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 – 133333Input: 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 <bits/stdc++.h>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
<script>// Javascript implementation to find// minimum number which// have a maximum length// Function to find the// minimum number which// have maximum lengthfunction max_number(arr,sum)
{ let frac = new Array(9);
let maxi = Number.MIN_VALUE;
let ans = [];
let pos = 0;
// Find Maximum length
// of number
for(let i = 0; i < 9; i++)
{
frac[i] = Math.floor(sum / arr[i]);
if (frac[i] > maxi)
{
pos = i;
maxi = frac[i];
}
}
for(let i = 0; i < maxi; i++)
{
ans.push(String.fromCharCode((pos + 1) + '0'.charCodeAt(0)));
}
sum -= maxi * arr[pos];
// Find minimum number WHich
// have maximum length
for(let i = 0; i < maxi; i++)
{
for(let j = 1; j <= 9; j++)
{
if (sum + arr[pos] - arr[j - 1] >= 0)
{
ans[i] = String.fromCharCode((j + '0'.charCodeAt(0)));
sum += arr[pos] - arr[j - 1];
break;
}
}
}
if (maxi == 0)
{
return "0";
}
else
{
return ans.join("");
}
}// Driver Codelet arr = [3, 4, 2, 4, 6, 5, 4, 2, 3];
let s = 13;document.write(max_number(arr, s));// This code is contributed by unknown2108</script> |
Output:
133333