## Problem:

Given a string containing parantheses ( (,).{,},[,]) check if the parantheses match or not.

For example,

The string “()[]{}” is a valid string because the opening parentheses are matched by closing ones.

The string “{[()]}” is also a valid expression since all the opening parentheses have corresponding closing parentheses

The string “(]” is invalid .

The string “({)}” is also invalid though there are matching parantheses because they are not in the correct positions.

For every opening parentheses , whatever parentheses comes in between the opening and closing parentheses should be closed.

Try out the solution here:

https://leetcode.com/problems/valid-parentheses/

## Solution:

## Hint:

- Use a stack data structure

## ALGORITHM:

**STEP 1:** Initialize a stack to store the parenthesis

**STEP 2:** For every character in the string , check the top character in the stack.

**STEP 2a:** If the current character is the closing parenthesis of the top character then pop it out, else add the character to the stack.

**STEP 3:** If the stack is empty at the end , then the given string is valid , return true . Else return false.

## Code:

class Solution { public boolean isValid(String s) { Stack<Character> stack = new Stack<Character>(); for(char c: s.toCharArray()){ boolean popped = false; if(!stack.empty()){ char top = stack.peek(); if(top == '(' && c == ')' || top == '{' && c == '}' || top == '[' && c == ']'){ popped = true; stack.pop(); } } if(!popped){ stack.push(c); } } if(stack.empty()){ return true; } return false; } }

Time complexity is O(n) where n is the length of the string and space complexity is also O(n) to store the n characters in the stack.

That’s it!