Greedy Validation + String Simulation DSA Challenge

This title was summarized by AI from the post below.

🚀 DSA Consistency Challenge – Day 76 Today’s focus was on Greedy Validation + String Simulation, where the real challenge wasn’t coding… but thinking in the right direction 🧠⚡ 🔴 1️⃣ Check if a Parentheses String Can Be Valid (Medium) 💡 Problem Twist: You’re allowed to change some characters — so instead of building the exact string, the real question is: 👉 Is it even possible to make it valid? 🧠 Approaches: 🔹 Brute Force Try all combinations for unlocked positions Validate each string ⛔ Exponential → Not practical 🔹 Stack-Based Approach Use stack for '(' and another for flexible indices Try to match ')' with available '(' or flexible positions Finally match remaining '(' with flexible positions (order matters!) ⏱️ O(n) time | O(n) space 🔹 Greedy (Optimal 💯) ✨ Key Insight: Validate instead of constructing Left → Right: Assume unlocked can act as '(' → Avoid excess ) Right → Left: Assume unlocked can act as ')' → Avoid excess ( 🔥 If both passes are valid → answer is TRUE ⏱️ O(n) time | O(1) space 🔴 2️⃣ Remove All Occurrences of a Substring (Medium) 💡 Core Idea: Removing a substring can create new occurrences → requires careful simulation 🧠 Approaches: 🔹 Brute Force (Find + Erase) Keep removing leftmost occurrence using find() ⛔ Inefficient due to repeated scanning ⏱️ O(n²) 🔹 Stack / String Simulation (Efficient ✅) Build result string step-by-step After each character: Check last m characters If match → remove instantly ✨ Works like a hidden stack ⏱️ O(n * m) | 🧠 O(n) 💭 Key Takeaways: Greedy isn’t always about max/min — sometimes it’s about validating feasibility Bidirectional traversal can simplify constraint-heavy problems Many string problems secretly use stack patterns Optimization often comes from avoiding repeated work

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