# Count of n digit numbers whose sum of digits equals to given sum

Given two integers ‘n’ and ‘sum’, find count of all n digit numbers with sum of digits as ‘sum’. Leading 0’s are not counted as digits.
1 <= n <= 100 and 1 <= sum <= 500

Example:

```Input:  n = 2, sum = 2
Output: 2
Explanation: Numbers are 11 and 20

Input:  n = 2, sum = 5
Output: 5
Explanation: Numbers are 14, 23, 32, 41 and 50

Input:  n = 3, sum = 6
Output: 21
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

The idea is simple, we subtract all values from 0 to 9 from given sum and recur for sum minus that digit. Below is recursive formula.

```    countRec(n, sum) = ∑countRec(n-1, sum-x)
where 0 =< x <= 9 and
sum-x >= 0

One important observation is, leading 0's must be
handled explicitly as they are not counted as digits.
So our final count can be written as below.
finalCount(n, sum) = ∑countRec(n-1, sum-x)
where 1 =< x <= 9 and
sum-x >= 0```

Below is a simple recursive solution based on above recursive formula.

## C++

```// A recursive program to count numbers with sum
// of digits as given 'sum'
#include<bits/stdc++.h>
using namespace std;

// Recursive function to count 'n' digit numbers
// with sum of digits as 'sum'. This function
// considers leading 0's also as digits, that is
// why not directly called
unsigned long long int countRec(int n, int sum)
{
// Base case
if (n == 0)
return sum == 0;

unsigned long long int ans = 0;

// Traverse through every digit and count
// numbers beginning with it using recursion
for (int i=0; i<=9; i++)
if (sum-i >= 0)
ans += countRec(n-1, sum-i);

return ans;
}

// This is mainly a wrapper over countRec. It
// explicitly handles leading digit and calls
// countRec() for remaining digits.
unsigned long long int finalCount(int n, int sum)
{
unsigned long long int ans = 0;

// Traverse through every digit from 1 to
// 9 and count numbers beginning with it
for (int i = 1; i <= 9; i++)
if (sum-i >= 0)
ans += countRec(n-1, sum-i);

return ans;
}

// Driver program
int main()
{
int n = 2, sum = 5;
cout << finalCount(n, sum);
return 0;
}
```

## Java

```class sum_dig
{
// Recursive function to count 'n' digit numbers
// with sum of digits as 'sum'. This function
// considers leading 0's also as digits, that is
// why not directly called
static int countRec(int n, int sum)
{
// Base case
if (n == 0)
return sum == 0 ?1:0;

int ans = 0;

// Traverse through every digit and count
// numbers beginning with it using recursion
for (int i=0; i<=9; i++)
if (sum-i >= 0)
ans += countRec(n-1, sum-i);

return ans;
}

// This is mainly a wrapper over countRec. It
// explicitly handles leading digit and calls
// countRec() for remaining digits.
static int finalCount(int n, int sum)
{
int ans = 0;

// Traverse through every digit from 1 to
// 9 and count numbers beginning with it
for (int i = 1; i <= 9; i++)
if (sum-i >= 0)
ans += countRec(n-1, sum-i);

return ans;
}

/* Driver program to test above function */
public static void main (String args[])
{
int n = 2, sum = 5;
System.out.println(finalCount(n, sum));
}
}/* This code is contributed by Rajat Mishra */
```

Output:
`5`

The time complexity of above solution is exponential. If we draw the complete recursion tree, we can observer that many subproblems are solved again and again. For example, if we start with n = 3 and sum = 10, we can reach n = 1, sum = 8, by considering digit sequences 1,1 or 2, 0.
Since same suproblems are called again, this problem has Overlapping Subprolems property. So min square sum problem has both properties (see this and this) of a dynamic programming problem.

Below is Memoization based the implementation.

## C++

```// A memoization based recursive program to count
// numbers with sum of n as given 'sum'
#include<bits/stdc++.h>
using namespace std;

// A lookup table used for memoization
unsigned long long int lookup[101][501];

// Memoizatiob based implementation of recursive
// function
unsigned long long int countRec(int n, int sum)
{
// Base case
if (n == 0)
return sum == 0;

// If this subproblem is already evaluated,
// return the evaluated value
if (lookup[n][sum] != -1)
return lookup[n][sum];

unsigned long long int ans = 0;

// Traverse through every digit and
// recursively count numbers beginning
// with it
for (int i=0; i<10; i++)
if (sum-i >= 0)
ans += countRec(n-1, sum-i);

return lookup[n][sum] = ans;
}

// This is mainly a wrapper over countRec. It
// explicitly handles leading digit and calls
// countRec() for remaining n.
unsigned long long int finalCount(int n, int sum)
{
// Initialize all entries of lookup table
memset(lookup, -1, sizeof lookup);

unsigned long long int ans = 0;

// Traverse through every digit from 1 to
// 9 and count numbers beginning with it
for (int i = 1; i <= 9; i++)
if (sum-i >= 0)
ans += countRec(n-1, sum-i);
return ans;
}

// Driver program
int main()
{
int n = 3, sum = 5;
cout << finalCount(n, sum);
return 0;
}
```

## Java

```class sum_dig
{
// A lookup table used for memoization
static int lookup[][] = new int[101][501];

// Memoizatiob based implementation of recursive
// function
static int countRec(int n, int sum)
{
// Base case
if (n == 0)
return sum == 0 ? 1 : 0;

// If this subproblem is already evaluated,
// return the evaluated value
if (lookup[n][sum] != -1)
return lookup[n][sum];

int ans = 0;

// Traverse through every digit and
// recursively count numbers beginning
// with it
for (int i=0; i<10; i++)
if (sum-i >= 0)
ans += countRec(n-1, sum-i);

return lookup[n][sum] = ans;
}

// This is mainly a wrapper over countRec. It
// explicitly handles leading digit and calls
// countRec() for remaining n.
static int finalCount(int n, int sum)
{
// Initialize all entries of lookup table
for(int i = 0;i<=100;++i){
for(int j=0;j<=50000;++j){
lookup[i][j] = -1;
}
}

int ans = 0;

// Traverse through every digit from 1 to
// 9 and count numbers beginning with it
for (int i = 1; i <= 9; i++)
if (sum-i >= 0)
ans += countRec(n-1, sum-i);
return ans;
}

/* Driver program to test above function */
public static void main (String args[])
{
int n = 2, sum = 5;
System.out.println(finalCount(n, sum));
}
}/* This code is contributed by Rajat Mishra */
```

Output:

`15`

Thanks to Gaurav Ahirwar for suggesting above solution.

# GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
3.5 Average Difficulty : 3.5/5.0
Based on 65 vote(s)