Member Avatar for dflor

I need to find the smallest number formed from the same digits of a given natural number n, using arrays.

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Three practical approaches were discussed in the thread: digit-frequency counting (the working solution posted by ), repeated-selection (the idea from ), and in-place sorting (what was trying). Trade-offs: counting is O(d) with fixed extra space (d = digit count), sorting a string is simplest and fine for typical inputs (O(d log d)), and repeated-selection is O(d^2) and usually unnecessary.

Why zero the array? A local int a[10]; starts with indeterminate values. Incrementing a[digit]++ without initializing reads and updates garbage, producing wrong counts. Safe alternatives: initialize on declaration (int a[10] = {0};), use std::vector<int> a(10);, or call std::fill_n. The important part is that the array is zeroed before any counting happens.

What is j doing? It’s just a repetition counter: for each digit value i, that inner loop emits digit i exactly a[i] times into the result. The same logic can be written more readably as while (a[i]-- > 0) { append digit i } or for (int k=0; k<a[i]; ++k) { ... }.

A few practical notes and fixes from the thread: ’s arrange loop compares list[i] with list[i+1] while iterating up to i < length, which accesses list[length] (out of bounds); use i < length-1 or simply std::sort. For debugging, printing the array first (as suggested) reveals uninitialized data quickly. Also consider large inputs: converting back to an integer may overflow — for very large numbers it’s safer to return or print the arranged digits as a string.

Example (string+sort approach):

std::string s = std::to_string(n);
std::sort(s.begin(), s.end());
if (s.front() == '0') {
    auto it = std::find_if(s.begin(), s.end(), [](char c){ return c != '0'; });
    if (it != s.end()) std::swap(s.front(), *it);
}
std::cout << s << '\n';

This handles leading zeros, is concise, and avoids manual indexing pitfalls.

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Member Avatar for alexchen 
#include <fstream>
#include <math.h>


int *list;
bool arrange(int length){

  for(int i=0;i<length;i++)
  {
	if(list[i]>list[i+1])
		swap(list[i],list[i+1]);

	

    }

  for(int i=0;i<length;i++)cout<<"my new list contains "<<list [i]<<endl;


return true;
}
bool file(char* Filename, int length)
{

	

	ifstream file;
	file.open(Filename);



	list=new int[length];
  for(int i=0;i<length;i++) 
  {
	file>>list[i];
	cout<<"the list contain: "<<list[i]<<endl;

    }
	arrange(length);

return true;
}
Edited by alexchen because: n/a
Member Avatar for SgtMe

Say you have the number: 251 The smallest number from those digits will be: 125 Therefore: digit1 = smallest number, digit2 = next smallest, etc.

Seperate the number into individual digits using an array. Find the smallest value and put this in the first index of a new array. Repeat this, removing the number you use each time, until you have no numbers left in the original array. Put all of the new array's values into a string and convert that to an int. This is your new number!

EDIT: You may want to use lists not an array.
EDIT EDIT: Oops got beaten too it. Ah well :)

Edited by SgtMe because: n/a
Member Avatar for dflor

I've found a code that works very well:

#include<iostream>
using namespace std;
int main()
{
    int n,a[10],i,j;
    for(i=0;i<=9;i++) a[i]=0;
    cin>>n;
    while(n>0)
    {
        a[n%10]++;
        n=n/10;
    }
    i=1;
    while(a[i]==0) i++;
    n=i;
    a[i]--;
    for(i=0;i<=9;i++)
        for(j=1;j<=a[i];j++)
            n=n*10+i;
    cout<<n;
    return 0;
}

I think it's still difficult for me to understand why for(i=0;i<=9;i++) a[i]=0;
before cin>>n and what is the significance of j.
Otherwise, the code looks like you are trying to explain.

Edited by mike_2000_17 because: Fixed formatting
Member Avatar for SgtMe

Next time, please press the code button before you paste your code.

for (i=0; i<=9; i++){}

Ever done a while loop before? Consider a loop which is while (i<=9){} . At the end of the loop, you would have i++; to increment the value of i .
Basically (in a for loop): i=0 makes a new variable called i which is set to 0 . i<=9 means that we keep looping while i is less than or equal to 9. i++ means that we increment the value of i once each loop.

Edited by SgtMe because: n/a
Member Avatar for dflor

I didn't understand why for(...) at the beginning, a has to be 0.

Member Avatar for WaltP

I didn't understand why for(...) at the beginning, a has to be 0.

Who says it has to be 0? Test it. First thing in the program just print out the array.

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