Q:

Given the number of digits n, find the count of total non-decreasing numbers with n digits

mahmoud2506
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2022-05-01
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Total number of non-decreasing numbers with n digits using dynamic programming

Given the number of digits n, find the count of total non-decreasing numbers with n digits.

A number is non-decreasing if every digit (except the first one) is greater than or equal to the previous digit. For example, 22,223, 45567, 899, are non-decreasing numbers whereas 321 or 322 are not.

Input:

The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. The first line of each test case contains the integer n.

Output:

Print the count of total non-decreasing numbers with n digits for each test case in a new line. You may need to use unsigned long long int as the count can be very large.

Constraints:

1 <= T <= 100
1 <= n <= 200

Example:

Input:
No of test cases, 3

n=1
n=2
n=3

Output:
For n=1
Total count is: 10

For n=2
Total count is: 55

For n=3
Total count is: 220

Explanation:

For n=1,
The non-decreasing numbers are basically 0 to 9, 
counting up to 10

For, n=2,
The non-decreasing numbers can be
00
01
02
..
11
12
13
14
15
.. so on total 55
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All Answers

mahmoud2506
2022-05-01
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The solution can be recursive. In recursion our strategy will be to build the string (read: number) such that at each recursive call the number to be appended would be necessarily bigger (or equal) than(to) the last one.

Say, at any recursion call,

The number already constructed is x1x2...xi where i<n, So at the recursive call we are allowed to append digits only which are equal to greater to xi.

So, let's formulate the recursion

Say, the recursive function is computerecur(int index, int last, int n)

Where,

  • index = current position to append digit
  • Last = the previous digit
  • n = number of total digits
unsigned long long int computerecur (int index,int last,int n){
    // base case
    if(index has reached the last digit)
        return 1;
    
    unsigned long long int sum=0;
    for digit to append at current index,i=0 to 9
        if(i>=last)
            // recur if I can be appended
            sum + =computerecur(index+1,i,n); 
    end for
    return sum    
End function

So the basic idea to the recursion is that if it's a valid digit to append (depending on the last digit) then append it and recur for the remaining digits.

Now the call from the main() function should be done by appending the first digit already (basically we will keep that first digit as last digit position for the recursion call).

So at main,

We will call as,

unsigned long long int result=0;
    for i=0 to 9 //starting digits
        // index=0, starting digit assigned as 
        // last digit for recursion
        result+=computerecur(0,i,n); 
end for

The result is the ultimate result.

I would suggest you draw the recursion tree to have a better understanding, Take n=3 and do yourself.

For, n=2, I will brief the tree below

For starting digit 0
Computerecur(0,0,2) 

Index!=n-1
So
Goes to the loop
And then 
It calls to
Computerecur(1,0,2) // it's for number 00
Computerecur(1,1,2) // it's for number 01
Computerecur(1,2,2) // it's for number 02
Computerecur(1,3,2) // it's for number 03
Computerecur(1,4,2) // it's for number 04
Computerecur(1,5,2) // it's for number 05
Computerecur(1,6,2) // it's for number 06
Computerecur(1,7,2) // it's for number 07
Computerecur(1,8,2) // it's for number 08
Computerecur(1,9,2) // it's for number 09
So on
...

Now, it's pretty easy to infer that it leads to many overlapping sub-problem and hence we need dynamic programming to store the results of overlapping sub-problems. That's why I have used the memoization technique to store the already computed sub-problem results. See the below implementation to understand the memorization part.

C++ Implementation:

#include <bits/stdc++.h>
using namespace std;

unsigned long long int dp[501][10];

unsigned long long int my(int index, int last, int n)
{

    if (index == n - 1)
        return 1;

    // memorization, don't compute again what is already computed
    if (dp[index][last] != -1) 
        return dp[index][last];

    unsigned long long int sum = 0;
    for (int i = 0; i <= 9; i++) {
        if (i >= last)
            sum += my(index + 1, i, n);
    }

    dp[index][last] = sum;
    return dp[index][last];
}

unsigned long long int compute(int n)
{
    unsigned long long int sum = 0;
    for (int i = 0; i <= 9; i++) {
        sum += my(0, i, n);
    }
    return sum;
}
int main()
{

    int t, n, item;
    cout << "enter number of testcase\n";
    scanf("%d", &t);

    for (int i = 0; i < t; i++) {
        cout << "Enter the number of digits,n:\n";
        scanf("%d", &n);
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= 9; j++) {
                dp[i][j] = -1;
            }
        }

        cout << "number of non-decreasing number with " << n;
        cout << " digits are: " << compute(n) << endl;
    }

    return 0;
}

Output:

enter number of testcase
3
Enter the number of digits,n:
2
number of non-decreasing number with 2 digits are: 55
Enter the number of digits,n:
3
number of non-decreasing number with 3 digits are: 220
Enter the number of digits,n:
6
number of non-decreasing number with 6 digits are: 5005

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