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:
- Convert the number to its Binary Coded Decimal form.
- Concatenate all the resulting nibbles to form a binary number.
- Remove the least significant N bits from the number.
- Convert this obtained binary number to its decimal form.
Input: N = 4
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
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:
- Each nibble in BCD does not increase beyond 1001 which is 9 in binary form, since the maximum single digit decimal number is 9.
- 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.
- 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.
- 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.
- 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.
- 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:
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.