✨ (solution): Problem 20. Valid Parentheses

solutions/20. Valid Parentheses/README.md

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+---
+comments: true
+difficulty: easy
+# Follow `Topics` tags
+tags:
+    - String
+    - Stack
+---
+
+# [20. Valid Parentheses](https://leetcode.com/problems/valid-parentheses/description/)
+
+## Description
+
+You are given a string s that consists only of the characters `(`, `)`, `{`, `}`, `[`, and `]`. Your task is to determine whether the string is valid.
+
+A string is considered valid if it meets the following conditions:
+
+1. Every opening bracket has a matching closing bracket of the same type.
+2. Brackets must close in the correct order (i.e., the most recently opened bracket must be the first one to close).
+3. There should not be any unmatched brackets.
+
+
+**Example 1:**
+```
+Input: s = "{}"
+Output: true
+```
+
+**Example 2:**
+```
+Input: s = "{]"
+Output: false
+```
+
+
+**Constraints:**
+`1 <= s.length <= 104`
+`s` consists of parentheses only `'()[]{}'`.
+
+## Solution
+
+As known valid combination:
+```
+mapping = { ')': '(', '}': '{', ']': '[' }
+```
+
+Let stack has only open parentheses. When a close parenthesis comes, use it as a key to get valid open parenthesis in the mapping. If mapped parentheses and stack top parenthese sare not valid combination, we should return False.
+
+At last, if stack is empty, we should return True.
+
+```java
+class Solution {
+    public boolean isValid(String s) {
+        Stack<Character> stack = new Stack<>();
+        for (char c : s.toCharArray()) {
+            if (c == '(' || c == '{' || c == '[') {
+                stack.push(c);
+            } else {
+                if (stack.isEmpty()) {
+                    return false;
+                }
+                char top = stack.pop();
+                if ((c == ')' && top != '(') || (c == '}' && top != '{') || (c == ']' && top != '[')) {
+                    return false;
+                }
+            }
+        }
+        return stack.isEmpty();
+    }
+}
+```
+
+```python
+class Solution:
+    def isValid(self, s: str) -> bool:
+        stack = []
+        mapping = {')': '(', '}': '{', ']': '['}
+        for char in s:
+            if char in mapping.values():
+                stack.append(char)
+            elif char in mapping:
+                if stack == [] or mapping[char] != stack.pop():
+                    return False
+        return not stack
+```
+
+## Complexity
+
+- Time complexity: $$O(n)$$
+<!-- Add time complexity here, e.g. $$O(n)$$ -->
+
+- Space complexity: $$O(n)$$
+<!-- Add space complexity here, e.g. $$O(n)$$ -->

solutions/20. Valid Parentheses/Solution.java

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+import java.util.Stack;
+
+class Solution {
+    public boolean isValid(String s) {
+        Stack<Character> stack = new Stack<>();
+        for (char c : s.toCharArray()) {
+            if (c == '(' || c == '{' || c == '[') {
+                stack.push(c);
+            } else {
+                if (stack.isEmpty()) {
+                    return false;
+                }
+                char top = stack.pop();
+                if ((c == ')' && top != '(') || (c == '}' && top != '{') || (c == ']' && top != '[')) {
+                    return false;
+                }
+            }
+        }
+        return stack.isEmpty();
+    }
+}

solutions/20. Valid Parentheses/Solution.py

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+class Solution:
+    def isValid(self, s: str) -> bool:
+        stack = []
+        mapping = {')': '(', '}': '{', ']': '['}
+        for char in s:
+            if char in mapping.values():
+                stack.append(char)
+            elif char in mapping:
+                if stack == [] or mapping[char] != stack.pop():
+                    return False
+        return not stack