Given a positive integers N, the task is to find the smallest number whose sum of digits is N.
Example:
Input: N = 10
Output: 19
Explanation: 1 + 9 = 10 = N
Input: N = 18
Output: 99
Explanation: 9 + 9 = 18 = N
Naive Approach:
- A Naive approach is to run a loop of i starting from 0 and find Sum of digits of i and check if it’s equal to N or not.
Below is the implementation of the above approach.
C++
#include <iostream>
#include <math.h>
using namespace std;
int getSum(int n)
{
int sum = 0;
while (n != 0) {
sum = sum + n % 10;
n = n / 10;
}
return sum;
}
void smallestNumber(int N)
{
int i = 1;
while (1) {
if (getSum(i) == N) {
cout << i;
break;
}
i++;
}
}
int main()
{
int N = 10;
smallestNumber(N);
return 0;
}
|
Java
class GFG{
static int getSum(int n)
{
int sum = 0;
while (n != 0)
{
sum = sum + n % 10;
n = n / 10;
}
return sum;
}
static void smallestNumber(int N)
{
int i = 1;
while (1 != 0)
{
if (getSum(i) == N)
{
System.out.print(i);
break;
}
i++;
}
}
public static void main(String[] args)
{
int N = 10;
smallestNumber(N);
}
}
|
Python3
def getSum(n):
sum1 = 0;
while (n != 0):
sum1 = sum1 + n % 10;
n = n // 10;
return sum1;
def smallestNumber(N):
i = 1;
while (1):
if (getSum(i) == N):
print(i);
break;
i += 1;
N = 10;
smallestNumber(N);
|
C#
using System;
class GFG{
static int getSum(int n)
{
int sum = 0;
while (n != 0)
{
sum = sum + n % 10;
n = n / 10;
}
return sum;
}
static void smallestNumber(int N)
{
int i = 1;
while (1 != 0)
{
if (getSum(i) == N)
{
Console.Write(i);
break;
}
i++;
}
}
public static void Main(String[] args)
{
int N = 10;
smallestNumber(N);
}
}
|
Javascript
<script>
function getSum(n)
{
let sum = 0;
while (n != 0) {
sum = sum + n % 10;
n = Math.floor(n / 10);
}
return sum;
}
function smallestNumber(N)
{
let i = 1;
while (1) {
if (getSum(i) == N) {
document.write(i);
break;
}
i++;
}
}
let N = 10;
smallestNumber(N);
</script>
|
Time Complexity: O(N).
Auxiliary space: O(1)
Efficient Approach:
- An efficient approach to this problem is an observation. Let’s see some examples.
- If N = 10, then ans = 19
- If N = 20, then ans = 299
- If N = 30, then ans = 3999
- So, it is clear that the answer will have all digits as 9 except the first one so that we get the smallest number.
- So, the Nth term will be =
Below is the implementation of the above approach
C++
#include <iostream>
#include <math.h>
using namespace std;
void smallestNumber(int N)
{
cout << (N % 9 + 1)
* pow(10, (N / 9))
- 1;
}
int main()
{
int N = 10;
smallestNumber(N);
return 0;
}
|
Java
class GFG{
static void smallestNumber(int N)
{
System.out.print((N % 9 + 1) *
Math.pow(10, (N / 9)) - 1);
}
public static void main(String[] args)
{
int N = 10;
smallestNumber(N);
}
}
|
Python3
def smallestNumber(N):
print((N % 9 + 1) * pow(10, (N // 9)) - 1)
N = 10
smallestNumber(N)
|
C#
using System;
class GFG{
static void smallestNumber(int N)
{
Console.WriteLine((N % 9 + 1) *
Math.Pow(10, (N / 9)) - 1);
}
public static void Main()
{
int N = 10;
smallestNumber(N);
}
}
|
Javascript
<script>
function smallestNumber(N)
{
document.write( (N % 9 + 1) * Math.pow(10, parseInt(N / 9, 10)) - 1 );
}
let N = 10;
smallestNumber(N);
</script>
|
Time complexity: O(logN) since inbuilt pow function is being used
Auxiliary space: O(1), since no extra space has been taken.