Skip to content
Related Articles

Related Articles

Minimum N-Digit number required to obtain largest N-digit number after performing given operations
  • Difficulty Level : Medium
  • Last Updated : 12 Apr, 2021

Given a positive integer N, the task is to find the minimum N-digit number such that performing the following operations on it in the following order results into the largest N-digit number:

  1. Convert the number to its Binary Coded Decimal form.
  2. Concatenate all the resulting nibbles to form a binary number.
  3. Remove the least significant N bits from the number.
  4. Convert this obtained binary number to its decimal form.

Examples:

Input: N = 4
Output: 9990
Explanation: 
Largest 4 digit number = 9999 
BCD of 9999 = 1001 1001 1001 1001 
Binary form = 1001100110011001 
Replacing last 4 bits by 0000: 1001 1001 1001 0000 = 9990 
Therefore, the minimum N-digit number that can generate 9999 is 9990

Input: N = 5
Output: 99980
Explanation: 
Largest 5 digit number = 99999 
BCD of 99999 = 1001 1001 1001 1001 1001 
Binary for = 10011001100110011001 
Replacing last 5 bits by 00000: 10011001100110000000 = 99980 
Therefore, the minimum N-digit number that can generate 99999 is 99980

Approach: The probelm can be solved based on the following observations of BCD numbers. Follow the steps below to solve the problem: 
 



  1. Each nibble in BCD does not increase beyond 1001 which is 9 in binary form, since the maximum single digit decimal number is 9.
  2. Thus, it can be concluded that the maximum binary number that can be obtained by bringing N nibbles together is 1001 concatenated N times, whose decimal representation is have to be the digit 9 concatenated N times.
  3. The last N LSBs from this binary form is required to be removed. Thus the values of these bits will not contribute in making the result larger. Therefore, keeping the last N bits as 9 is not necessary as we need the minimum number producing the maximum result.
  4. The value of floor(N/4) will give us the number of nibbles that will be completely removed from the number. Assign these nibbles the value of 0000 to minimize the number.
  5. The remainder of N/4 gives us the number of digits that would be switched to 0 from the LSB of the last non-zero nibble after having performed the previous step.
  6. This BCD formed by performing the above steps, when converted to decimal, generates the required maximized N digit number.

Below is the implementation of the above approach:

C++




// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
void maximizedNdigit(int n)
{
 
    int count0s, count9s;
    // If n is divisible by 4
    if (n % 4 == 0) {
 
        count0s = n / 4;
        count9s = n - n / 4;
    }
 
    // Otherwise
    else {
 
        count0s = n / 4 + 1;
        count9s = n - count0s;
        count0s--;
    }
 
    while (count9s--)
        cout << '9';
 
    if (n % 4 != 0)
        cout << '8';
 
    while (count0s--)
        cout << '0';
    cout << endl;
}
 
// Driver Code
int main()
{
    int n = 5;
    maximizedNdigit(n);
}

Java




// Java program to implement
// the above approach
import java.io.*;
 
class GFG{
 
static void maximizedNdigit(int n)
{
    int count0s, count9s;
     
    // If n is divisible by 4
    if (n % 4 == 0)
    {
        count0s = n / 4;
        count9s = n - n / 4;
    }
 
    // Otherwise
    else
    {
        count0s = n / 4 + 1;
        count9s = n - count0s;
        count0s--;
    }
 
    while (count9s != 0)
    {
        count9s--;
        System.out.print('9');
    }
 
    if (n % 4 != 0)
        System.out.print('8');
 
    while (count0s != 0)
    {
        count0s--;
        System.out.print('0');
    }
     
    System.out.println();
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 5;
     
    maximizedNdigit(n);
}
}
 
// This code is contributed by sanjoy_62

Python3




# Python3 program to implement
# the above approach
def maximizedNdigit(n):
 
    # If n is divisible by 4
    if (n % 4 == 0):
        count0s = n // 4
        count9s = n - n // 4
     
    # Otherwise
    else:
        count0s = n // 4 + 1
        count9s = n - count0s
        count0s -= 1
     
    while (count9s):
        print('9', end = "")
        count9s -= 1
 
    if (n % 4 != 0):
        print('8', end = "")
 
    while (count0s):
        print('0', end = "")
        count0s -= 1
         
    print()
 
# Driver Code
if __name__ == "__main__":
 
    n = 5
    maximizedNdigit(n)
 
# This code is contributed by chitranayal

C#




// C# program to implement
// the above approach
using System;
 
class GFG{
 
static void maximizedNdigit(int n)
{
    int count0s, count9s;
     
    // If n is divisible by 4
    if (n % 4 == 0)
    {
        count0s = n / 4;
        count9s = n - n / 4;
    }
 
    // Otherwise
    else
    {
        count0s = n / 4 + 1;
        count9s = n - count0s;
        count0s--;
    }
 
    while (count9s != 0)
    {
        count9s--;
        Console.Write('9');
    }
 
    if (n % 4 != 0)
        Console.Write('8');
 
    while (count0s != 0)
    {
        count0s--;
        Console.Write('0');
    }
         
    Console.WriteLine();
}
 
// Driver Code
public static void Main()
{
    int n = 5;
     
    maximizedNdigit(n);
}
}
 
// This code is contributed by sanjoy_62

Javascript




<script>
 
// JavaScript Program to implement
// the above approach
 
function maximizedNdigit(n)
{
 
    let count0s, count9s;
    // If n is divisible by 4
    if (n % 4 == 0) {
 
        count0s = Math.floor(n / 4);
        count9s = n - Math.floor(n / 4);
    }
 
    // Otherwise
    else {
 
        count0s = Math.floor(n / 4) + 1;
        count9s = n - count0s;
        count0s--;
    }
 
    while (count9s--)
        document.write('9');
 
    if (n % 4 != 0)
        document.write('8');
 
    while (count0s--)
        document.write('0');
    document.write("<br>");
}
 
// Driver Code
 
    let n = 5;
    maximizedNdigit(n);
 
 
// This code is contributed by Surbhi Tyagi.
 
</script>
Output: 
99980

 

Time Complexity: O(N) 
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live and Geeks Classes Live USA

My Personal Notes arrow_drop_up
Recommended Articles
Page :