Problem
A number with k digits is an Armstrong number if the sum of each digit raised to the power k equals the number. For example 153 = 1³ + 5³ + 3³.
Implement is_armstrong(n).
Input
n: A non-negative integer.
Output
Return true if n is an Armstrong number.
Constraints
0 ≤ n ≤ 10¹⁸
Examples
Example 1
Input:
n = 153
Expected output:
expected = TrueInput:
long long n = 153;
Expected output:
bool expected = true;Input:
long n = 153;
Expected output:
bool expected = true;Explanation
1³ + 5³ + 3³ = 1 + 125 + 27 = 153.
Example 2
Input:
n = 10
Expected output:
expected = FalseInput:
long long n = 10;
Expected output:
bool expected = false;Input:
long n = 10;
Expected output:
bool expected = false;Explanation
1² + 0² = 1, which is not 10.
How to approach it
Try on your own first, then open one hint at a time.
Hint 1
Count how many digits the number has before summing powers.
Hint 2
Compare the original number with the sum of each digit raised to that digit count.
Hint 3
Remember that 0 is a special one-digit case.
Expected complexity
O(k) time, O(1) extra space
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.