# Number of decimal numbers of length k, that are strict monotone

• Difficulty Level : Easy
• Last Updated : 26 Jul, 2018

We 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 strict monotone. Number can’t start with 0.

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Examples :

Input : 2
Output : 36
Numbers are 12, 13, ... 19, 23
24, ... 29, .... 89.

Input : 3
Output : 84


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

Explanations of this problem follows the same rules as applied on:
Number of decimal numbers of length k, that are monotone

The only difference is that now we cannot take duplicates, so previously computed values are the one on the left and left top diagonal.

## C++

 // CPP program to count numbers of k// digits that are strictly monotone.#include #include   int static const DP_s = 9;  int getNumStrictMonotone(int len){    // DP[i][j] is going to store monotone    // numbers of length i+1 considering    // j+1 digits (1, 2, 3, ..9)    int DP[len][DP_s];    memset(DP, 0, sizeof(DP));       // Unit length numbers    for (int i = 0; i < DP_s; ++i)         DP[i] = i + 1;          // Building dp[] in bottom up    for (int i = 1; i < len; ++i)         for (int j = 1; j < DP_s; ++j)             DP[i][j] = DP[i - 1][j - 1] + DP[i][j - 1];                  return DP[len - 1][DP_s - 1];}  // Driver codeint main(){    std::cout << getNumStrictMonotone(2);     return 0;}

## Java

 // Java program to count numbers of k// digits that are strictly monotone.import java.io.*;import java.util.*;  class GFG {      static int DP_s = 9;          static int getNumStrictMonotone(int len)     {        // DP[i][j] is going to store monotone        // numbers of length i+1 considering        // j+1 digits (1, 2, 3, ..9)        int[][] DP = new int[len][DP_s];              // Unit length numbers        for (int i = 0; i < DP_s; ++i)        DP[i] = i + 1;              // Building dp[] in bottom up        for (int i = 1; i < len; ++i)             for (int j = 1; j < DP_s; ++j)                DP[i][j] = DP[i - 1][j - 1]                              + DP[i][j - 1];              return DP[len - 1][DP_s - 1];    }    public static void main(String[] args)     {        int n = 2;        System.out.println(getNumStrictMonotone(n));    }}  // This code is contributed by Gitanjali.

## Python3

 # Python3 program to count numbers of k# digits that are strictly monotone.  DP_s = 9  def getNumStrictMonotone(ln):          # DP[i][j] is going to store monotone    # numbers of length i+1 considering    # j+1 digits (1, 2, 3, ..9)    DP = [ * DP_s for _ in range(ln)]      # Unit length numbers    for i in range(DP_s):        DP[i] = i + 1       # Building dp[] in bottom up    for i in range(1, ln):                  for j in range(1, DP_s):                          DP[i][j] = DP[i - 1][j - 1] + DP[i][j - 1]               return DP[ln - 1][DP_s - 1]  # Driver codeprint(getNumStrictMonotone(2))    # This code is contributed by Ansu Kumari.

## C#

 // C# program to count numbers of k// digits that are strictly monotone.using System;  class GFG {      static int DP_s = 9;          static int getNumStrictMonotone(int len)     {        // DP[i][j] is going to store monotone        // numbers of length i+1 considering        // j+1 digits (1, 2, 3, ..9)        int[,] DP = new int[len,DP_s];              // Unit length numbers        for (int i = 0; i < DP_s; ++i)        DP[0,i] = i + 1;              // Building dp[] in bottom up        for (int i = 1; i < len; ++i)            for (int j = 1; j < DP_s; ++j)                DP[i,j] = DP[i - 1,j - 1]                             + DP[i,j - 1];              return DP[len - 1,DP_s - 1];    }          // Driver code    public static void Main()     {        int n = 2;        Console.WriteLine(getNumStrictMonotone(n));    }}  // This code is contributed by vt_m.

## PHP

 

Output :
36


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