n-th number whose sum of digits is ten
Last Updated :
17 Aug, 2023
Given an integer value n, find out the n-th positive integer whose sum is 10.
Examples:
Input: n = 2
Output: 28
The first number with sum of digits as
10 is 19. Second number is 28.
Input: 15
Output: 154
Method 1 (Simple):
We traverse through all numbers. For every number, we find the sum of digits. We stop when we find the n-th number with the sum of digits as 10.
C++
#include <bits/stdc++.h>
using namespace std;
int findNth( int n)
{
int count = 0;
for ( int curr = 1;; curr++) {
int sum = 0;
for ( int x = curr; x > 0; x = x / 10)
sum = sum + x % 10;
if (sum == 10)
count++;
if (count == n)
return curr;
}
return -1;
}
int main()
{
printf ( "%d\n" , findNth(5));
return 0;
}
|
Java
import java.util.*;
import java.lang.*;
public class GFG {
public static int findNth( int n)
{
int count = 0 ;
for ( int curr = 1 ;; curr++) {
int sum = 0 ;
for ( int x = curr; x > 0 ; x = x / 10 )
sum = sum + x % 10 ;
if (sum == 10 )
count++;
if (count == n)
return curr;
}
}
public static void main(String[] args)
{
System.out.print(findNth( 5 ));
}
}
|
Python3
import itertools
def findNth(n):
count = 0
for curr in itertools.count():
sum = 0
x = curr
while (x):
sum = sum + x % 10
x = x / / 10
if ( sum = = 10 ):
count = count + 1
if (count = = n):
return curr
return - 1
if __name__ = = '__main__' :
print (findNth( 5 ))
|
C#
using System;
class GFG {
public static int findNth( int n)
{
int count = 0;
for ( int curr = 1;; curr++) {
int sum = 0;
for ( int x = curr; x > 0; x = x / 10)
sum = sum + x % 10;
if (sum == 10)
count++;
if (count == n)
return curr;
}
}
static public void Main()
{
Console.WriteLine(findNth(5));
}
}
|
Javascript
<script>
function findNth(n)
{
let count = 0;
for (let curr = 1;; curr++) {
let sum = 0;
for (let x = curr; x > 0; x = Math.floor(x / 10))
sum = sum + x % 10;
if (sum == 10)
count++;
if (count == n)
return curr;
}
return -1;
}
document.write(findNth(5));
</script>
|
PHP
<?php
function findNth( $n )
{
$count = 0;
for ( $curr = 1; ; $curr ++)
{
$sum = 0;
for ( $x = $curr ;
$x > 0; $x = $x / 10)
$sum = $sum + $x % 10;
if ( $sum == 10)
$count ++;
if ( $count == $n )
return $curr ;
}
return -1;
}
echo findNth(5);
?>
|
Time Complexity: O(N*log10(N))
Auxiliary Space: O(1)
Method 2 (Efficient):
If we take a closer look, we can notice that all multiples of 9 are present in arithmetic progression 19, 28, 37, 46, 55, 64, 73, 82, 91, 100, 109,… However, there are numbers in the above series whose sum of digits is not 10, for example, 100. So instead of checking one by one, we start with 19 and increment by 9.
C++
#include <bits/stdc++.h>
using namespace std;
int findNth( int n)
{
int count = 0;
for ( int curr = 19;; curr += 9) {
int sum = 0;
for ( int x = curr; x > 0; x = x / 10)
sum = sum + x % 10;
if (sum == 10)
count++;
if (count == n)
return curr;
}
return -1;
}
int main()
{
printf ( "%d\n" , findNth(5));
return 0;
}
|
Java
import java.util.*;
import java.lang.*;
public class GFG {
public static int findNth( int n)
{
int count = 0 ;
for ( int curr = 19 ;; curr += 9 ) {
int sum = 0 ;
for ( int x = curr; x > 0 ; x = x / 10 )
sum = sum + x % 10 ;
if (sum == 10 )
count++;
if (count == n)
return curr;
}
}
public static void main(String[] args)
{
System.out.print(findNth( 5 ));
}
}
|
Python3
def findNth(n):
count = 0 ;
curr = 19 ;
while ( True ):
sum = 0 ;
x = curr;
while (x > 0 ):
sum = sum + x % 10 ;
x = int (x / 10 );
if ( sum = = 10 ):
count + = 1 ;
if (count = = n):
return curr;
curr + = 9 ;
return - 1 ;
print (findNth( 5 ));
|
C#
using System;
class GFG {
public static int findNth( int n)
{
int count = 0;
for ( int curr = 19;; curr += 9) {
int sum = 0;
for ( int x = curr;
x > 0; x = x / 10)
sum = sum + x % 10;
if (sum == 10)
count++;
if (count == n)
return curr;
}
}
static public void Main()
{
Console.WriteLine(findNth(5));
}
}
|
Javascript
<script>
function findNth(n)
{
let count = 0;
for (let curr = 19; ;curr += 9)
{
let sum = 0;
for (let x = curr; x > 0;
x = parseInt(x / 10))
sum = sum + x % 10;
if (sum == 10)
count++;
if (count == n)
return curr;
}
return -1;
}
document.write(findNth(5));
</script>
|
PHP
<?php
function findNth( $n )
{
$count = 0;
for ( $curr = 19; ; $curr += 9)
{
$sum = 0;
for ( $x = $curr ; $x > 0;
$x = (int) $x / 10)
$sum = $sum + $x % 10;
if ( $sum == 10)
$count ++;
if ( $count == $n )
return $curr ;
}
return -1;
}
echo findNth(5);
?>
|
Time Complexity: O(N*log10(N))
Auxiliary Space: O(1)
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...