# Number of n-digits non-decreasing integers

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

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

Naive Approach : We generate all possible n-digit numbers and then for each number we check whether it is non-decreasing or not.

Time Complexity : (n*10^n), where 10^n is for generating all possible n-digits number and n is for checking whether a particular number is non-decreasing or not.

Dynamic Programming :
If we fill digits one by one from left to right, following conditions hold.

1. If current last digit is 9, we can fill only 9s in remaining places. So only one solution is possible if current last digit is 9.
2. If current last digit is less than 9, then we can recursively compute count using following formula.
a[i][j] = a[i-1][j] + a[i][j + 1]
For every digit j smaller than 9.

We consider previous length count and count
to be increased by all greater digits.

We build a matrix a[][] where a[i][j] = count of all valid i-digit non-decreasing integers with j or greater than j as the leading digit. The solution is based on below observations. We fill this matrix column-wise, first calculating a[1][9] then using this value to compute a[2][8] and so on.
At any instant if we wish to calculate a[i][j] means number of i-digits non-decreasing integers with leading digit as j or digit greater than j, we should add up a[i-1][j] (number of i-1 digit integers which should start from j or greater digit, because in this case if we place j as its left most digit then our number will be i-digit non-decreasing number) and a[i][j+1] (number of i-digit integers which should start with digit equals to greater than j+1). So, a[i][j] = a[i-1][j] + a[i][j+1].

## C/C++

 // C++ program for counting n digit numbers with // non decreasing digits #include using namespace std;    // Returns count of non- decreasing numbers with // n digits. int nonDecNums(int n) {     /* a[i][j] = count of all possible number        with i digits having leading digit as j */     int a[n + 1][10];        // Initialization of all 0-digit number     for (int i = 0; i <= 9; i++)         a[0][i] = 1;        /* Initialization of all i-digit       non-decreasing number leading with 9*/     for (int i = 1; i <= n; i++)         a[i][9] = 1;        /* for all digits we should calculate       number of ways depending upon leading       digits*/     for (int i = 1; i <= n; i++)         for (int j = 8; j >= 0; j--)             a[i][j] = a[i - 1][j] + a[i][j + 1];        return a[n][0]; }    // driver program int main() {     int n = 2;     cout << "Non-decreasing digits = "          << nonDecNums(n) << endl;     return 0; }

## Java

 // Java program for counting n digit numbers with // non decreasing digits import java.io.*;    class GFG {        // Function that returns count of non- decreasing numbers     // with n digits     static int nonDecNums(int n)     {         // a[i][j] = count of all possible number         // with i digits having leading digit as j         int[][] a = new int[n + 1][10];            // Initialization of all 0-digit number         for (int i = 0; i <= 9; i++)             a[0][i] = 1;            // Initialization of all i-digit         // non-decreasing number leading with 9         for (int i = 1; i <= n; i++)             a[i][9] = 1;            // for all digits we should calculate         // number of ways depending upon leading         // digits         for (int i = 1; i <= n; i++)             for (int j = 8; j >= 0; j--)                 a[i][j] = a[i - 1][j] + a[i][j + 1];            return a[n][0];     }        // driver program     public static void main(String[] args)     {         int n = 2;         System.out.println("Non-decreasing digits = " + nonDecNums(n));     } }    // Contributed by Pramod Kumar

## Python3

 # Python3 program for counting n digit  # numbers with non decreasing digits  import numpy as np    # Returns count of non- decreasing  # numbers with n digits.  def nonDecNums(n) :                # a[i][j] = count of all possible number      # with i digits having leading digit as j      a = np.zeros((n + 1, 10))         # Initialization of all 0-digit number      for i in range(10) :         a[0][i] = 1        # Initialization of all i-digit      # non-decreasing number leading with 9     for i in range(1, n + 1) :          a[i][9] = 1        # for all digits we should calculate      # number of ways depending upon      # leading digits     for i in range(1, n + 1) :         for j in range(8, -1, -1) :              a[i][j] = a[i - 1][j] + a[i][j + 1]        return int(a[n][0])     # Driver Code  if __name__ == "__main__" :         n = 2     print("Non-decreasing digits = ",                         nonDecNums(n))    # This code is contributed by Ryuga

## C#

 // C# function to find number of diagonals // in n sided convex polygon using System;    class GFG {            // Function that returns count of non-      // decreasing numbers with n digits     static int nonDecNums(int n)     {         // a[i][j] = count of all possible number         // with i digits having leading digit as j         int[, ] a = new int[n + 1, 10];            // Initialization of all 0-digit number         for (int i = 0; i <= 9; i++)             a[0, i] = 1;            // Initialization of all i-digit         // non-decreasing number leading with 9         for (int i = 1; i <= n; i++)             a[i, 9] = 1;            // for all digits we should calculate         // number of ways depending upon leading         // digits         for (int i = 1; i <= n; i++)             for (int j = 8; j >= 0; j--)                 a[i, j] = a[i - 1, j] + a[i, j + 1];            return a[n, 0];     }        // driver program     public static void Main()     {         int n = 2;         Console.WriteLine("Non-decreasing digits = " +                                         nonDecNums(n));     } }    // This code is contributed by Sam007

## PHP

 = 0; \$j--)             \$a[\$i][\$j] = \$a[\$i - 1][\$j] +                           \$a[\$i][\$j + 1];        return \$a[\$n][0]; }    // Driver Code \$n = 2; echo "Non-decreasing digits = ",             nonDecNums(\$n),"\n";    // This code is contributed by m_kit ?>

Output :

Non-decreasing digits = 55

Time Complexity : O(10*n) equivalent to O(n).