# Total number of non-decreasing numbers with n digits

• Difficulty Level : Medium
• Last Updated : 17 Feb, 2022

A number is non-decreasing if every digit (except the first one) is greater than or equal to previous digit. For example, 223, 4455567, 899, are non-decreasing numbers.
So, given the number of digits n, you are required to find the count of total non-decreasing numbers with n digits.
Examples:

```Input:  n = 1
Output: count  = 10

Input:  n = 2
Output: count  = 55

Input:  n = 3
Output: count  = 220```

We strongly recommend you to minimize your browser and try this yourself first.
One way to look at the problem is, count of numbers is equal to count n digit number ending with 9 plus count of ending with digit 8 plus count for 7 and so on. How to get count ending with a particular digit? We can recur for n-1 length and digits smaller than or equal to the last digit. So below is recursive formula.

```Count of n digit numbers = (Count of (n-1) digit numbers Ending with digit 9) +
(Count of (n-1) digit numbers Ending with digit 8) +
.............................................+
.............................................+
(Count of (n-1) digit numbers Ending with digit 0) ```

Let count ending with digit ‘d’ and length n be count(n, d)

```count(n, d) = ∑(count(n-1, i)) where i varies from 0 to d

Total count = ∑count(n-1, d) where d varies from 0 to n-1```

The above recursive solution is going to have many overlapping subproblems. Therefore, we can use Dynamic Programming to build a table in bottom up manner.
Below is the implementation of above idea :

## C++

 `// C++ program to count non-decreasing number with n digits``#include``using` `namespace` `std;` `long` `long` `int` `countNonDecreasing(``int` `n)``{``    ``// dp[i][j] contains total count of non decreasing``    ``// numbers ending with digit i and of length j``    ``long` `long` `int` `dp[n+1];``    ``memset``(dp, 0, ``sizeof` `dp);` `    ``// Fill table for non decreasing numbers of length 1``    ``// Base cases 0, 1, 2, 3, 4, 5, 6, 7, 8, 9``    ``for` `(``int` `i = 0; i < 10; i++)``        ``dp[i] = 1;` `    ``// Fill the table in bottom-up manner``    ``for` `(``int` `digit = 0; digit <= 9; digit++)``    ``{``        ``// Compute total numbers of non decreasing``        ``// numbers of length 'len'``        ``for` `(``int` `len = 2; len <= n; len++)``        ``{``            ``// sum of all numbers of length of len-1``            ``// in which last digit x is <= 'digit'``            ``for` `(``int` `x = 0; x <= digit; x++)``                ``dp[digit][len] += dp[x][len-1];``        ``}``    ``}` `    ``long` `long` `int` `count = 0;` `    ``// There total nondecreasing numbers of length n``    ``// won't be dp[n] +  dp[n] ..+ dp[n]``    ``for` `(``int` `i = 0; i < 10; i++)``        ``count += dp[i][n];` `    ``return` `count;``}` `// Driver program``int` `main()``{``    ``int` `n = 3;``    ``cout << countNonDecreasing(n);``    ``return` `0;``}`

## Java

 `class` `NDN``{``    ``static` `int` `countNonDecreasing(``int` `n)``    ``{``        ``// dp[i][j] contains total count of non decreasing``        ``// numbers ending with digit i and of length j``        ``int` `dp[][] = ``new` `int``[``10``][n+``1``];``     ` `        ``// Fill table for non decreasing numbers of length 1``        ``// Base cases 0, 1, 2, 3, 4, 5, 6, 7, 8, 9``        ``for` `(``int` `i = ``0``; i < ``10``; i++)``            ``dp[i][``1``] = ``1``;``     ` `        ``// Fill the table in bottom-up manner``        ``for` `(``int` `digit = ``0``; digit <= ``9``; digit++)``        ``{``            ``// Compute total numbers of non decreasing``            ``// numbers of length 'len'``            ``for` `(``int` `len = ``2``; len <= n; len++)``            ``{``                ``// sum of all numbers of length of len-1``                ``// in which last digit x is <= 'digit'``                ``for` `(``int` `x = ``0``; x <= digit; x++)``                    ``dp[digit][len] += dp[x][len-``1``];``            ``}``        ``}``     ` `        ``int` `count = ``0``;``     ` `        ``// There total nondecreasing numbers of length n``        ``// won't be dp[n] +  dp[n] ..