Total number of non-decreasing numbers with n digits - GeeksforGeeks
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.
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.
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
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;
}
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
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;
}
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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