Armstrong Number - Practice Coding Problems

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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Understanding the Visualization

Input Number

Given integer n to check

Calculate Powers

Sum each digit raised to power of digit count

Compare

Check if sum equals original number

Key Takeaway

🎯 Key Insight: An Armstrong number equals the sum of its digits each raised to the power of the digit count

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

Approach	Time	Space	Notes
✓ Brute Force - Convert to String	O(log n)	O(log n)	Convert number to string, extract digits, calculate powers and sum
Mathematical Approach	O(log n)	O(1)	Use mathematical operations to extract digits without string conversion

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 <string.h>
#include <math.h>
#include <stdbool.h>

bool solution(int n) {
    if (n < 0) return false;
    
    char numStr[20];
    sprintf(numStr, "%d", n);
    int numDigits = strlen(numStr);
    int digitSum = 0;
    
    for (int i = 0; i < numDigits; i++) {
        int digit = numStr[i] - '0';
        digitSum += (int)pow(digit, numDigits);
    }
    
    return digitSum == n;
}

int main() {
    int n;
    scanf("%d", &n);
    bool result = solution(n);
    printf(result ? "true\n" : "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

⚡ Linearithmic Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Convert to String — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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