Capgemini Exceller Technical Interview Questions for Analyst

#include <stdio.h>

int checkHarshad(int num){
    
    int sum = 0;
    int temp = num;
    
    while(temp != 0){
        sum = sum + temp % 10;
        temp /= 10;
    }
    
    // will return 1 if num is divisible by sum, else 0
    return num % sum == 0;
}

int main ()
{
    int num = 153;
    
    if(checkHarshad(num))
        printf("%d is Harshad's Number", num);
    else
        printf("%d is not Harshad's Number", num);

    return 0;
}
// Time complexity: O(N)
// Space complexity: O(1)

#include <iostream>
using namespace std;

int checkHarshad(int num){
    
    int sum = 0;
    int temp = num;
    
    while(temp != 0){
        sum = sum + temp % 10;
        temp /= 10;
    }
    
    // will return 1 if num is divisible by sum, else 0
    return num % sum == 0;
}

int main ()
{
    int n = 153;
    
    if(checkHarshad(n))
        cout << n << " is a Harshad's number";
    else
        cout << n << " is not a Harshad's number";

    return 0;
}
// Time complexity: O(N)
// Space complexity: O(1)
// Where N is the number of digits in number

public class Main {
    public static void main(String[] args)
	{
		//make a copy of original number
		int n = 47;
		//declare a variable to store sum of digits
		int result = 0;
		//perform logic for calculating sum of digits of a number
		while(n != 0)
		{
			int pick_last = n % 10;
			result = result + pick_last;
			n = n / 10;
		}
		/*use condition to check whether the number entered by  
		user is completely divisible by its sum of digits or not*/
      if(n % result == 0)
      System.out.println("Harshad Number");
          else
        System.out.println("Not a Harshad Number");
}
}

n = 21
p=n
l=[]
sum1=0
while(n>0):
    x=n%10
    l.append(x)
    n=n//10
sum1=sum(l)
if(p%sum1==0):
    print("Harshad number")
else:
    print("Not harshad number")