Expanding use of Javascript array.reduce helper method

Question

Background

I am following a course on Udemy which goes over all of the ES6 features. In one of the lessons the instructor talks about using the reduce helper method to solve the popular balance parenthesis interview question.

I can solve this without the reduce method. Although with the reduce method it does get the job done in less code. I have been asked to find the depth of the parenthesis in an interview before and was wondering if this could all be done in the same method using reduce.

I do not know why this addition to the question confuses me so much but I would like to learn.

Problem

I have been trying to figure it out for a while and it might be my lack of understanding how reduce works.

Example

This uses reduce to return true of false regarding if the parenthesis or open and closed evenly.

function balanceParens(string) {
    return !string.split("").reduce((counter, char) => {
        // Handle if parens open and close out of order
        if (counter < 0) { return counter; }
        // Add 1 for each open in order
        if (char === "(") { return ++counter; }
        // subtract 1 for each close in order
        if (char === ")") { return --counter; }
        // handle use case if char is not a paren
        return counter;
    }, 0);
}
console.log(balanceParens("((()))"));

Question

How would I return the max depth of the parenthesis using the reduce helper method.

How would you solve the problem without `reduce`? If you can post your loop code, we can help you transform it into `reduce`. — Bergi, Jul 06 '17 at 18:01
Yes it was introduced in es5. But the course if over es6 and this was just something he was talking about in the es6 course. — wuno, Jul 06 '17 at 18:03
@Bergi when I get home I can post a solution not using reduce. But it will be a few hours. — wuno, Jul 06 '17 at 18:04
Generally when you have something that fits the pattern of `.reduce()` but you need multiple values, you just use an object as the reduce target. — Pointy, Jul 06 '17 at 18:05

score 3 · Accepted Answer · answered Jul 06 '17 at 18:16

You could maintain current depth and max depth while reducing.

function maxDepth(string) {
    return string.split("").reduce(({current, max}, char) => {
        // Handle if parens open and close out of order
        if (current < 0) return {current, max}
        // Add 1 for each open in order
        if (char === "(") return { current: current + 1, max: Math.max(max, current + 1)}
        // subtract 1 for each close in order
        if (char === ")") return { current: current - 1, max}
        return {current, max}
    }, {current: 0, max: 0}).max;
}
console.log(maxDepth("(((()))(((())))()(((((()))))))"));

Thanks a lot! This is perfect can now see my problem. Man I was so close. — wuno, Jul 06 '17 at 20:50

score 1 · Answer 2 · answered Jul 06 '17 at 18:58

Here is a compact version that returns NaN when the parentheses are not balanced. It uses nested functions in a functional style:

function maxDepth(string) {
    return ( ([depth, max]) => depth ? NaN : max )
        ([...string].reduce(([depth, max], ch) => 
            (newDepth => [newDepth, newDepth < 0 ? NaN : Math.max(max, newDepth)])
                 (depth + (ch === "(") - (ch === ")"))
        , [0, 0]));
}
console.log(maxDepth("(((()))(((())))()(((((()))))))"));

score 0 · Answer 3 · answered Jul 06 '17 at 18:09

0

This should answer it!

function balanceParens(string) {
    let max = 0;
    let res = string.split("").reduce((counter, char) => {
        // Handle if parens open and close out of order
        if (counter < 0) { 
          return counter;
        }
        // Add 1 for each open in order
        if (char === "(") {
          if(++counter > max) {
            max = counter;
          }
          return counter;
        }
        // subtract 1 for each close in order
        if (char === ")") {
          return --counter;
        }
        // handle use case if char is not a paren
        return counter;
    }, 0);
    console.log("Max depth was :", max);
    return !res;
}
console.log(balanceParens("((()(((())))))((((()))))"));

answered Jul 06 '17 at 18:09

Aftab Khan

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What do you mean? `const` variables where? – Aftab Khan Jul 06 '17 at 18:15
1

I like this answer. It seems more consistent with the question example. – Rick Jul 06 '17 at 18:20
@AftabKhan For `max` specifically – Bergi Jul 06 '17 at 18:23
@Arrow Not sure what you mean by "consistent" – Bergi Jul 06 '17 at 18:23
@Bergi Why should I make it const when it's value is changing? – Aftab Khan Jul 06 '17 at 18:24
@AftabKhan I'm suggesting that you should not change it when using `reduce` – Bergi Jul 06 '17 at 18:26
Any specific reason why? – Aftab Khan Jul 06 '17 at 18:27
@Arrow Mutating some state outside of a reducer is not really "consistent" :) – Yury Tarabanko Jul 06 '17 at 18:27
@Bergi because this solution simply increments a counter while the other increments a counter and uses Math.max. – Rick Jul 06 '17 at 18:38
@AftabKhan Because immutability is the whole point of using `reduce` for the `counter` as well – Bergi Jul 06 '17 at 18:42
I didn't understand that. Counter is one single accumulator, whose value keep changing for every iteration. – Aftab Khan Jul 06 '17 at 18:47
@Bergi but reduce has an array callback which can mutate the original arrray. You can't say that the whole point of reducing is immutability unless I'm missing something I am unaware of? – Rick Jul 06 '17 at 18:49
1

Asked a question [here](https://stackoverflow.com/questions/44956739/array-reduce-mutating-state-outside-is-not-recommended) if anyone is interested to give an elaborate answer – Aftab Khan Jul 06 '17 at 18:51
@Bergi your argument also seems absurd because reduce takes an initial argument which it frequently mutates. – Rick Jul 06 '17 at 19:03
@Arrow No, the accumulator (or the array) is *not* mutated. The `counter` doesn't change. It's a new one for every invocation of the callback, which itself is a pure function. – Bergi Jul 06 '17 at 19:17

Expanding use of Javascript array.reduce helper method

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