Given an integer n > 0, which denotes the number of digits, the task to find total number of n-digit positive integers which are non-decreasing in nature.
A non-decreasing integer is a one in which all the digits from left to right are in non-decreasing form. ex: 1234, 1135, ..etc.
Note :Leading zeros also count in non-decreasing integers such as 0000, 0001, 0023, etc are also non-decreasing integers of 4-digits.
Examples:
Input : n = 1
Output : 10
Numbers are 0, 1, 2, ...9.
Input : n = 2
Output : 55
Input : n = 4
Output : 715
where n<10^3
Good day to you,
Imho some DP could be used for this purposes. The parameters might be [DigitsRemaining <=10^3][LastUsedDigit <10]. In body, we could simply try all digits betweer <LastUsedDigit,9>.
We call such DP with [n][0].
If they want the output "by modulo" then it shall be easy like this with complexity A(N*10*10). In other case some efficient BigInteger structure shall be used (but It might also pass, since there won't be more than 10^(10^3) numbers :P)
Good Luck & Have Nice Day!
can you explain more sir?
What part exactly?
It might look like following pseudocode: (+/-)
So some recursion on such base.
GL
Let's observe that the final number will have some digits 0 (possibly zero), some digits 1 (possibly zero), some digits 2 (possibly zero), etc.. In total, there will be n digits and they must be in non-decreasing order. For any combination of digits, there will be only one number that satisfies the non-decreasing condition, so let's see the problem in this way: We have 10 boxes and we have to distribute n balls into them. This is a fairly known problem and the answer in this case is .