A number is non-decreasing if every digit (except the first one) is greater than or equal to the 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
Method 1: 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 the above idea :
// 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
// won't 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;
} |
import java.io.*;
public 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[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 */
|
# 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[0][n] +
# dp[1][n] ..+ dp[9][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# 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[0][n]
// + dp[1][n] ..+ dp[9][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 // PHP program to count non-decreasing number with n digits function countNonDecreasing( $n )
{ // dp[i][j] contains total count of non decreasing
// numbers ending with digit i and of length j
$dp = array_fill (0,10, array_fill (0, $n +1,NULL));
// Fill table for non decreasing numbers of length 1
// Base cases 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
for ( $i = 0; $i < 10; $i ++)
$dp [ $i ][1] = 1;
// Fill the table in bottom-up manner
for ( $digit = 0; $digit <= 9; $digit ++)
{
// Compute total numbers of non decreasing
// numbers of length 'len'
for ( $len = 2; $len <= $n ; $len ++)
{
// sum of all numbers of length of len-1
// in which last digit x is <= 'digit'
for ( $x = 0; $x <= $digit ; $x ++)
$dp [ $digit ][ $len ] += $dp [ $x ][ $len -1];
}
}
$count = 0;
// There total nondecreasing numbers of length n
// won't be dp[0][n] + dp[1][n] ..+ dp[9][n]
for ( $i = 0; $i < 10; $i ++)
$count += $dp [ $i ][ $n ];
return $count ;
} // Driver program $n = 3;
echo countNonDecreasing( $n );
return 0;
?> |
<script> function countNonDecreasing(n)
{
// dp[i][j] contains total count of non decreasing
// numbers ending with digit i and of length j
let dp= new Array(10);
for (let i=0;i<10;i++)
{
dp[i]= new Array(n+1);
}
for (let i=0;i<10;i++)
{
for (let j=0;j<n+1;j++)
{
dp[i][j]=0;
}
}
// Fill table for non decreasing numbers of length 1
// Base cases 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
for (let i = 0; i < 10; i++)
dp[i][1] = 1;
// Fill the table in bottom-up manner
for (let digit = 0; digit <= 9; digit++)
{
// Compute total numbers of non decreasing
// numbers of length 'len'
for (let len = 2; len <= n; len++)
{
// sum of all numbers of length of len-1
// in which last digit x is <= 'digit'
for (let x = 0; x <= digit; x++)
dp[digit][len] += dp[x][len-1];
}
}
let count = 0;
// There total nondecreasing numbers of length n
// won't be dp[0][n] + dp[1][n] ..+ dp[9][n]
for (let i = 0; i < 10; i++)
count += dp[i][n];
return count;
}
let n = 3;
document.write(countNonDecreasing(n));
// This code is contributed by avanitrachhadiya2155
</script> |
220
Thanks to Gaurav Ahirwar for suggesting above method.
Method 2: 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++ program to count non-decreasing number 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;
} |
// java program to count non-decreasing // number with n digits import java.io.*;
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. |
# 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# 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 // PHP program to count non-decreasing // number with n digits function countNonDecreasing( $n )
{ $N = 10;
// Compute value of N*(N+1)/2*(N+2)/3* ...
// ....*(N+n-1)/n
$count = 1;
for ( $i = 1; $i <= $n ; $i ++)
{
$count *= ( $N + $i - 1);
$count /= $i ;
}
return $count ;
} // Driver Code
$n = 3;
echo countNonDecreasing( $n );
// This code is contributed by Sam007 ?> |
<script> // javascript program to count non-decreasing // number with n digits function countNonDecreasing(n)
{
let N = 10;
// Compute value of N * (N+1)/2 *
// (N+2)/3 * ....* (N+n-1)/n
let count = 1;
for (let i = 1; i <= n; i++)
{
count *= (N + i - 1);
count = Math.floor(count/ i);
}
return count;
}
// Driver code
let n = 3;
document.write(countNonDecreasing(n));
// This code is contributed by rag2127.
</script> |
220
Time Complexity: O(n)
Auxiliary Space: O(n)
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.