My Approach: Used backtracking with two key decisions at each step: Pick: Include current element and explore further (stay at same index for reuse) Skip: Move to next element without including current one Base case: When we've explored all elements, check if target reached 0 Optimization: Only pick if element ≤ remaining target 📊 Complexity Analysis: ⏳ Time: O(2^t) where t = target/min(candidates) — worst case recursion tree 💾 Space: O(t) — maximum recursion depth 📌 Example: Input: candidates = [2,3,6,7], target = 7 Output: [[2,2,3],[7]] Explanation: - 2+2+3 = 7 (reused 2 twice!) - 7 = 7 (single element) 🎯 Key Takeaway: The beauty of this problem lies in handling unlimited reuse. By staying at the same index when picking an element, we elegantly allow repetition without creating duplicates. This pick/skip pattern is fundamental to combinatorial backtracking! Real-world analogy: Like making change for a dollar — you can use the same coin denomination multiple times! 💰 Day 67/100 complete. Mastering backtracking, one problem at a time! 💪 #LeetCode #Backtracking #DynamicProgramming #DSA #Algorithms #CodingInterview #100DaysOfCode #SoftwareEngineering #ProblemSolving #TechCareer

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