20. Valid Parentheses LeetCode Solution

In this guide we will provide 20. Valid Parentheses LeetCode Solution with best time and space complexity. The solution to Valid Parentheses problem is provided in various programming languages like C++, Java and python. This will be helpful for you if you are preparing for placements, hackathon, interviews or practice purposes. The solutions provided here are very easy to follow and with detailed explanations.

Problem Statement

Complexity Analysis

Valid Parentheses solution in C++
Valid Parentheses soution in Java
Valid Parentheses solution Python
Additional Resources

20. Valid Parentheses LeetCode Solution image

Problem Statement of Valid Parentheses

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

Example 4:

Input: s = “([])”
Output: true

Constraints:

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

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

20. Valid Parentheses LeetCode Solution in C++

class Solution {
 public:
  bool isValid(string s) {
    stack<char> stack;

    for (const char c : s)
      if (c == '(')
        stack.push(')');
      else if (c == '{')
        stack.push('}');
      else if (c == '[')
        stack.push(']');
      else if (stack.empty() || pop(stack) != c)
        return false;

    return stack.empty();
  }

 private:
  int pop(stack<char>& stack) {
    const int c = stack.top();
    stack.pop();
    return c;
  }
};
/* code provided by PROGIEZ */

20. Valid Parentheses LeetCode Solution in Java

class Solution {
  public boolean isValid(String s) {
    Deque<Character> stack = new ArrayDeque<>();

    for (final 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();
  }
}
// code provided by PROGIEZ

20. Valid Parentheses LeetCode Solution in Python

class Solution:
  def isValid(self, s: str) -> bool:
    stack = []

    for c in s:
      if c == '(':
        stack.append(')')
      elif c == '{':
        stack.append('}')
      elif c == '[':
        stack.append(']')
      elif not stack or stack.pop() != c:
        return False

    return not stack
 #code by PROGIEZ