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

Last Updated : 16 Jun, 2021

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 lengthstring max_number(int arr[], int sum){    int frac[9];    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 Codeint main(){    int arr[9] = { 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 lengthclass GFG{ // Function to find the// minimum number which// have maximum lengthstatic String max_number(int arr[], int sum) {    int frac[] = new int[9];    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 Codepublic 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 lengthdef max_number(arr, sum):         frac = [0]*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 Codeif __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 lengthusing System;using System.Text;class GFG{ // Function to find the// minimum number which// have maximum lengthstatic String max_number(int []arr,                         int sum) {  int []frac = new int[9];  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 Codepublic 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

## Javascript

 

Output:
133333

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