# Number of n-digits non-decreasing integers

Given an integer n > 0, which denotes the number of digits, the task to find the total number of n-digit positive integers which are non-decreasing in nature.

A non-decreasing integer is 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 numbers 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, the 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++ 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` `?>` |

## Javascript

`<script>` ` ` `// Javascript program for counting n digit` ` ` `// numbers with non decreasing digits` ` ` ` ` `// Function that 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` ` ` `let a = ` `new` `Array(n + 1)` ` ` `for` `(let i = 0; i < n + 1; i++)` ` ` `{` ` ` `a[i] = ` `new` `Array(10);` ` ` `}` ` ` ` ` `// Initialization of all 0-digit number` ` ` `for` `(let i = 0; i <= 9; i++)` ` ` `a[0][i] = 1;` ` ` ` ` `// Initialization of all i-digit` ` ` `// non-decreasing number leading with 9` ` ` `for` `(let i = 1; i <= n; i++)` ` ` `a[i][9] = 1;` ` ` ` ` `// for all digits we should calculate` ` ` `// number of ways depending upon leading` ` ` `// digits` ` ` `for` `(let i = 1; i <= n; i++)` ` ` `for` `(let j = 8; j >= 0; j--)` ` ` `a[i][j] = a[i - 1][j] + a[i][j + 1];` ` ` ` ` `return` `a[n][0];` ` ` `}` ` ` ` ` `let n = 2;` ` ` `document.write(` ` ` `"Non-decreasing digits = "` `+ nonDecNums(n)` ` ` `);` ` ` `</script>` |

**Output :**

Non-decreasing digits = 55

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

**Another Approach: **

If we observe, we can see that 0 has to be placed before 1-9, 1 has to be placed before 2-9 and so on. As we are asked to find non-decreasing integers, 111223 is a valid non-decreasing integer which means same digit can occur conscuetively.

__Example 1____:__ When N=2, we have 11C9, which is equal to **55.**

** Example 2**: When N=5, we have 14C9, which is equal to

**2002.**

## C++

`// CPP program To calculate Number of n-digits non-decreasing integers` `//Contributed by Parishrut Kushwaha//` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Returns factorial of n` `long` `long` `int` `fact(` `int` `n)` `{` ` ` `long` `long` `int` `res = 1;` ` ` `for` `(` `int` `i = 2; i <= n; i++)` ` ` `res = res * i;` ` ` `return` `res;` `}` ` ` `// returns nCr` `long` `long` `int` `nCr(` `int` `n, ` `int` `r)` `{` ` ` `return` `fact(n) / (fact(r) * fact(n - r));` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 2;` ` ` `cout <<` `"Number of Non-Decreasing digits: "` `<< nCr(n+9,9);` ` ` `return` `0;` `}` |

**Output:**

Number of Non-Decreasing digits: 24310

**Time Complexity : **O(n).

**Space Complexity: **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 review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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