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# Valid Parentheses

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#### Problem Statement

Given a string 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.
Note that an empty string is also considered valid.

#### Examples

``````Example 1:
Input: "([)]"
Output: false``````
``````Example 2:
Input: "{[]}"
Output: true``````
``````Example 3:
Input: "()[]{}"
Output: true``````

#### Solution

``````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();
}
}``````