LeetCode Problem 22: Generate Parentheses with Backtracking

Solved “Letter Combinations of a Phone Number” on LeetCode 💡 What looks simple on the surface is actually a clean exercise in recursion + backtracking. 🔍 Key takeaways: - Mapping digits to characters is straightforward, but generating all combinations efficiently is where the thinking comes in - Backtracking helps explore all possibilities while keeping the solution clean and scalable - Important to control recursion depth and maintain state (temp) correctly 💭 What I focused on: - Building combinations incrementally - Reverting state after each recursive call (classic backtracking pattern) - Keeping the solution readable and modular This problem reinforced how powerful recursion is when combined with proper state management. On to the next one 🚀 #DSA #LeetCode #Backtracking #Recursion #ProblemSolving #CodingJourney

🚀 Day 80 of LeetCode Problem Solving Journey — 100 Days LeetCode Challenge Today, I solved LeetCode #3467 — Transform Array by Parity using C++, under the guidance of Trainer NEKAL SINGH SALARIA at REGex Software Services. 🔍 Problem Summary: Given an integer array nums, perform the following operations: Replace each even number with 0 Replace each odd number with 1 Sort the modified array in non-decreasing order Return the resulting array 🧠 Approach Used (Transformation + Sorting): I transformed the array and then sorted it: Traverse the array If element is even → replace with 0 If element is odd → replace with 1 Sort the array to arrange values in non-decreasing order This approach directly follows the problem steps. 📚 Key Learnings of the Day ✔ Array transformation simplifies problem requirements ✔ Modulo operation helps identify even/odd numbers ✔ Sorting ensures correct final arrangement ✔ Breaking problems into steps improves clarity ⏱ Complexity • Time Complexity: O(n log n) • Space Complexity: O(1) 💡 Optimization Insight: Instead of sorting, we can count number of 0s and 1s and rebuild the array in O(n) time, avoiding sorting overhead. #Day80 #100DaysLeetCodeChallenge #LeetCode #RegexSoftwareServices #NekalSingh #ProblemSolving #DSA #CPlusPlus #CodingChallenge #ProgrammingJourney #Arrays #KeepGrowing

🚀 Day 10/100 — LeetCode Challenge Solved 3Sum The brute force approach is simple: Try all triplets → O(n³) But that’s not scalable. 💡 Optimized Approach: 1) First, sort the array 2) Fix one element 3) Use two pointers to find the remaining pair 👉 This reduces the complexity significantly. Also handled duplicates carefully to avoid repeated triplets. 🧠 Time Complexity: O(n²) 💾 Space Complexity: O(1) (ignoring output) 💡 What I learned: Sometimes optimization is not about complex logic, but about: 1) Sorting 2) Using patterns like two pointers 3) Eliminating unnecessary work This problem felt like a step up from previous ones. Staying consistent. #LeetCode #DSA #100DaysOfCode #Cpp #TwoPointers #CodingJourney

🧠 Cracking LeetCode Became Easier Once I Understood Patterns Earlier, I used to approach every problem from scratch — which was slow and frustrating. Then I realized most problems follow common patterns: 🔁 Sliding Window – for subarrays/substrings 🐢🐇 Two Pointers – for sorted arrays & linked lists 🧩 Hashing – for fast lookups 🔍 Binary Search – for efficient searching 🔄 Backtracking – for combinations & recursion 💡 Key Insight: You don’t need to memorize solutions — just recognize the pattern. Once I started focusing on patterns, problem-solving became faster and more structured. Currently practicing consistently and improving my approach 🚀 #LeetCode #DSA #CodingInterview #SDET #ProblemSolving

🔥 Day 171 of My LeetCode Journey Problem 112: Path Sum 💡 Problem Insight: Today’s problem was about checking whether a binary tree has a root-to-leaf path whose sum equals a target value. The key detail here is root-to-leaf — not any path. Missing that leads to wrong answers. 🧠 Concept Highlight: The solution is a clean use of DFS (recursion): Subtract the current node’s value from the target Move to left and right subtrees When you reach a leaf, check if the remaining sum is zero This ensures you explore all valid paths without unnecessary work. 💪 Key Takeaway: Tree problems often reduce to accumulating state along a path. Instead of storing paths, update the condition as you traverse. ✨ Daily Reflection: This problem reinforced how recursion naturally fits tree traversal. Once you think in terms of paths and state, the solution becomes straightforward. #Day171 #LeetCode #BinaryTree #DFS #PathSum #ProblemSolving #DSA #CodingJourney #Consistency

