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

**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.

- 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.
- 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 <bits/stdc++.h> ` `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

`<?php ` `// PHP program for counting ` `// n digit numbers with ` `// non decreasing digits ` ` ` `// Returns count of non- ` `// decreasing numbers with ` `// n digits. ` ` ` `function` `nonDecNums(` `$n` `) ` `{ ` ` ` `/* a[i][j] = count of ` ` ` `all possible number ` ` ` `with i digits having ` ` ` `leading digit as j */` ` ` ` ` `// Initialization of ` ` ` `// all 0-digit number ` ` ` `for` `(` `$i` `= 0; ` `$i` `<= 9; ` `$i` `++) ` ` ` `$a` `[0][` `$i` `] = 1; ` ` ` ` ` `/* Initialization of all ` ` ` `i-digit non-decreasing ` ` ` `number leading with 9*/` ` ` `for` `(` `$i` `= 1; ` `$i` `<= ` `$n` `; ` `$i` `++) ` ` ` `$a` `[` `$i` `][9] = 1; ` ` ` ` ` `/* for all digits we should ` ` ` `calculate number of ways ` ` ` `depending upon leading digits*/` ` ` `for` `(` `$i` `= 1; ` `$i` `<= ` `$n` `; ` `$i` `++) ` ` ` `for` `(` `$j` `= 8; ` `$j` `>= 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).

This article is contributed by **Shivam Pradhan (anuj_charm)**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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