Valid Parentheses LeetCode Solution

Last updated on August 5th, 2024 at 02:27 am

Here, We see Valid Parentheses LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc., with different approaches.

List of all LeetCode Solution

Topics

Stack, String

Companies

Airbnb, Amazon, Bloomberg, Facebook, Google, Microsoft, Twitter, Zenefits

Level of Question

Easy

Valid Parentheses LeetCode Solution

Valid Parentheses LeetCode Solution

Problem Statement

Given a string s containing just the characters ‘(‘‘)’‘{‘‘}’‘[‘ and ‘]’, determine if the input string is valid.

An input string is valid if:

  • Open brackets must be closed by the same type of brackets.
  • Open brackets must be closed in the correct order.
  • Every close bracket has a corresponding open bracket of the same type.
Example 1:
Input: s = "()"
Output: true

Example 2:
Input: s = "()[]{}"
Output: true

Example 3:
Input: s = "(]"
Output: false

1. Valid Parentheses Leetcode Solution C++

class Solution {
public:
    bool isValid(string s) {
        stack<char> st;
    for(char c : s){
        if(c == '('|| c == '{' || c == '['){
            st.push(c);
        }else{
            if(st.empty()) return false;
            if(c == ')' && st.top() != '(') return false;
            if(c == '}' && st.top() != '{') return false;
            if(c == ']' && st.top() != '[') return false;
            st.pop();
        }
    }
    return st.empty();
    }
};

2. Valid Parentheses Leetcode Solution Java

class Solution {
    public boolean isValid(String s) {
        HashMap map = new HashMap();
        map.put('(',')');
        map.put('[',']');
        map.put('{','}');
        Stack<Character> stack = new Stack<>();

        for(int i = 0;i < s.length();i++){
            char c = s.charAt(i);
            if(c == '(' || c == '{' || c == '['){
                stack.push(c);
            }else{
                if(stack.isEmpty()){
                    return false;
                }
                if(map.get(stack.pop()).equals(c)){
                    continue;
                }else{
                    return false;
                }
            }
        }
        return stack.isEmpty();
    }
}

3. Valid Parentheses Leetcode Solution JavaScript

var isValid = function(s) {
    const stack = [];
    
    for (let i = 0 ; i < s.length ; i++) {
        let c = s.charAt(i);
        switch(c) {
            case '(': stack.push(')');
                break;
            case '[': stack.push(']');
                break;
            case '{': stack.push('}');
                break;
            default:
                if (c !== stack.pop()) {
                    return false;
                }
        }
    }
    
    return stack.length === 0;
};

4. Valid Parentheses Leetcode Solution Python

class Solution(object):
    def isValid(self, s):
        opcl = dict(('()', '[]', '{}'))
        stack = []
        for idx in s:
            if idx in '([{':
                stack.append(idx)
            elif len(stack) == 0 or idx != opcl[stack.pop()]:
                return False
        return len(stack) == 0
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