added validate-parenthesis_en

problem/validate-parenthesis_en.md

@@ -1,4 +1,47 @@
+# Valid Parentheses Checker
 
-There is no solution yet.
+This code implements a solution to check if a given string containing different types of brackets is valid or not. It uses a stack data structure to maintain the order of opening and closing brackets and ensure their validity.
 
-Would you like to [contribute to the solution](https://github.com/BFEdev/BFE.dev-solutions/blob/main/problem/validate-parenthesis_en.md)? [Contribute guideline](https://github.com/BFEdev/BFE.dev-solutions#how-to-contribute)
+You can watch a video explanation of this solution [here](https://youtu.be/lwN4SDGTBaY?si=HTgiwVLHvQjaxMuK).
+
+## Explanation
+
+The solution employs a stack to manage the opening and closing brackets. The algorithm iterates through the characters of the input string, and for each character:
+
+1. If it's an opening bracket, it's pushed onto the stack.
+2. If it's a closing bracket, the stack's top element is popped and compared to the expected opening bracket based on the `mapping` dictionary.
+   - If they don't match, the brackets are not valid.
+3. If the character is neither an opening nor a closing bracket, it's ignored.
+
+After iterating through all characters, the stack should be empty if all brackets are properly matched. Otherwise, it's not a valid string.
+
+```javascript
+function isValid(s) {
+    const stack = [];  // Initialize an empty stack to store opening brackets
+    const mapping = { ')': '(', '}': '{', ']': '[' };  // Mapping of closing brackets to their corresponding opening brackets
+
+    for (const char of s) {  // Iterate through each character in the input string
+        if (mapping[char]) {  // If the character is a closing bracket
+            const top = stack.pop();  // Pop the top element from the stack (last opening bracket)
+            if (mapping[char] !== top) {  // Check if the closing bracket corresponds to the popped opening bracket
+                return false;  // Brackets are not valid
+            }
+        } else {  // If the character is an opening bracket
+            stack.push(char);  // Push it onto the stack
+        }
+    }
+
+    return stack.length === 0;  // If the stack is empty, all brackets are valid; otherwise, the string is not valid
+}
+
+// Test cases
+console.log(isValid("()"));        // Output: true
+console.log(isValid("()[]{}"));    // Output: true
+console.log(isValid("(]"));        // Output: false
+```
+
+## Time and Space Complexity
+
+The time complexity of this solution is O(n), where 'n' represents the length of the input string. The algorithm iterates through each character in the string exactly once, and the operations performed within each iteration are constant time operations.
+
+The space complexity is also O(n). The space used by the stack is proportional to the length of the input string in the worst case, as it stores opening brackets. The mapping dictionary occupies a constant amount of space, irrespective of the input size.