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Counting numbers of n digits that are monotone
• Difficulty Level : Medium
• Last Updated : 22 May, 2019

Call decimal number a monotone if:

.

Write a program which takes positive number n on input and returns number of decimal numbers of length n that are monotone. Numbers can’t start with 0.

Examples :

Input : 1
Output : 9
Numbers are 1, 2, 3, ... 9

Input : 2
Output : 45
Numbers are 11, 12, 13, .... 22, 23
...29, 33, 34, ... 39.
Count is 9 + 8 + 7 ... + 1 = 45


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Explanation: Let’s start by example of monotone numbers:

All those numbers are monotone as each digit on higher place is than the one before it.
What are the monotone numbers are of length 1 and digits 1 or 2? It is question to ask yourself at the very beginning. We can see that possible numbers are:

That was easy, now lets expand the question to digits 1, 2 and 3:

Now different question, what are the different monotone numbers consisting of only 1 and length 3 are there?

Lets try now draw this very simple observation in 2 dimensional array for number of length 3, where first column is the length of string and first row is possible digits:

Let’s try to fill 3rd row 3rd column(number of monotone numbers consisting from numbers 1 or 2 with length 2). This should be:
If we will look closer we already have subsets of this set i.e:
– Monotone numbers that has length 2 and consist of 1 or 2
– Monotone numbers of length 2 and consisting of number 2

We just need to add previous values to get the longer one.
Final matrix should look like this:

## C++

 // CPP program to count numbers of n digits// that are  monotone.#include #include   // Considering all possible digits as// {1, 2, 3, ..9}int static const DP_s = 9;  int getNumMonotone(int len){      // DP[i][j] is going to store monotone    // numbers of length i+1 considering    // j+1 digits.    int DP[len][DP_s];    memset(DP, 0, sizeof(DP));      // Unit length numbers    for (int i = 0; i < DP_s; ++i)        DP[0][i] = i + 1;      // Single digit numbers    for (int i = 0; i < len; ++i)        DP[i][0] = 1;      // Filling rest of the entries in bottom    // up manner.    for (int i = 1; i < len; ++i)        for (int j = 1; j < DP_s; ++j)            DP[i][j] = DP[i - 1][j] + DP[i][j - 1];      return DP[len - 1][DP_s - 1];}  // Driver code.int main(){    std::cout << getNumMonotone(10);    return 0;}

## Java

 // Java program to count numbers // of n digits that are monotone.  class GFG {    // Considering all possible     // digits as {1, 2, 3, ..9}    static final int DP_s = 9;          static int getNumMonotone(int len)    {              // DP[i][j] is going to store         // monotone numbers of length         // i+1 considering j+1 digits.        int[][] DP = new int[len][DP_s];              // Unit length numbers        for (int i = 0; i < DP_s; ++i)            DP[0][i] = i + 1;              // Single digit numbers        for (int i = 0; i < len; ++i)            DP[i][0] = 1;              // Filling rest of the entries         // in bottom up manner.        for (int i = 1; i < len; ++i)            for (int j = 1; j < DP_s; ++j)                DP[i][j] = DP[i - 1][j]                            + DP[i][j - 1];              return DP[len - 1][DP_s - 1];    }          // Driver code.    public static void main (String[] args)     {        System.out.println(getNumMonotone(10));    }}  // This code is contributed by Ansu Kumari.

## Python3

 # Python3 program to count numbers of n # digits that are monotone.  # Considering all possible digits as# {1, 2, 3, ..9}DP_s = 9  def getNumMonotone(ln):      # DP[i][j] is going to store monotone    # numbers of length i+1 considering    # j+1 digits.    DP = [[0]*DP_s for i in range(ln)]      # Unit length numbers    for i in range(DP_s):        DP[0][i] = i + 1      # Single digit numbers    for i in range(ln):        DP[i][0] = 1      # Filling rest of the entries      # in bottom up manner.    for i in range(1, ln):          for j in range(1, DP_s):            DP[i][j] = DP[i - 1][j] + DP[i][j - 1]      return DP[ln - 1][DP_s - 1]    # Driver codeprint(getNumMonotone(10))    # This code is contributed by Ansu Kumari

## C#

 // C# program to count numbers // of n digits that are monotone.using System;  class GFG {    // Considering all possible     // digits as {1, 2, 3, ..9}    static int DP_s = 9;          static int getNumMonotone(int len)    {              // DP[i][j] is going to store         // monotone numbers of length         // i+1 considering j+1 digits.        int[,] DP = new int[len,DP_s];              // Unit length numbers        for (int i = 0; i < DP_s; ++i)            DP[0,i] = i + 1;              // Single digit numbers        for (int i = 0; i < len; ++i)            DP[i,0] = 1;              // Filling rest of the entries         // in bottom up manner.        for (int i = 1; i < len; ++i)            for (int j = 1; j < DP_s; ++j)                DP[i,j] = DP[i - 1,j]                         + DP[i,j - 1];              return DP[len - 1,DP_s - 1];    }          // Driver code.    public static void Main ()     {        Console.WriteLine(getNumMonotone(10));    }}  // This code is contributed by vt_m.

## PHP

 

Output :
43758


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