🚀 𝐆𝐅𝐆 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝐒𝐨𝐥𝐯𝐞𝐝: 𝐂𝐨𝐮𝐧𝐭 𝐭𝐡𝐞 𝐑𝐞𝐯𝐞𝐫𝐬𝐚𝐥𝐬 Solved an interesting problem on stacks and string balancing that really tests logical thinking 🔍 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝐒𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭: Given a string consisting of only { and }, find the minimum number of reversals required to make the string balanced. 💡 𝐊𝐞𝐲 𝐈𝐧𝐬𝐢𝐠𝐡𝐭: A valid (balanced) string must have: • Equal number of opening { and closing } brackets. • Proper ordering. 👉 If the string length is odd, it can never be balanced ❌ ⚙️ 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡 𝐈 𝐔𝐬𝐞𝐝: 𝟏. 𝐂𝐡𝐞𝐜𝐤 𝐎𝐝𝐝 𝐋𝐞𝐧𝐠𝐭𝐡: • If n % 2 != 0 → return -1 𝟐. 𝐔𝐬𝐞 𝐒𝐭𝐚𝐜𝐤 𝐟𝐨𝐫 𝐏𝐫𝐨𝐜𝐞𝐬𝐬𝐢𝐧𝐠: • Traverse the string • Push { into stack • For }: a. If stack top is { → pop (valid pair) b. Else → push } 𝟑. 𝐇𝐚𝐧𝐝𝐥𝐞 𝐑𝐞𝐦𝐚𝐢𝐧𝐢𝐧𝐠 𝐔𝐧𝐛𝐚𝐥𝐚𝐧𝐜𝐞𝐝 𝐁𝐫𝐚𝐜𝐤𝐞𝐭𝐬: • After traversal, stack contains only invalid pairs 𝟒. 𝐂𝐨𝐮𝐧𝐭 𝐑𝐞𝐯𝐞𝐫𝐬𝐚𝐥𝐬: • Pop two elements at a time: a. If both same ({{ or }}) → 1 reversal b. If different (}{) → 2 reversals 🧠𝐖𝐡𝐲 𝐓𝐡𝐢𝐬 𝐖𝐨𝐫𝐤𝐬: The stack helps eliminate already balanced pairs. What remains are only invalid combinations, which we fix optimally by counting minimum reversals. 🧾𝐂𝐨𝐫𝐞 𝐋𝐨𝐠𝐢𝐜: if(a == b) ans += 1; else ans += 2; 📈 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: • Time Complexity: O(n) • Space Complexity: O(n) ✨ 𝐖𝐡𝐚𝐭 𝐈 𝐋𝐞𝐚𝐫𝐧𝐞𝐝: • Stack is very useful for bracket-related problems • Eliminating valid pairs simplifies the problem • Edge cases like odd length are very important 🔥 Small problems, big learning! #DSA #Stack #GeeksforGeeks #Coding #CPP #ProblemSolving #Algorithms

🔥 Day 173 of My LeetCode Journey Problem 113: Path Sum II 💡 Problem Insight: Today’s problem extends Path Sum — instead of just checking existence, you must return all root-to-leaf paths whose sum equals the target. This shift from a Boolean list makes the problem more complex. 🧠 Concept Highlight: The solution uses DFS + backtracking: Traverse the tree while maintaining the current path Subtract node values from the target sum When a valid leaf is reached, store the path Backtrack to explore other possibilities Backtracking is essential to avoid mixing paths. 💪 Key Takeaway: Whenever a problem asks for all possible paths/combinations, backtracking is the go-to technique. Managing state correctly is more important than traversal itself. ✨ Daily Reflection: This problem reinforced the need for discipline when storing results. Without proper backtracking, solutions become incorrect quickly. #Day173 #LeetCode #BinaryTree #DFS #Backtracking #PathSum #ProblemSolving #DSA #CodingJourney #Consistency

🚀 Day 91 of LeetCode Problem Solving Journey — 100 Days LeetCode Challenge Today, I solved LeetCode #209 — Minimum Size Subarray Sum using C++, under the guidance of Trainer NEKAL SINGH SALARIA Singh at REGex Software Services. 🔍 Problem Summary: Given an array of positive integers nums and a target value target, return the minimal length of a contiguous subarray whose sum is greater than or equal to target. If no such subarray exists, return 0. 🧠 Approach Used (Sliding Window): I used the sliding window technique to efficiently find the minimum length: Initialize two pointers → left = 0 and sum = 0 Expand the window by moving right and adding elements to sum While sum >= target: Update minimum length Shrink the window from the left by subtracting nums[left] Move left forward This approach dynamically adjusts the window to find the smallest valid subarray. 📚 Key Learnings of the Day ✔ Sliding window is powerful for subarray problems ✔ Expanding and shrinking window helps optimize search ✔ Works efficiently when all numbers are positive ✔ Avoids nested loops and reduces complexity ⏱ Complexity • Time Complexity: O(n) • Space Complexity: O(1) 💡 Optimization Insight: The brute-force approach takes O(n²), but sliding window reduces it to O(n) by maintaining a dynamic range. Getting closer to the finish line — see you on Day 92 🚀 #Day91 #100DaysLeetCodeChallenge #LeetCode #RegexSoftwareServices #NekalSingh #ProblemSolving #DSA #CPlusPlus #CodingChallenge #ProgrammingJourney #SlidingWindow #KeepGrowing

Day 30 of #100DaysOfCode 💻 Today’s focus: Permutation Problem Worked on generating all permutations of an array using recursion + backtracking. This problem really tested my understanding of decision trees and tracking visited elements. What stood out today: Every choice creates a new path — and managing those choices efficiently is the key. It’s interesting how a problem that looks simple at first quickly turns into a deep exercise in logic building and control over recursion. Slowly getting more comfortable with backtracking patterns 🚀 #DSA #Backtracking #Recursion #CodingJourney #LeetCode #Consistency #Day30