Longest Valid Parentheses

Author: gouthampradhan

README.md

@@ -250,6 +250,7 @@ My accepted leetcode solutions to some of the common interview problems.
 - [Exclusive Time of Functions](problems/src/stack/ExclusiveTimeOfFunctions.java) (Medium)
 - [Basic Calculator](problems/src/stack/BasicCalculator.java) (Hard)
 - [Decode String](problems/src/stack/DecodeString.java) (Medium)
+- [Longest Valid Parentheses](problems/src/stack/LongestValidParentheses.java) (Hard)
 
 
 #### [String](problems/src/string)

problems/src/stack/LongestValidParentheses.java

@@ -0,0 +1,75 @@
+package stack;
+
+import java.util.*;
+
+/**
+ * Created by gouthamvidyapradhan on 08/07/2018.
+ * Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed)
+ * parentheses substring.
+
+ Example 1:
+
+ Input: "(()"
+ Output: 2
+ Explanation: The longest valid parentheses substring is "()"
+ Example 2:
+
+ Input: ")()())"
+ Output: 4
+ Explanation: The longest valid parentheses substring is "()()"
+
+ Solution: O(N) Iterate through each of the parentheses and if '(' is encountered push it to stack else check the top
+ of the stack to see if there is a matching parentheses, if yes pop it and then take the length (currIndex - index at
+ top of the stack). Maintain a max length and return this as the answer.
+ */
+public class LongestValidParentheses {
+
+    private class Node{
+        char c;
+        int i;
+        Node(char c, int i){
+            this.c = c;
+            this.i = i;
+        }
+    }
+
+    /**
+     *
+     * @param args
+     * @throws Exception
+     */
+    public static void main(String[] args) throws Exception{
+        System.out.println(new LongestValidParentheses().longestValidParentheses("((()()(((())))))"));
+    }
+
+    public int longestValidParentheses(String s) {
+        Stack<Node> stack = new Stack<>();
+        int max = 0;
+        for(int i = 0, l = s.length(); i < l; i ++){
+            char c = s.charAt(i);
+            switch (c){
+                case '(':
+                    stack.push(new Node(c, i));
+                    break;
+
+                case ')':
+                    if(!stack.isEmpty()){
+                        if(stack.peek().c == '('){
+                            stack.pop();
+                            if(stack.isEmpty()){
+                                max = Math.max(max, i + 1);
+                            } else {
+                                max = Math.max(max, i - stack.peek().i);
+                            }
+                        } else{
+                            stack.push(new Node(c, i));
+                        }
+                    } else{
+                        stack.push(new Node(c, i));
+                    }
+            }
+        }
+        return max;
+    }
+
+}