## Problem

https://leetcode.com/problems/valid-parentheses/

Given a string 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.

Note that an empty string is also considered valid.

Example:

```
Input: "()"
Output: true
Example 2:
Input: "()[]{}"
Output: true
Example 3:
Input: "(]"
Output: false
Example 4:
Input: "([)]"
Output: false
Example 5:
Input: "{[]}"
Output: true
```

## Solution

```
fun isValid(s: String): Boolean {
val stack = Stack<Char>()
val map = mapOf(
'}' to '{',
')' to '(',
']' to '['
)
s.forEach {
stack.push(it)
if (map.containsKey(it)) {
if (stack.size < 2) {
return false
}
if (stack[stack.size - 2] != map[it]) {
return false
}
stack.pop()
stack.pop()
}
}
return stack.isEmpty()
}
```

```
fun isValid(s: String): Boolean {
val stack = Stack<Char>()
val range = (1..2)
s.forEach { c ->
when {
stack.empty() -> stack.push(c)
(c - stack.peek()) in range -> stack.pop()
else -> stack.push(c)
}
}
return stack.empty()
}
```

```
fun isValid(s: String): Boolean {
val stack = Stack<Char>()
val map = mapOf(
'(' to ')', ')' to '(',
'[' to ']', ']' to '[',
'{' to '}', '}' to '{'
)
s.forEach { c ->
when {
stack.empty() -> stack.push(c)
stack.peek() == map[c] -> stack.pop()
else -> stack.push(c)
}
}
return stack.empty()
}
```

## Test code

```
@Test
fun isValid() {
assertTrue(solution.isValid("()"))
assertTrue(solution.isValid("[]"))
assertTrue(solution.isValid("{}"))
assertTrue(solution.isValid("{()}"))
assertTrue(solution.isValid("{([])}"))
assertFalse(solution.isValid("{{"))
assertFalse(solution.isValid("{)"))
assertFalse(solution.isValid(")"))
assertFalse(solution.isValid(")("))
}
```

## Model solution

https://leetcode.com/problems/valid-parentheses/solution/

### Approach 1: Stacks

```
class Solution {
// Hash table that takes care of the mappings.
private HashMap<Character, Character> mappings;
// Initialize hash map with mappings. This simply makes the code easier to read.
public Solution() {
this.mappings = new HashMap<Character, Character>();
this.mappings.put(')', '(');
this.mappings.put('}', '{');
this.mappings.put(']', '[');
}
public boolean isValid(String s) {
// Initialize a stack to be used in the algorithm.
Stack<Character> stack = new Stack<Character>();
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
// If the current character is a closing bracket.
if (this.mappings.containsKey(c)) {
// Get the top element of the stack. If the stack is empty, set a dummy value of '#'
char topElement = stack.empty() ? '#' : stack.pop();
// If the mapping for this bracket doesn't match the stack's top element, return false.
if (topElement != this.mappings.get(c)) {
return false;
}
} else {
// If it was an opening bracket, push to the stack.
stack.push(c);
}
}
// If the stack still contains elements, then it is an invalid expression.
return stack.isEmpty();
}
}
```

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