+ dp[n]``        ``for` `(``int` `i = ``0``; i < ``10``; i++)``            ``count += dp[i][n];``     ` `        ``return` `count;``    ``}``    ``public` `static` `void` `main(String args[])``    ``{``       ``int` `n = ``3``;``       ``System.out.println(countNonDecreasing(n));``    ``}``}``/* This code is contributed by Rajat Mishra */`

## Python3

 `# Python3 program to count``# non-decreasing number with n digits``def` `countNonDecreasing(n):``    ` `    ``# dp[i][j] contains total count``    ``# of non decreasing numbers ending``    ``# with digit i and of length j``    ``dp ``=` `[[``0` `for` `i ``in` `range``(n ``+` `1``)]``             ``for` `i ``in` `range``(``10``)]``             ` `    ``# Fill table for non decreasing``    ``# numbers of length 1.``    ``# Base cases 0, 1, 2, 3, 4, 5, 6, 7, 8, 9` `    ``for` `i ``in` `range``(``10``):``        ``dp[i][``1``] ``=` `1` `    ``# Fill the table in bottom-up manner``    ``for` `digit ``in` `range``(``10``):``        ` `        ``# Compute total numbers of non``        ``# decreasing numbers of length 'len'``        ``for` `len` `in` `range``(``2``, n ``+` `1``):``            ` `            ``# sum of all numbers of length``            ``# of len-1 in which last``            ``# digit x is <= 'digit'``            ``for` `x ``in` `range``(digit ``+` `1``):``                ``dp[digit][``len``] ``+``=` `dp[x][``len` `-` `1``]``    ``count ``=` `0``    ` `    ``# There total nondecreasing numbers``    ``# of length n won't be dp[n] +``    ``# dp[n] ..+ dp[n]``    ``for` `i ``in` `range``(``10``):``        ``count ``+``=` `dp[i][n]``    ``return` `count``    ` `# Driver Code``n ``=` `3``print``(countNonDecreasing(n))` `# This code is contributed``# by sahilshelangia`

## C#

 `// C# program to print sum``// triangle for a given array``using` `System;` `class` `GFG {``    ` `    ``static` `int` `countNonDecreasing(``int` `n)``    ``{``        ``// dp[i][j] contains total count``        ``// of non decreasing numbers ending``        ``// with digit i and of length j``        ``int` `[,]dp = ``new` `int``[10,n + 1];``    ` `        ``// Fill table for non decreasing``        ``// numbers of length 1 Base cases``        ``// 0, 1, 2, 3, 4, 5, 6, 7, 8, 9``        ``for` `(``int` `i = 0; i < 10; i++)``            ``dp[i, 1] = 1;``    ` `        ``// Fill the table in bottom-up manner``        ``for` `(``int` `digit = 0; digit <= 9; digit++)``        ``{``            ` `            ``// Compute total numbers of non decreasing``            ``// numbers of length 'len'``            ``for` `(``int` `len = 2; len <= n; len++)``            ``{``                ` `                ``// sum of all numbers of length of len-1``                ``// in which last digit x is <= 'digit'``                ``for` `(``int` `x = 0; x <= digit; x++)``                    ``dp[digit, len] += dp[x, len - 1];``            ``}``        ``}``    ` `        ``int` `count = 0;``    ` `        ``// There total nondecreasing numbers``        ``// of length n won't be dp[n]``        ``// + dp[n] ..+ dp[n]``        ``for` `(``int` `i = 0; i < 10; i++)``            ``count += dp[i, n];``    ` `        ``return` `count;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 3;``        ``Console.WriteLine(countNonDecreasing(n));``    ``}``}` `// This code is contributed by Sam007.`

## PHP

 ``

## Javascript

 ``

Output:

`220`

Thanks to Gaurav Ahirwar for suggesting above method.
Another method is based on below direct formula

```Count of non-decreasing numbers with n digits =
N*(N+1)/2*(N+2)/3* ....*(N+n-1)/n
Where N = 10```

Below is the program to compute count using above formula.

