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# Count of all n digit numbers with digits in non-decreasing order

Published on 11 Mar 2020

A non-decreasing number is a number whose digits are in non-decreasing order such as 11 12 22 34 33 etc., Given a number n which is the length of the number, print the count of all n digit numbers with the digits in non-decreasing order.

For example,

Input: n = 1

Output: count = 10

Explanation: The numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

Input: n = 5

Output: count = 2002

## Algorithm to find the count of all n digit numbers with digits in non-decreasing order

• One way to investigate the problem is to have the number of numbers equal to the number of n-digit numbers ending with 9 plus the number with the number 8 and the number for 7, and so on.

• To get the count with a specific number return for n-1 length and digits less than or equal to the last digit. Below is the recursive formula of the approach, • The above recursive solution contains overlapping subproblems increasing the time complexity.

• Therefore, use Dynamic Programming approach to build a table in a bottom-up manner.

• The solution using Dynamic Programming is given below.

C++

Output
Input - 5 Output - 2002

Test Case 1

I/P

5

O/P

2002

Test Case 2

I/P

10

O/P

92378

Test Case 3

I/P

25

O/P

52451256

Test Case 4

I/P

4

O/P

715

Test Case 5

I/P

1

O/P

10

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