TCS Coding Practice Question | Check Armstrong Number

Last Updated : 28 Apr, 2023

Given a number, the task is to check if this number is Armstrong or not using Command Line Arguments. A positive integer of n digits is called an Armstrong number of order n (order is the number of digits) if.

abcd... = pow(a, n) + pow(b, n) + pow(c, n) + pow(d, n) + ....

Example:

Input: 153
Output: Yes
153 is an Armstrong number.
1*1*1 + 5*5*5 + 3*3*3 = 153

Input: 120
Output: No
120 is not a Armstrong number.
1*1*1 + 2*2*2 + 0*0*0 = 9

Input: 1253
Output: No
1253 is not a Armstrong Number
1*1*1*1 + 2*2*2*2 + 5*5*5*5 + 3*3*3*3 = 723

Input: 1634
Output: Yes
1*1*1*1 + 6*6*6*6 + 3*3*3*3 + 4*4*4*4 = 1634

Approach:

Since the number is entered as Command line Argument, there is no need for a dedicated input line
Extract the input number from the command line argument
This extracted number will be in string type.
Convert this number into integer type and store it in a variable, say num
Count the number of digits (or find the order) of the number num and store it in a variable, say n.
For every digit r in input number num, compute rⁿ.
If the sum of all such values is equal to num
If they are not same, the number is not Armstrong
If they are the same, the number is an Armstrong Number.

Program:

C++

#include <iostream>
#include <cstdlib> /* atoi */
 
// Function to calculate x raised to the power y
int power(int x, unsigned int y)
{
  if (y == 0)
    return 1;
  if (y % 2 == 0)
    return power(x, y / 2) * power(x, y / 2);
  return x * power(x, y / 2) * power(x, y / 2);
}
 
// Function to calculate order of the number
int order(int x)
{
  int n = 0;
  while (x) {
    n++;
    x = x / 10;
  }
  return n;
}
 
// Function to check whether the given number is
// Armstrong number or not
int isArmstrong(int x)
{
  // Calling order function
  int n = order(x);
  int temp = x, sum = 0;
  while (temp) {
    int r = temp % 10;
    sum += power(r, n);
    temp = temp / 10;
  }
 
  // If satisfies Armstrong condition
  if (sum == x)
    return 1;
  else
    return 0;
}
 
// Driver code
int main(int argc, char* argv[])
{int num, res = 0;
 
 // Check if the length of args array is 1
 if (argc == 1)
   std::cout << "No command line arguments found." << std::endl;
 
 else {
 
   // Get the command line argument and
   // Convert it from string type to integer type
   // using function "atoi( argument)"
   num = atoi(argv[1]);
 
   // Check if it is Armstrong
   res = isArmstrong(num);
 
   // Check if res is 0 or 1
   if (res == 0)
     // Print No
     std::cout << "No" << std::endl;
   else
     // Print Yes
     std::cout << "Yes" << std::endl;
 }
 return 0;
}

C

// C program to check if a number is Armstrong
// using command line arguments
 
#include <stdio.h>
#include <stdlib.h> /* atoi */
 
// Function to calculate x raised to the power y
int power(int x, unsigned int y)
{
    if (y == 0)
        return 1;
    if (y % 2 == 0)
        return power(x, y / 2) * power(x, y / 2);
    return x * power(x, y / 2) * power(x, y / 2);
}
 
// Function to calculate order of the number
int order(int x)
{
    int n = 0;
    while (x) {
        n++;
        x = x / 10;
    }
    return n;
}
 
// Function to check whether the given number is
// Armstrong number or not
int isArmstrong(int x)
{
    // Calling order function
    int n = order(x);
    int temp = x, sum = 0;
    while (temp) {
 
        int r = temp % 10;
        sum += power(r, n);
        temp = temp / 10;
    }
 
    // If satisfies Armstrong condition
    if (sum == x)
        return 1;
    else
        return 0;
}
 
// Driver code
int main(int argc, char* argv[])
{
 
    int num, res = 0;
 
    // Check if the length of args array is 1
    if (argc == 1)
        printf("No command line arguments found.\n");
 
    else {
 
        // Get the command line argument and
        // Convert it from string type to integer type
        // using function "atoi( argument)"
        num = atoi(argv[1]);
 
        // Check if it is Armstrong
        res = isArmstrong(num);
 
        // Check if res is 0 or 1
        if (res == 0)
            // Print No
            printf("No\n");
        else
            // Print Yes
            printf("Yes\n");
    }
    return 0;
}

Java

// Java program to check if a number is Armstrong
// using command line arguments
 
class GFG {
 
    // Function to calculate x
    // raised to the power y
    public static int power(int x, long y)
    {
        if (y == 0)
            return 1;
        if (y % 2 == 0)
            return power(x, y / 2) * power(x, y / 2);
        return x * power(x, y / 2) * power(x, y / 2);
    }
 
    // Function to calculate order of the number
    public static int order(int x)
    {
        int n = 0;
        while (x != 0) {
            n++;
            x = x / 10;
        }
        return n;
    }
 
    // Function to check whether the given number is
    // Armstrong number or not
    public static int isArmstrong(int x)
    {
        // Calling order function
        int n = order(x);
        int temp = x, sum = 0;
        while (temp != 0) {
            int r = temp % 10;
            sum = sum + power(r, n);
            temp = temp / 10;
        }
 
        // If satisfies Armstrong condition
        if (sum == x)
            return 1;
        else
            return 0;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        // Check if length of args array is
        // greater than 0
        if (args.length > 0) {
 
            // Get the command line argument and
            // Convert it from string type to integer type
            int num = Integer.parseInt(args[0]);
 
            // Get the command line argument
            // and check if it is Armstrong
            int res = isArmstrong(num);
 
            // Check if res is 0 or 1
            if (res == 0)
                // Print No
                System.out.println("No\n");
            else
                // Print Yes
                System.out.println("Yes\n");
        }
        else
            System.out.println("No command line "
                               + "arguments found.");
    }
}

Output:

In C:
In Java:

Time Complexity: O (log N)
Auxiliary Space: O (1)

C++ Program To Find Armstrong Numbers Between Two Integers

RishabhPrabhu

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TCS Coding Practice Question | Check Armstrong Number

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