Stack

Valid Closing Parenthesis - Easy

Can use stacks to check for the ClosingParenthesis problem. One loop through (O(n)) and Stacks make this fast albeit a teeny bit memory hungry.

1. Iterate through, if current char is an opening parenthesis/bracket, then add it to the stack (put).
2. Else if the current is a closing parenthesis, check if its a pair with the opener at the top of the stack (peak)
   1. If not pair or stack size is zero then return False,
   2. else pop the opener from the stack because they’re a pair.
3. If stack size is zero then return true (cuz means all openers have been paired), else false.

class Solution {
 
    public boolean pair(Character chr, MyStack stack) {
        switch (stack.peak()) {
            case '(':
                if (chr == ')') {
                    return true;
                }
                break;
            case '{':
                if (chr == '}') {
                    return true;
                }
                break;
            case '[':
                if (chr == ']') {
                    return true;
                }
            default:
                return false;
        }
        return false;
    }
 
    public boolean isValid(String input) {
        MyStack stack = new MyStack();
        for (int i = 0; i < input.length(); i++) {
            if (input.charAt(i) == '(' || input.charAt(i) == '{' || input.charAt(i) == '[') {
                stack.push(input.charAt(i));
            } else if (input.charAt(i) == ')' || input.charAt(i) == '}' || input.charAt(i) == ']') {
                if (stack.size() == 0 || !(pair(input.charAt(i), stack))) {
                    return false;
                } else {
                    stack.pop();
                }
            }
        }
        if (stack.size() == 0) {
            return true;
        } else {
            return false;
        }
    }
}