dev.to staff

Posted on Feb 11, 2020

Daily Challenge #184 - Form the Minimum

#challenge

Setup

Given an array of integers, implement a function that will return the lowest possible number that can be formed from these digits. You cannot use the same number more than once, even if it appears in the array multiple times.

Examples

minValue({1, 3, 1}) ==> return 13

minValue({5, 7, 5, 9, 7}) ==> return 579

Tests

minValue({1, 2, 3, 4})

minValue({1, 1, 7, 0})

minValue({9, 2, 1, 4, 3, 9, 5})

Good luck!

This challenge comes from MrZizoScream 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)

Craig McIlwrath • Feb 11 '20

Haskell. Also works for empty lists, returns 0.

import Data.List (nub, sort)
import Data.Maybe (fromMaybe) 
import Text.Read (readMaybe) 

minVal :: [Int] -> Int
minVal = fromMaybe 0 . readMaybe . concatMap show . sort . nub

@nobody • Feb 11 '20

this is why Haskell scares me :o
that looks like Overlord Wizardry compared
to Clojure

Valts Liepiņš • Feb 11 '20 • Edited

Ruby:

def minValue arr
    set = arr.uniq.sort
    set[0..1] = set[0..1].rotate if set[0] == 0 # Edge case for sets with 0
    set.join.to_i
end

Edit: Fixed the 0 edge case. Instead of adding 0 to the end of the number, put it in the 2. position.

Simon Landry • Feb 12 '20

Wouldn't rotating the array bring the biggest number at the front?

I think swapping the 1st and 2nd number would be ideal.

Also is having a leading zero really against the rules?

Valts Liepiņš • Feb 12 '20

Yeah, swapping seems to be better, would keep the large digits in the lesser valuable positions.

I interpreted the rules this way, because it didn't seem right to lose a digit in the resulting number. That would be like assuming input set [1,2,0] as equivelent to [1,2].

Viktor • Feb 12 '20

My try to solve this challenge :)

In my solution I don't include 0 in end result.

var minValue = arr => parseInt(arr.sort().map((e,i,a) => {if( e != a[i+1])return e;}).filter(x=>x).toString().replace(/,/g, ""));

matej • Feb 13 '20

Java

public String minValue(int[] ints) {
   return Arrays.stream(ints)
      //.filter(i -> i > 0) //does not include zeroes
      .distinct()
      .sorted()
      .mapToObj(i -> String.valueOf(i))
      .collect(Collectors.joining());
}

emadsaber • Feb 12 '20

Javascript

function getMinimum(arr){
let sorted = arr.sort();
let distinct = '';

sorted.forEach(function(element){
if(distinct.indexOf(element) < 0){
distinct = ${distinct}${element};
}
});

console.log(parseInt(distinct));
return parseInt(distinct);
}

Vidit Sarkar • Feb 11 '20 • Edited

Here is my C++ code

vector<int> minValueHelper(vector<int> v){
    set<int> s;
    bool presentZero = false;
    for(int i : v){
        if(i==0)
            presentZero = true;
        s.insert(i);
    }
    if(presentZero){
        // because 0 cannot be in front of any number
        // minValue({1, 1, 7, 0}) should not return 017 but should return 107
        vector<int> v(s.begin(),s.end());

        // if only zero is given in vector return zero
        // else swaping
        if(v.size() > 1)
            swap(v[0],v[1]);
        return v;
    }
    return vector<int>(s.begin(),s.end());
}

int minValue(vector<int> v){
    vector<int> minVector = minValueHelper(v);
    string s = "";
    for(int i : minVector)
        s += to_string(i);
    return stoi(s);
}

divyansh-pratap • Feb 11 '20 • Edited

this is solution in c . I have first sorted the array then solved it using simple logics . suggestions are welcome .

include

int sort(int A[] , int n );
int min(int A[] , int n);
int main()
{
int A[30];
int a , i , s;
printf("enter the number of elements you want to enter");
scanf("%d" ,&a);

for(i=0;i<a;i++)
{
    printf("enter the %dth element" , i+1);
    scanf("%d" , &A[i]);
}
s = min(A , a);
printf("/n %d" , s);

}
int sort(int A[] , int n )
{
int i;
int temp , j;
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
if(A[j]<A[i])
{
temp=A[i];
A[i]=A[j];
A[j]=temp;
}
}
}
}

int min(int A[] ,int n)
{
int s , j=0 , k , i , l=1;

  sort(A , n);
  s=A[0];
  for(i=1;i<n;i++)
  {

    j=A[i];

    for(k=0;k<i;k++)
    {
        if(A[i]==A[k])
        {
         l=0;
          }
      }
    if(l==1)
    {
        s=s*10 + j;
      }

     l=1 ;

  }

  return s;

}

Matt Ellen-Tsivintzeli • Feb 12 '20 • Edited

const minValue = ns => parseInt([...(new Set(ns))].sort().join(''));

George WL • Feb 13 '20 • Edited

Even works for [7,0,2], neat

SavagePixie • Feb 11 '20 • Edited

JavaScript

const formMin = arr => +Array
  .from(new Set(arr))
  .sort((a, b) => a - b)
  .join('')

Rohit Prasad • Feb 14 '20

Python

var = input("Enter the numbers: ")
var = list(set(var))
var.sort()
var = ''.join(var)
print(var)

View full discussion (14 comments)

DEV Community

Daily Challenge #184 - Form the Minimum

Setup

Examples

Tests

Top comments (14)

include

Read next

Why Global Standards of API Design Save Your Team Time

Unlocking Docker BuildKit for Faster and More Secure Builds

Docker Distributed Storage: GlusterFS vs. Ceph for Persistent Container Data

Building and Distributing Multi-Architecture Docker Images