# Form smallest number using indices of numbers chosen from Array with sum less than S

Given an array arr[] and an integer S, the task is to choose the maximum count of numbers from the array such that the sum of numbers is less than S and form the smallest possible number using their indices
Note: Any element can be chosen any number of times.

Examples:

Input: arr[] = {3, 4, 2, 4, 6, 5, 4, 2, 3}, S = 13
Output: 133333
Explanation:
Elements chosen – 3 + 2 + 2 + 2 + 2 + 2 = 13
Therefore, Concatenation of indices – 133333

Input: arr[] = {18, 21, 22, 51, 13, 14, 17, 15, 17}, S = 50
Output: 115

Approach: The idea is to find the maximum count of the elements that can be chosen which can be computed for the number using Finally, the minimum indices that can choose multiple times are computed by taking the minimum digit in the number for each digit place.

Below is the implementation of the above approach:

## C++

 // C++ implementation to find // minimum number which // have a maximum length   #include  using namespace std;   // Function to find the // minimum number which // have maximum length string max_number(int arr[], int sum) {     int frac;     int maxi = INT_MIN;     string ans;     int pos;       // Find Maximum length     // of number     for (int i = 0; i < 9; i++) {         frac[i] = sum / arr[i];         if (frac[i] > maxi) {             pos = i;             maxi = frac[i];         }     }       ans.insert(0,                string(maxi,                       (pos + 1) + '0'));     sum -= maxi * arr[pos];       // Find minimum number WHich     // have maximum length     for (int i = 0; i < maxi; i++) {         for (int j = 1; j <= 9; j++) {               if (sum                     + arr[pos]                     - arr[j - 1]                 >= 0) {                   ans[i] = (j + '0');                 sum += arr[pos]                        - arr[j - 1];                 break;             }         }     }       if (maxi == 0) {         return 0;     }     else {         return ans;     } }   // Driver Code int main() {     int arr = { 3, 4, 2, 4, 6,                    5, 4, 2, 3 };     int s = 13;     cout << max_number(arr, s);     return 0; }

## Java

 // Java implementation to find // minimum number which // have a maximum length class GFG{   // Function to find the // minimum number which // have maximum length static String max_number(int arr[], int sum)  {     int frac[] = new int;     int maxi = Integer.MIN_VALUE;           StringBuilder ans = new StringBuilder();     int pos = 0;       // Find Maximum length     // of number     for(int i = 0; i < 9; i++)     {         frac[i] = sum / arr[i];         if (frac[i] > maxi)         {             pos = i;             maxi = frac[i];         }     }       for(int i = 0; i < maxi; i++)     {         ans.append((char)((pos + 1) + '0'));     }       sum -= maxi * arr[pos];       // Find minimum number WHich     // have maximum length     for(int i = 0; i < maxi; i++)     {         for(int j = 1; j <= 9; j++)         {             if (sum + arr[pos] - arr[j - 1] >= 0)             {                 ans.setCharAt(i, (char)(j + '0'));                           sum += arr[pos] - arr[j - 1];                 break;             }         }     }           if (maxi == 0)     {         return "0";     }      else     {         return ans.toString();     } }   // Driver Code public static void main(String str[])  {     int arr[] = { 3, 4, 2, 4, 6,                   5, 4, 2, 3 };     int s = 13;           System.out.println(max_number(arr, s)); } }   // This code is contributed by rutvik_56

## Python3

 # Python3 implementation to find # minimum number which # have a maximum length   # Function to find the # minimum number which # have maximum length def max_number(arr, sum):           frac = *9     maxi = -10**9       pos = 0           # Find Maximum length     # of number     for i in range(9):         frac[i] = sum // arr[i]                   if (frac[i] > maxi):             pos = i             maxi = frac[i]       an = str((pos + 1)) * maxi           #print(an)     sum -= maxi * arr[pos]       ans = [i for i in an]       # Find minimum number WHich     # have maximum length     for i in range(maxi):         for j in range(1, 10):             if (sum + arr[pos] - arr[j - 1] >= 0):                 ans[i] = str(j)                 sum += arr[pos] - arr[j - 1]                 break       if (maxi == 0):         return 0     else:         return "".join(ans)   # Driver Code if __name__ == '__main__':           arr = [ 3, 4, 2, 4, 6,             5, 4, 2, 3 ]     s = 13           print(max_number(arr, s))   # This code is contributed by mohit kumar 29

## C#

 // C# implementation to find // minimum number which // have a maximum length using System; using System.Text; class GFG{   // Function to find the // minimum number which // have maximum length static String max_number(int []arr,                          int sum)  {   int []frac = new int;   int maxi = int.MinValue;   StringBuilder ans =                  new StringBuilder();   int pos = 0;     // Find Maximum length   // of number   for(int i = 0; i < 9; i++)   {     frac[i] = sum / arr[i];     if (frac[i] > maxi)     {       pos = i;       maxi = frac[i];     }   }     for(int i = 0; i < maxi; i++)   {     ans.Append((char)((pos + 1) + '0'));   }   sum -= maxi * arr[pos];     // Find minimum number WHich   // have maximum length   for(int i = 0; i < maxi; i++)   {     for(int j = 1; j <= 9; j++)     {       if (sum + arr[pos] -            arr[j - 1] >= 0)       {         ans[i] = (char)(j + '0');           sum += arr[pos] - arr[j - 1];         break;       }     }   }     if (maxi == 0)   {     return "0";   }    else   {     return ans.ToString();   } }   // Driver Code public static void Main(String []str)  {   int []arr = {3, 4, 2, 4, 6,                5, 4, 2, 3};   int s = 13;   Console.WriteLine(max_number(arr, s)); } }   // This code is contributed by 29AjayKumar

Output:

133333



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