Valid Parentheses LeetCode Solution

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

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

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(); } };
Code language: C++ (cpp)

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(); } }
Code language: Java (java)

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; };
Code language: JavaScript (javascript)

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
Code language: Python (python)