Valid Parentheses LeetCode Solution

Problem – Valid Parentheses LeetCode Solution

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

An input string is valid if:

1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.

Example 1:

Input: s = "()"
Output: true

Example 2:

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

Example 3:

Input: s = "(]"
Output: false

Constraints:

• 1 <= s.length <= 104
• s consists of parentheses only '()[]{}'.

Valid Parentheses LeetCode Solution in Java

public boolean isValid(String s) {
	Stack<Character> stack = new Stack<Character>();
	for (char c : s.toCharArray()) {
		if (c == '(')
			stack.push(')');
		else if (c == '{')
			stack.push('}');
		else if (c == '[')
			stack.push(']');
		else if (stack.isEmpty() || stack.pop() != c)
			return false;
	}
	return stack.isEmpty();
}

Valid Parentheses LeetCode Solution in Python

class Solution:
    # @return a boolean
    def isValid(self, s):
        stack = []
        dict = {"]":"[", "}":"{", ")":"("}
        for char in s:
            if char in dict.values():
                stack.append(char)
            elif char in dict.keys():
                if stack == [] or dict[char] != stack.pop():
                    return False
            else:
                return False
        return stack == []

Valid Parentheses LeetCode Solution in C++

class Solution {
public:
    bool isValid(string s) {
        stack<char> st;  //taking stack for keep tracking the order of the brackets..
        for(auto i:s)  //iterate over each and every elements
        {
            if(i=='(' or i=='{' or i=='[') st.push(i);  //if current element of the string will be opening bracket then we will just simply push it into the stack
            else  //if control comes to else part, it means that current element is a closing bracket, so check two conditions  current element matches with top of the stack and the stack must not be empty...
            {
                if(st.empty() or (st.top()=='(' and i!=')') or (st.top()=='{' and i!='}') or (st.top()=='[' and i!=']')) return false;
                st.pop();  //if control reaches to that line, it means we have got the right pair of brackets, so just pop it.
            }
        }
        return st.empty();  //at last, it may possible that we left something into the stack unpair so return checking stack is empty or not..
    }
};

Valid Parentheses LeetCode Solution in 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;
};

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