Coding problem · Easy · Basics

Armstrong Number

Easy Algorithms & data structures

An Armstrong number (narcissistic number) equals the sum of its digits, each raised to the power of the number of digits. 153 = 1³ + 5³ + 3³. Return true if n (non-negative) is one.

Example 1

Input: n = 153
Output: true

Example 2

Input: n = 123
Output: false

Example 3

Input: n = 9474
Output: true
Explanation: 9⁴ + 4⁴ + 7⁴ + 4⁴ = 9474

Hints

Hint 1

Count the digits first (the length of the number as a string).

Hint 2

Add up digit^count for every digit and compare with n.

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Solution and explanation

Show the solution

Time O(digits) · Space O(1)

Get the digits, let k be how many there are, sum d ** k, and compare with the original number.

Reference solution

function isArmstrong(n) {
  const digits = String(n).split('').map(Number);
  return digits.reduce((sum, d) => sum + d ** digits.length, 0) === n;
}

function isArmstrong(n: number): boolean {
  const digits = String(n).split('').map(Number);
  return digits.reduce((sum, d) => sum + d ** digits.length, 0) === n;
}

def is_armstrong(n: int) -> bool:
    digits = [int(ch) for ch in str(n)]
    return sum(d ** len(digits) for d in digits) == n

<?php

function isArmstrong(int $n): bool
{
    $digits = str_split((string) $n);
    $k = count($digits);
    return array_sum(array_map(fn ($d) => ((int) $d) ** $k, $digits)) === $n;
}

class Solution {
    public boolean isArmstrong(int n) {
        String s = String.valueOf(n);
        long sum = 0;
        for (char ch : s.toCharArray()) sum += (long) Math.pow(ch - '0', s.length());
        return sum == n;
    }
}

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