Hello Everyone,
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.
// Simple CPP program to find n-th number
// with sum of digits as 10.
#include <bits/stdc++.h>
using namespace std;
int findNth( int n)
{
int count = 0;
for ( int curr = 1;; curr++) {
// Find sum of digits in current no.
int sum = 0;
for ( int x = curr; x > 0; x = x / 10)
sum = sum + x % 10;
// If sum is 10, we increment count
if (sum == 10)
count++;
// If count becomes n, we return current
// number.
if (count == n)
return curr;
}
return -1;
}
int main()
{
printf ( "%d\n" , findNth(5));
return 0;
}
Output
55
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.
// Simple CPP program to find n-th number
// with sum of digits as 10.
#include <bits/stdc++.h>
using namespace std;
int findNth( int n)
{
int count = 0;
for ( int curr = 19;; curr += 9) {
// Find sum of digits in current no.
int sum = 0;
for ( int x = curr; x > 0; x = x / 10)
sum = sum + x % 10;
// If sum is 10, we increment count
if (sum == 10)
count++;
// If count becomes n, we return current
// number.
if (count == n)
return curr;
}
return -1;
}
int main()
{
printf ( "%d\n" , findNth(5));
return 0;
}
Output
55