# Total number of non-decreasing numbers with n digits

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 Dynamic programming based C++ program.

## C++

```// C++ program to count non-decreasing number with n digits
#include<bits/stdc++.h>
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[10][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] = 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
// wiint be dp[0][n] +  dp[1][n] ..+ dp[9][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
// wiint be dp[0][n] +  dp[1][n] ..+ dp[9][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 */
```

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 a C++ program to compute count using above formula.

```// C++ program to count non-decreasing numner with n digits
#include<bits/stdc++.h>
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;
}
```

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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