Valid Parentheses

Valid Parentheses - LeetCode

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.
3. Every close bracket has a corresponding open bracket of the same type.

Example 1:

Input: s = "()"
Output: true

Example 2:

Input: s = "()[]{}"
Output: true

Example 3:

Input: s = "(]"
Output: false

APPROACH

I am going to use HashMap to store the brackets and use Stack to check if the brackets are valid.

First, we will initialize a character HashMap with brackets

private HashMap<Character, Character> brackets;

public Solution(){

}

SOLUTION

class Solution {
  // Hash table that takes care of the mappings.
  private HashMap<Character,Character> brackets;
  // Initialize hash map with mappings. This simply makes the code easier to read.
  public Solution() {
    this.brackets = new HashMap < Character, Character > ();
    this.brackets.put(')', '(');
    this.brackets.put('}', '{');
    this.brackets.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.brackets.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.brackets.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();
  }
}