Check if the given number is Harshad(Or Niven) Number

Problem Statement: Check if the number is a Harshad(or Niven) number or not.

Examples:

Example 1:
Input: 378
Output: Yes it is a Harshad number.
Explanation: 3+7+8=18. 378 is divisible by 18. Therefore 378 is a harshad number.

Example 2:
Input: 379
Output: No
 it is not a Harshad number.
Explanation: 3+7+9=19. 379 is not divisible by 19. Therefore 379 is a harshad number.

Solution

Disclaimer: Don’t jump directly to the solution, try it out yourself first.

Solution 1:

Intuition: If the sum of digits is divisible by the number then it is called Harshad number.

Approach:

Maintain a variable sum to store sum of digits of the number.
Now check if n is divisible by sum or not.
If it is divisible print yes,else print no.

Code:

C++ Code

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int n = 378;
	int temp = n;
	int sum = 0;
	while (temp!=0) {
		sum += temp % 10;
		temp /= 10;
	}
	if (n % sum == 0) {
		cout << "YES it is Harshad Number" << "\n";
	}
	else {
		cout << "NO it is not Harshad Number" << "\n";
	}

}

Output:

YES it is Harshad Number

Time Complexity: O(N), where N is the number of digits

Space Complexity: O(1).

Java Code

public class Main {
  public static void main(String args[]) {
    int n = 378;
    int temp = n;
    int sum = 0;
    while (temp != 0) {
      sum += temp % 10;
      temp /= 10;
    }
    if (n % sum == 0) {
      System.out.println("YES it is Harshad Number");
    } else {
      System.out.println("NO it is Harshad Number");
    }

  }
}

Output:

YES it is Harshad Number

Time Complexity: O(N), where N is the number of digits

Space Complexity: O(1).

Solution 2: Using string

Intuition: In solution1 we needed an extra variable temp to store the value of n, but if we use string there won’t be any need for the extra variable.

Approach:

Convert the intger to string.
Traverse through the string and calculate the sum of all the digits.
Now check if the number is divisible by the sum of digits or not.
If the number is divisible, print yes, else print no.

Code:

C++ Code

#include<bits/stdc++.h>
using namespace std;
int main()
{   
    int num=378;
	string s = to_string(num);
	int sum = 0;
	for (int i = 0; i < s.length(); i++) {
		sum += s[i] - '0';
	}
	if (num % sum == 0) {
		cout << "YES it is Harshad Number" << "\n";
	}
	else {
		cout << "NO it is not a Harshad Number" << "\n";
	}

}

Output:

YES it is Harshad Number

Time Complexity: O(N), where N is the number of digits

Space Complexity: O(1).

Java Code

public class Main {
  public static void main(String args[]) {
    int num=378;
    String s = Integer.toString(num);
    int sum = 0;
    for (int i = 0; i < s.length(); i++) {
      sum += s.charAt(i) - '0';
    }
    if (num % sum == 0) {
      System.out.print("YES it is a Harshad Number");
    } else {
      System.out.print("NO it is Not a Harshad Number");
    }
  }
}

Output:

YES it is Harshad Number

Time Complexity: O(N), where N is the number of digits

Space Complexity: O(1).

Special thanks to Pranav Padawe for contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam, please check out this article