# Data Structures & Algorithms in Java – Strings – Valid Parantheses

## 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/

## 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!

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