Valid Parentheses LeetCode Problem | LeetCode Problems For Beginners | LeetCode Problems & Solutions | Improve Problem Solving Skills | LeetCode Problems Java | LeetCode Problems C++
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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 Problem – Valid 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:
- Open brackets must be closed by the same type of brackets.
- 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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