269308300None of these
Answer : C
Solution : Before 1000 there are one digit, two digits and three digits numbers.
Number of times 3 appear in one digit number `= 20 xx 9`
Number of times 3 appear in two digit number `= 11 xx 9`
Number of times 3 appear in three digit number = 21
Hence total number of times the digit 3 appear while writing the integers from 1 to 1000
`= 180 + 99 + 21 = 300`
Given a number n, we need to find the sum of its digits such that:
If n < 10 digSum[n] = n Else digSum[n] = Sum[digSum[n]]Examples :
Input : 1234 Output : 1 Explanation : The sum of 1+2+3+4 = 10, digSum[x] == 10 Hence ans will be 1+0 = 1 Input : 5674 Output : 4A brute force approach is to sum all the digits until the sum < 10.
Flowchart:
Below is the brute force program to find the sum.
C++
#include
using namespace std;
int digSum[int n]
{
int sum = 0;
while[n > 0 || sum > 9]
{
if[n == 0]
{
n = sum;
sum = 0;
}
sum += n % 10;
n /= 10;
}
return sum;
}
int main[]
{
int n = 1234;
cout 0 || sum > 9]
{
if [n == 0] {
n = sum;
sum = 0;
}
sum += n % 10;
n /= 10;
}
return sum;
}
public static void main[String argc[]]
{
int n = 1234;
System.out.println[digSum[n]];
}
}
Python
import math
def digSum[ n]:
sum = 0
while[n > 0 or sum > 9]:
if[n == 0]:
n = sum
sum = 0
sum += n % 10
n /= 10
return sum
n = 1234
print [digSum[n]]
C#
using System;
class GFG {
static int digSum[int n]
{
int sum = 0;
while [n > 0 || sum > 9]
{
if [n == 0]
{
n = sum;
sum = 0;
}
sum += n % 10;
n /= 10;
}
return sum;
}
public static void Main[]
{
int n = 1234;
Console.Write[digSum[n]];
}
}
PHP
Javascript
let n = 1234;
function getSum[n] {
let sum = 0;
while [n > 0 || sum > 9] {
if[n == 0] {
n = sum;
sum = 0;
}
sum = sum + n % 10;
n = Math.floor[n / 10];
}
return sum;
}
document.write[getSum[n]];
C
#include
int digSum[int n]
{
int sum = 0;
while[n > 0 || sum > 9]
{
if[n == 0]
{
n = sum;
sum = 0;
}
sum += n % 10;
n /= 10;
}
return sum;
}
int main[]
{
int n = 1234;
printf["%d",digSum[n]];
return 0;
}
Output :
1Time Complexity: O[log[n]].
Auxiliary Space: O[1]
So, another challenge is “Could you do it without any loop/recursion in O[1] runtime?”
YES!!
There exists a simple and elegant O[1] solution for this too. The answer is given simply:-
How does the above logic works?
The logic behind this approach is :
To check if a number is divisible by 9, add the digits of the number and check if the sum is divisible by 9 or not. If yes, is the case, the number is divisible by 9, otherwise, it’s not.
let’s
take 27 i.e [2+7 = 9] hence divisible by 9.
If a number n is divisible by 9, then the sum of its digit until the sum becomes a single digit is always 9. For example,
Let, n = 2880
Sum of digits = 2 + 8 + 8 = 18: 18 = 1 + 8 = 9
Therefore,
A number can be of the form 9x or 9x + k. For the first case, the answer is always 9. For the second case, and is always k which is the remainder left.
The problem is widely known as the digit root problem.
You may find this Wikipedia article useful. -> //en.wikipedia.org/wiki/Digital_root
Below is the implementation of the above idea :
C++
#include
using namespace std;
int digSum[int n]
{
if [n == 0]
return 0;
return [n % 9 == 0] ? 9 : [n % 9];
}
int main[]
{
int n = 9999;
cout
Javascript
function digSum[n]
{
if [n == 0]
return 0;
return [n % 9 == 0] ? 9 : [n % 9];
}
n = 9999;
document.write[digSum[n]];
Output:
9Time Complexity: O[1]
Auxiliary Space: O[1]
Related Post :
//www.geeksforgeeks.org/digital-rootrepeated-digital-sum-given-integer/
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