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++14
// 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;
}
Java
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
# 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#
// 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
// 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;
?>
Javascript
<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>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 = 10Below is the program to compute count using above formula.
C++
// 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
// 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.
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
<?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
?>
Javascript
<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>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.
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