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 maximum count of numbers from the array such that sum of numbers is less than S and form the smallest possible number using their indices

Note: Any element can be choosen 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 \frac{Sum}{X}. Finally, the minimum indices that can be choose multiple times is computed by taking the minimum digit in the number for each digit place.



Below is the implementation of the above approach:

C++

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// C++ implementation to find
// minimum number which
// have a maximum length
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the
// minimum number which
// have maximum length
string 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 Code
int main()
{
    int arr[9] = { 3, 4, 2, 4, 6,
                   5, 4, 2, 3 };
    int s = 13;
    cout << max_number(arr, s);
    return 0;
}

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Java

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// 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[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 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

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Output:

133333

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