Armstrong Number

Problem Statement

Armstrong numbers, or plenary numbers, are fascinating numerical entities that exhibit a unique property: a number of k digits is deemed an Armstrong number if the sum of its digits, each raised to the power of k, equals the number itself. For example, the number 153, which is a 3-digit number, equals (1^3 + 5^3 + 3^3). The task is to determine if a given integer n is an Armstrong number by verifying if it meets the conditions of this definition.

Examples

Example 1

Input:

n = 153

Output:

true

Explanation:

153 is a 3-digit number, and 153 = 13 + 53 + 33.

Example 2

Input:

n = 123

Output:

false

Explanation:

123 is a 3-digit number, and 123 != 13 + 23 + 33 = 36.

Constraints

• 1 <= n <= 108

Approach and Intuition

To verify if a given integer n is an Armstrong number, one must perform the following steps:

1. Determine the number of digits k in n. This can typically be calculated by converting the number to a string and counting its length.
2. Decompose the number into its individual digits.
3. Compute the sum of the kth powers of each digit.
4. Compare the result of this computation with n. If they are equal, then n is an Armstrong number; otherwise, it is not.

By using these steps, the Armstrong property of any number within the provided constraints can be assessed.

• Compute Length: Start by determining the number of digits k in the number n.
• Decompose and Summarize: Break down n into its constituent digits and, for each digit d, compute (d^k). Sum all these values.
• Validation: Check if the sum obtained is identical to the original number n. If so, it confirms that n is an Armstrong number.

This step-by-step approach breaks down the problem into manageable parts and clarifies the computational requirements to categorize a number based on its Armstrong status, which aligns precisely with the constraints (1 \leq n \leq 10^8).

Solutions

class Solution {
public:
    int sumDigitPowers(int number, int power) {
        int sum = 0;
        while(number != 0) {
            sum += pow(number % 10, power);
            number /= 10;
        }
        return sum;
    }
    bool isArmstrongNumber(int number) {
        int digits = 0;
        int temp = number;
        while (temp) {
            digits++;
            temp /= 10;
        }
        return sumDigitPowers(number, digits) == number;
    }
};