## C++

 `// C++ program to count non-decreasing number with n digits``#include``using` `namespace` `std;` `long` `long` `int` `countNonDecreasing(``int` `n)``{``    ``int` `N = 10;` `    ``// Compute value of N*(N+1)/2*(N+2)/3* ....*(N+n-1)/n``    ``long` `long` `count = 1;``    ``for` `(``int` `i=1; i<=n; i++)``    ``{``        ``count *= (N+i-1);``        ``count /= i;``    ``}` `    ``return` `count;``}` `// Driver program``int` `main()``{``    ``int` `n = 3;``    ``cout << countNonDecreasing(n);``    ``return` `0;``}`

## Java

 `// java program to count non-decreasing``// number with n digits``public` `class` `GFG {``    ` `    ``static` `long` `countNonDecreasing(``int` `n)``    ``{``        ``int` `N = ``10``;``     ` `        ``// Compute value of N * (N+1)/2 *``        ``// (N+2)/3 * ....* (N+n-1)/n``        ``long` `count = ``1``;``         ` `        ``for` `(``int` `i = ``1``; i <= n; i++)``        ``{``            ``count *= (N + i - ``1``);``            ``count /= i;``        ``}``     ` `        ``return` `count;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String args[]) {``        ` `        ``int` `n = ``3``;``        ``System.out.print(countNonDecreasing(n));``    ``}  ``}` `// This code is contributed by Sam007.`

## Python3

 `# python program to count non-decreasing``# number with n digits` `def` `countNonDecreasing(n):``    ``N ``=` `10` `    ``# Compute value of N*(N+1)/2*(N+2)/3``    ``# * ....*(N+n-1)/n``    ``count ``=` `1``    ``for` `i ``in` `range``(``1``, n``+``1``):``        ``count ``=` `int``(count ``*` `(N``+``i``-``1``))``        ``count ``=` `int``(count ``/` `i )``        ` `    ``return` `count` `# Driver program``n ``=` `3``;``print``(countNonDecreasing(n))``    ` `# This code is contributed by Sam007`

## C#

 `// C# program to count non-decreasing``// number with n digits``using` `System;` `class` `GFG {``    ` `    ``static` `long` `countNonDecreasing(``int` `n)``    ``{``        ``int` `N = 10;``    ` `        ``// Compute value of N * (N+1)/2 *``        ``// (N+2)/3 * ....* (N+n-1)/n``        ``long` `count = 1;``        ` `        ``for` `(``int` `i = 1; i <= n; i++)``        ``{``            ``count *= (N + i - 1);``            ``count /= i;``        ``}``    ` `        ``return` `count;``    ``}` `    ` `    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 3;``        ` `        ``Console.WriteLine(countNonDecreasing(n));``    ``}``}` `// This code is contributed by Sam007.`

## PHP

 ``

## Javascript

 ``

Output:

`220`

Thanks to Abhishek Somani for suggesting this method.
How does this formula work?

```N * (N+1)/2 * (N+2)/3 * .... * (N+n-1)/n
Where N = 10 ```

Let us try for different values of n.

```For n = 1, the value is N from formula.
Which is true as for n = 1, we have all single digit
numbers, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

For n = 2, the value is N(N+1)/2 from formula
We can have N numbers beginning with 0, (N-1) numbers
beginning with 1, and so on.
So sum is N + (N-1) + .... + 1 = N(N+1)/2

For n = 3, the value is N(N+1)/2(N+2)/3 from formula
We can have N(N+1)/2 numbers beginning with 0, (N-1)N/2
numbers beginning with 1 (Note that when we begin with 1,
we have N-1 digits left to consider for remaining places),
(N-2)(N-1)/2 beginning with 2, and so on.
Count = N(N+1)/2 + (N-1)N/2 + (N-2)(N-1)/2 +
(N-3)(N-2)/2 .... 3 + 1
[Combining first 2 terms, next 2 terms and so on]
= 1/2[N2 + (N-2)2 + .... 4]
= N*(N+1)*(N+2)/6  [Refer this , putting n=N/2 in the
even sum formula]```

For general n digit case, we can apply Mathematical Induction. The count would be equal to count n-1 digit beginning with 0, i.e., N*(N+1)/2*(N+2)/3* ….*(N+n-1-1)/(n-1). Plus count of n-1 digit numbers beginning with 1, i.e., (N-1)*(N)/2*(N+1)/3* ….*(N-1+n-1-1)/(n-1) (Note that N is replaced by N-1) and so on.
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