The k-digit number N is an Armstrong number if and only if the k-th power of each digit sums to N. Given a positive integer N, return true if and only if it is an Armstrong number.

Example 1:
Input: 153
Output: true
Explanation:
153 is a 3-digit number, and 153 = 1^3 + 5^3 + 3^3.

Example 2:
Input: 123
Output: false
Explanation:
123 is a 3-digit number, and 123 != 1^3 + 2^3 + 3^3 = 36.

Note:
1 <= N <= 10^8

Python to Check if a Integer is Armstrong

Do what it says. The solution should be quite intuitive. Two things need to be resolved: get the number of digits which could be done via converting to string and get the length or using the log10 math function. The second is to compute a^b which in Python this could be done simply by a**b.

Then, we need to extract the right-most digit by do the moduluo ten, and do the integer division using // in Python..

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class Solution:
    def isArmstrong(self, N: int) -> bool:
        s = 0
        n = N
        l = len(str(N))
        while n > 0:
            t = n % 10
            n //= 10
            s = s + (t ** l)
        return s == N

class Solution:
    def isArmstrong(self, N: int) -> bool:
        s = 0
        n = N
        l = len(str(N))
        while n > 0:
            t = n % 10
            n //= 10
            s = s + (t ** l)
        return s == N

–EOF (The Ultimate Computing & Technology Blog) —