Valid Parentheses LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Valid Parentheses in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemValid Parentheses– LeetCode Problem

Valid Parentheses– LeetCode Problem

Problem:

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 Solutions
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;
  }
};
class Solution {
  public boolean isValid(String s) {
    Stack<Character> stack = new Stack<>();

    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();
  }
}
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

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