Armstrong Number
You are given an integer ‘NUM’ . Your task is to find out whether this number is an Armstrong number or not.

A k-digit number ‘NUM’ is an Armstrong number if and only if the k-th power of each digit sums to ‘NUM’.

Example

153 = 1^3 + 5^3 + 3^3. Therefore 153 is an Armstrong number.

Input Format:

The first line contains an integer ‘T’, which denotes the number of test cases to be run. Then, the ‘T’ test cases follow. The first line of each test case contains a single positive integer, ‘NUM’.

Output Format:

For each test case, print ‘YES’ if it is an Armstrong number otherwise print ‘NO’. The output of each test case will be printed in a separate line.

Note:

You do not need to print anything. It has already been taken care of. Just implement the given function.

Constraints:

1

Question

Armstrong Number

You are given an integer ‘NUM’ . Your task is to find out whether this number is an Armstrong number or not.

A k-digit number ‘NUM’ is an Armstrong number if and only if the k-th power of each digit sums to ‘NUM’.

Example

153 = 1^3 + 5^3 + 3^3.

Therefore 153 is an Armstrong number.

Input Format:

The first line contains an integer ‘T’, which denotes the number of test cases to be run. Then, the ‘T’ test cases follow.

The first line of each test case contains a single positive integer, ‘NUM’.

Output Format:

For each test case, print ‘YES’ if it is an Armstrong number otherwise print ‘NO’.

The output of each test case will be printed in a separate line.

Note:

You do not need to print anything. It has already been taken care of. Just implement the given function.

Constraints:

1 <= ‘T’ <= 100
1 <= ‘N’ <= 10^9

Time Limit: 1 sec

Accepted Answer

class Solution {
static String armstrongNumber(int n){
int temp = n;
int arm = 0;
while(n!=0) {
int rem = n%10;
arm += Math.pow(rem,3);
n = n/10;
}
if(arm == temp) {
return "Yes";
}
else {
return "No";
}
}
}

Accepted Answer

Maths

The first step will be to find the length of the number.
Initialize an integer k = 0, it will be used to find the length of the number.
Initialise an integer sum = 0.
Run a while loop till num is not equal to 0 (to calculate the length of number):
- Increment k by 1.
Run a while loop till num is not equal to 0:
- The current digit will be num%10.
- Add (current digit) ^ k into sum.
- Make num = num/10.
Check if ‘sum’ is equal to the initial number.

Space Complexity: O(1)Explanation:

O(1)

As we are using constant extra space.

Time Complexity: O(logn)Explanation:

O(log10(N)), where N is the number given to you.

We just need to traverse the digits of the number.

Java (SE 1.8)

/*

    Time Complexity O(log10(N))

    Space Complexity: O(1),



    where N is the number given to you.

*/



public class Solution {

	public static boolean isArmstrong(int num) {



		// This stores the length of the number.

		int len = 0;



		int temp = num;



		// Finding the length of the number using this while loop.

		while (temp > 0) {

			++len;

			temp /= 10;

		}



		temp = num;



		// 'sum' stores the sum of current_digit^length.

		int sum = 0;

		while (temp > 0) {

			sum += Math.pow(temp % 10, len);

			temp /= 10;

		}



		// If sum == num then it is an Armstrong number.

		return (sum == num ? true : false);

	}

}

C++ (g++ 5.4)

/*

    Time Complexity O(log10(N))

    Space Complexity: O(1),



    where N is the number given to you.

*/



bool isArmstrong(int num) {

    

	// This stores the length of the number.

	int len = 0;



	int temp = num;



	// Finding the length of the number using this while loop.

	while (temp) {

		++len;

		temp /= 10;

	}



	temp = num;



	// 'sum' stores the sum of current_digit^length.

	int sum = 0;

	while (temp) {

		sum += pow(temp % 10, len);

		temp /= 10;

	}



	// If sum == num then it is an Armstrong number.

	return sum == num;

}

Python (3.5)

'''

    Time Complexity O(log10(N))

    Space Complexity: O(1),



    where N is the number given to you.

'''



def isArmstrong(num):



    # This stores the length of the number.

    len1 = 0



    temp = num



    # Finding the length of the number using this while loop.

    while(temp):

        len1 += 1

        temp  = temp//10



    temp = num

    

    # 'sum1' stores the sum of current_digit^length.

    sum1 = 0

    while(temp):

        sum1 += pow(temp%10, len1)

        temp  = temp//10



    # If sum1 == num then it is an Armstrong number.

    return sum1 == num