Armstrong Number

Armstrong Number - Problem

Math Easy

Given an integer n, return true if and only if it is an Armstrong number.

The k-digit number n is an Armstrong number if and only if the kth power of each digit sums to n.

Example: 153 is a 3-digit number. 1³ + 5³ + 3³ = 1 + 125 + 27 = 153, so 153 is an Armstrong number.

Input & Output

Example 1 — Classic Armstrong Number

$ Input: n = 153

› Output: true

💡 Note: 153 is a 3-digit number. 1³ + 5³ + 3³ = 1 + 125 + 27 = 153, which equals the original number.

Example 2 — Single Digit

$ Input: n = 9

› Output: true

💡 Note: 9 is a 1-digit number. 9¹ = 9, which equals the original number.

Example 3 — Not Armstrong

$ Input: n = 123

› Output: false

💡 Note: 123 is a 3-digit number. 1³ + 2³ + 3³ = 1 + 8 + 27 = 36 ≠ 123.

Constraints

1 ≤ n ≤ 2³¹ - 1

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that an Armstrong number equals the sum of each digit raised to the power of the total digit count. The optimal approach uses mathematical operations to extract digits without string conversion. Time: O(log n), Space: O(1).

Common Approaches

✓ Brute Force - Convert to String

⏱️ Time: O(log n) Space: O(log n)

Convert the number to a string to easily count digits and extract each digit. Then raise each digit to the power of the total digit count and sum them up.

Mathematical Approach

⏱️ Time: O(log n) Space: O(1)

Count digits using logarithm, then extract each digit using modulo and division operations. This avoids string conversion overhead while maintaining clarity.

Brute Force - Convert to String — Algorithm Steps

Convert number to string to count digits
For each digit, calculate digit^k where k is digit count
Sum all powered digits and compare with original

Visualization

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Step-by-Step Walkthrough

Convert to String

Convert number to string to count digits

Process Each Digit

Raise each digit to the power of digit count

Sum and Compare

Sum all powers and compare with original number

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int power(int base, int exp) {
    int result = 1;
    for (int i = 0; i < exp; i++) {
        result *= base;
    }
    return result;
}

int countDigits(int n) {
    if (n == 0) return 1;
    int count = 0;
    while (n > 0) {
        count++;
        n /= 10;
    }
    return count;
}

int isArmstrong(int n) {
    if (n < 0) return 0;  // negative numbers are not Armstrong numbers
    
    int original = n;
    int digits = countDigits(n);
    int sum = 0;
    
    while (n > 0) {
        int digit = n % 10;
        sum += power(digit, digits);
        n /= 10;
    }
    
    return sum == original;
}

int main() {
    int n;
    scanf("%d", &n);
    
    if (isArmstrong(n)) {
        printf("true\n");
    } else {
        printf("false\n");
    }
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log n)

We process each digit once, and there are log₁₀(n) digits in number n

⚡ Linearithmic

Space Complexity

O(log n)

String representation requires space proportional to number of digits

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Convert to String — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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