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+# 1021. Remove Outermost Parentheses
+
+**Difficulty:** *Easy*  
+**Category:** *Strings, Stack, Greedy*  
+**Leetcode Link:** [Problem Link](https://leetcode.com/problems/remove-outermost-parentheses/)
+
+---
+
+## 📝 Introduction
+
+*Given a valid parentheses string, the task is to remove the outermost parentheses of every primitive substring. A primitive string is a non-empty valid parentheses string that cannot be split further into valid parts.*
+
+---
+
+## 💡 Approach & Key Insights
+
+*Each primitive parentheses string is balanced and starts/ends with outermost parentheses. If we track the nesting level using a counter, we can easily identify the start and end of each primitive and skip appending the outermost layer.*
+
+---
+
+## 🛠️ Breakdown of Approaches
+
+### 1️⃣ Brute Force / Naive Approach
+
+- **Explanation:** *Use a stack to build substrings for each primitive, detect primitives by ensuring stack becomes empty, then remove the first and last characters.*
+- **Time Complexity:** *O(n²) - Substring operations inside loop.*
+- **Space Complexity:** *O(n) - For temporary stack and result string.*
+- **Example/Dry Run:**
+
+Example input: "(()())(())"  
+Step 1 → Extract primitives: "(()())", "(())"  
+Step 2 → Remove outermost: "()()", "()"  
+Step 3 → Output: "()()()"
+
+### 2️⃣ Optimized Approach
+
+- **Explanation:** *Use a counter to track the nesting depth. While iterating, append characters only if they’re not the outermost (i.e., depth > 1 before ‘)’ or after ‘(’).*
+- **Time Complexity:** *O(n) - Single pass over string.*
+- **Space Complexity:** *O(n) - For output string.*
+- **Example/Dry Run:**
+
+Example input: "(()())(())"  
+Step 1 → Traverse:
+- ‘(’: depth becomes 1 → don’t add  
+- ‘(’: depth becomes 2 → add ‘(’  
+- ‘)’: depth becomes 1 → add ‘)’  
+- ‘)’: depth becomes 0 → don’t add  
+... and so on  
+Output: "()()()"
+
+---
+
+## 📊 Complexity Analysis
+
+| Approach      | Time Complexity | Space Complexity |
+| ------------- | --------------- | ---------------- |
+| Brute Force   | O(n²)           | O(n)             |
+| Optimized     | O(n)            | O(n)             |
+
+---
+
+## 📉 Optimization Ideas
+
+*The optimized solution is already linear in time and space. Further improvement would require in-place modification or custom output building if language permits.*
+
+---
+
+## 📌 Example Walkthroughs & Dry Runs
+
+```plaintext
+Example:
+Input: "(()())(())(()(()))"
+Process:
+Primitive 1: "(()())" → Output: "()()"
+Primitive 2: "(())" → Output: "()"
+Primitive 3: "(()(()))" → Output: "()(())"
+Final Output: "()()()()(())"
+```
+
+---
+
+## 🔗 Additional Resources
+
+- [Stack Data Structure](https://en.wikipedia.org/wiki/Stack_(abstract_data_type))
+- [Valid Parentheses Leetcode](https://leetcode.com/problems/valid-parentheses/)
+
+---
+
+Author: Abdul Wahab  
+Date: 19/07/2025
diff --git a/Arrays & Strings/#1021 - Remove outer most parantheses - Easy/removeOuterParentheses.cpp b/Arrays & Strings/#1021 - Remove outer most parantheses - Easy/removeOuterParentheses.cpp
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+++ b/Arrays & Strings/#1021 - Remove outer most parantheses - Easy/removeOuterParentheses.cpp	
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+class Solution {
+public:
+    string removeOuterParentheses(string s) {
+        string res;
+        int balance = 0, start = 0;
+        for (int i = 0; i < s.length(); ++i) {
+            if (s[i] == '(') balance++;
+            else balance--;
+            if (balance == 0) {
+                res += s.substr(start + 1, i - start - 1);
+                start = i + 1;
+            }
+        }
+        return res;
+    }
+};
diff --git a/Arrays & Strings/#1021 - Remove outer most parantheses - Easy/removeOuterParentheses.py b/Arrays & Strings/#1021 - Remove outer most parantheses - Easy/removeOuterParentheses.py
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+class Solution:
+    def removeOuterParentheses(self, s: str) -> str:
+        res = []
+        balance = 0
+        start = 0
+
+        for i, ch in enumerate(s):
+            if ch == '(':
+                balance += 1
+            else:
+                balance -= 1
+            if balance == 0:
+                res.append(s[start + 1:i])
+                start = i + 1
+        return ''.join(res)
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