# Super Simple DFS Java/C# solution

• ``````public class Solution
{
public int CountNumbersWithUniqueDigits(int n)
{
return CountNumbersWithUniqueDigitsHelper(n, 0) + 1;
}

public int CountNumbersWithUniqueDigitsHelper(int n, int currentNumDigits)
{
if(currentNumDigits == n) return 0;

int sum = 0;
for (int i = currentNumDigits; i < (currentNumDigits == 0 ? 9 : 10); i++)
{
sum += 1 + CountNumbersWithUniqueDigitsHelper(n, currentNumDigits + 1);
}
return sum;
}
}
``````

Every node of the DFS forms a number and when you go one level deeper you have one less digit to use. The first recursion call has a trick that you only should use 9 digits instead of 10. That's because the numbers starting with 0 will already be covered in other nodes. i.e: 01,02,03,04... etc is already covered by the call to 1,2,3,4... The only number missing will be the actual 0, that's why the final +1 sum on the recursion call.

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