# Daily Challenge #184 - Form the Minimum

### 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!

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### Discussion  Valts Liepiņš

Ruby:

``````def minValue arr
set = arr.uniq.sort
set[0..1] = set[0..1].rotate if set == 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. Valts Liepiņš

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]`. Michael Kohl

Naive Ruby solution, no consideration for performance:

``````def min_value(arr)
uarr = arr.uniq
uarr.permutation(uarr.size)
.map { |a| a.map(&:to_s).join.to_i }
.min
end

min_value([1,3,1])
#=> 13
min_value([5, 7, 5, 9, 7])
#=> 579
min_value([1, 2, 3, 4])
#=> 1234
min_value([1, 1, 7, 0])
#=> 17 (017)
min_value([9, 2, 1, 4, 3, 9, 5])
#=> 123459
`````` matej

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());
}
`````` 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

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,v);
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);
}

`````` Viktor

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, ""));
`````` divyansh-pratap

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;
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;
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;
``````

}  