Valid Parentheses LeetCode Solution | Easy Approach

Valid Parentheses | 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 Solutions

✅Time: O(n)
✅Space: O(n)

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;
  }
};

Java

class Solution {
  public String intToRoman(int num) {
    final int[] values = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
    final String[] symbols = {"M",  "CM", "D",  "CD", "C",  "XC", "L",
                              "XL", "X",  "IX", "V",  "IV", "I"};
    StringBuilder sb = new StringBuilder();

    for (int i = 0; i < values.length; ++i) {
      if (num == 0)
        break;
      while (num >= values[i]) {
        num -= values[i];
        sb.append(symbols[i]);
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();
  }
}

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

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