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Valid Parentheses

annawijetunga profile image Anna Wijetunga ・2 min read

I completed a technical interview this week and received specific feedback:

"Get faster."

I have a process, the process works, the process is slow.

The best way I know to get better at something is to practice, so I'm upping the ante.

Codewars, giddy up.

The Challenge:

Write a function called that takes a string of parentheses, and determines if the order of the parentheses is valid. The function should return true if the string is valid, and false if it's invalid.

Examples
"()"              =>  true
")(()))"          =>  false
"("               =>  false
"(())((()())())"  =>  true
Constraints
0 <= input.length <= 100

I carry a great deal of hesitation and doubt about getting started with any problem. What if what I write isn't perfect? What if it's embarrassingly wrong? So I freeze.

My shiny, new goal: put pen to paper and start writing as if there's a clock ticking. Progress not perfection.

So here goes.

My Soon-To-Be-Much-Faster Process:

1) Understand what's being asked. What's valid vs invalid?

// string of open AND closed parentheses = valid
// just open OR just closed = invalid

2) Think about the method - loops are handy, if you want to run the same code over and over again, each time with a different value. Let's use a loop!

Code newbie here, so I still have to check if my syntax is correct - so I check the documentation to confirm the template:

for (i = 0; i < 5; i++)

3) Let's add our own variables to customize the loop:

let n = 0;
for (let i = 0; i < parens.length; i++)

4) Now we have to specify what's valid/invalid in terms of our parentheses:

for (let i = 0; i < parens.length; i++) {
  if (parens[i] == '(') n++;
  if (parens[i] == ')') n--;
  if (n < 0) return false;
}

We're incrementing and decrementing when there are open AND closed parentheses, this equaling 0 - which in our case is valid/true.

If we end up with less than 0, we know we're missing a parenthesis, and we return false.

5) Put it all together, and don't forget our final return:

function validParentheses(parens){
  let n = 0;
  for (let i = 0; i < parens.length; i++) {
    if (parens[i] == '(') n++;
    if (parens[i] == ')') n--;
    if (n < 0) return false;
  }

  return n == 0;
}

It works! Happy with the results, but I have work to do.

I ended up going over my time limit. 15 minutes wasn't enough, but I got more accomplished than I expected. I guess a lil' fire under the buns helped.

After submitting my solution, I love browsing through other solutions. Here is one that stood out to me because of how concise it is:

Clever Solution:

function validParentheses(parens){
  let indent = 0;

  for (let i = 0; i < parens.length && indent >=0; i++) {
    indent += (parens[i] == '(') ? 1 : -1
    }

  return (indent == 0);
}

Always more to learn, and here's hoping this type of daily coding with a time limit helps me boost my confidence and gets my speed up to par.

Did you code along? Have a different solution or a tip for improvement? Sure would love to hear about it :).

Posted on May 23 by:

annawijetunga profile

Anna Wijetunga

@annawijetunga

Flatiron Grad and Software Engineer! Learning to code, coding to learn.

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