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Posted on Jan 31, 2020

Daily Challenge #177 - Supersize Me

#challenge

Setup

Implement a function that rearranges an integer into its largest possible value. If the integer is already in its maximum possible value, simply return it.

Examples

superSize(123456) //654321
superSize(105) //510

Tests

superSize(513)
superSize(2020)
superSize(4194)
superSize(608719)
superSize(123456789)
superSize(700000000001)
superSize(666666)

Good luck!

This challenge comes from ozymandias1 on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge idea for a future post? Email yo+challenge@dev.to with your suggestions!

Top comments (14)

SavagePixie • Jan 31 '20

JavaScript oneliners for the win!

const superSize = n => +n.toString().split('').sort((a, b) => b - a).join('')

George WL • Jan 31 '20

Had a feeling it was this simple.

Why the plus though?

Craig McIlwrath • Jan 31 '20

The argument to the unary plus operator is a string. The + is a very concise method of converting a string to a number, in this situation.

George WL • Jan 31 '20

Ah, I totes prefer the readability over conciseness approach

Lamonte • Feb 3 '20 • Edited

dart

int superSize(int number) {
  var data = number.toString();
  var dataInt = List<int>();
  for(var n = 0; n < data.length; n++) {
    var val = int.tryParse(data[n]);
    if(val != null) {
      dataInt.add(val);
    }
  }
  dataInt.sort((a, b) => b.compareTo(a));
  return int.tryParse(dataInt.join());
}

I'm sure this could be done shorter.

Dimitri Marion • Jan 31 '20

Javascript

const superSize = n => {
    const nArr  = Array.from(n.toString()).map(Number);

    const nArrSorted = nArr.sort((a, b) => b - a);

    return parseInt(nArrSorted.join(''));
};

Jehiel Martinez • Feb 1 '20

In JS, implementing Bubble Sort just because.

function superSize(num) {
    let str = [...num.toString()];
    let swap;
    do {
        swap = false
        for (i = 0; i <= str.length; i++) {
            if (str[i] < str[i + 1]) {
                const bigger = str[i + 1];
                str[i + 1] = str[i];
                str[i] = bigger;
                swap = true
            }
        }
    } while (swap);
    return(+str.join(''));
};

Andrei Visoiu • Feb 2 '20

in C++ using the standard template library:

typedef unsigned long long ull;
ull superSize(ull x) {
    vector<short> d;
    while(x > 0) {
        d.push_back(x%10);
        x /= 10;
    }
    sort(d.begin(), d.end(), greater<short>());
    ull r = 0;
    for(auto i: d)
        r = r*10 + i;

    return r;
}

Viktor • Jan 31 '20 • Edited

Javascript solution for superSize :)

function superSize(number){
let revNum = "";
let tempNum = number;
while(number != 0){
    revNum = `${revNum}${(number % 10)}`;
    number = parseInt(number / 10);
}
revNum = parseInt(revNum);
return revNum > tempNum ? revNum : tempNum;
}

Valts Liepiņš • Jan 31 '20

Yet another Ruby one liner:

def superSize int
    int.to_s.chars.sort.reverse!.join.to_i
end

Also neat how it reads almost literarly, no need to explain each step in the chain!

lusen / they / them 🏳️‍🌈🥑 • Feb 1 '20

def super_size(num):
  integers = [int(c) for c in str(num)]
  integers.sort(reverse=True)
  return ''.join([str(i) for i in integers])

zoe-j-m • Jan 31 '20 • Edited

Scala

def superSize = (_ : Int).toString.sorted.reverse.mkString

Mohamed ABDELLANI • Feb 1 '20 • Edited

def superSize(x)
 x.to_s.to_s.split("").sort{|x,y|-(x<=>y)}.join.to_i
end

View full discussion (14 comments)

DEV Community

Daily Challenge #177 - Supersize Me

Setup

Examples

Tests

Top comments (14)

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