Last updated on October 10th, 2024 at 01:59 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

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