# Valid Parentheses LeetCode Solution | Easy Approach

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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 `'()[]{}'`.

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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