6
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function dashesToParentheses(str) {
    var list = str.split('-');
    return str.replace(/-/g, '(') + repeatString(')', list.length - 1);
}

function repeatString(str, times) {
    if (times == 1)
        return str;

    return new Array(times + 1).join(str);
}

dashesToParentheses('a-b-c') // "a(b(c))"
dashesToParentheses('a-b') // "a(b)"
dashesToParentheses('a') // "a"
dashesToParentheses('') // ""

dashesToParentheses works correct. Can I make it simpler or/and faster?

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7
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Having split the string you can join it with brackets instead of replacing them. You could optionally choose to remove the repeatString function and your +/- 1, but it does make a lot of sense the way you have it.

function dashesToParentheses(str) {
    var list = str.split('-');
    return list.join('(') + Array(list.length).join(')');
}
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1
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Well, you could remove the RegExp. But whether that helps performance is anyone's guess.

var d2p = function(s){
    var one=[], two=[], a=s.split('-');
    one.push(a.shift());
    a.forEach(function(part){
        one.push('(' + part);
        two.push(')');
    });
    return one.join('') + two.join('');
};

If you don't mind having the result fully parenthesised, then you can do this:

var d2p = function(s){
    return s.split('-').reduceRight(function(whole, part){
        return '(' + part + whole + ')';
    }, '');
};
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0
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The first term is obvious: Just replace all dashes by opening parentheses. The second replaces all dashes by closing parentheses while dropping everything else.

function dashesToParentheses(str) {
    return str.replace(/-/g, "(") + str.replace(/[^-]*-[^-]*/g, ")");
}
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