**Problem Statement:** Given a number, check if it is Armstrong Number or Not.

**Examples:**

Example 1:Input:153Output:Yes, it is an Armstrong NumberExplanation:1^3 + 5^3 + 3^3 = 153Input:170Output:No, it is not an Armstrong NumberExplanation:1^3 + 7^3 + 0^3 != 170

**What are Armstrong Numbers?**

Armstrong Numbers are the numbers having the sum of digits raised to power no. of digits is equal to a given number. Examples- 0, 1, 153, 370, 371, 407, and 1634 are some of the Armstrong Numbers.

**Solution**

**Approach**:

Let us check one 3-digit Armstrong Number

n=153

sum=0

No. of digits = 3 so we need to cube every digit.

**In the First iteration**, extract digit 3 from 153 and cube it which becomes 27, add it to sum= 0+27=27 which becomes 27 now

**In Second iteration , **extract digit 5 from 15 and cube it which becomes 125 , add it to sum = 27 +125 = 152 which becomes 152 now

**In Third iteration , **extract digit 1 from 1 and cube it which becomes 1 , add it to sum = 152 + 1 = 153 which becomes 153 now.

The original Number was 153 and the sum of cubes = 153.

**This means it is an Armstrong Number.**

**Code:**

## C++ Code

```
#include <bits/stdc++.h>
using namespace std;
bool ArmstrongNumber(int n)
{
int originalno = n;
int count = 0;
int temp = n;
while (temp != 0)
{
count++;
temp = temp / 10;
}
int sumofpower = 0;
while (n != 0)
{
int digit = n % 10;
sumofpower += pow(digit,count);
n /= 10;
}
return (sumofpower == originalno);
}
int main()
{
int n1 = 153;
if (ArmstrongNumber(n1))
{
cout << "Yes, it is an Armstrong Number\n";
}
else
{
cout << "No, it is not an Armstrong Number\n";
}
return 0;
}
```

**Output:**

Yes, it is an Armstrong Number

**Time Complexity: **O(n) where n is the number of digits since we need to traverse every digit and add digits raised to power no. of digits to sum.

**Space Complexity:** O(1) since no extra space is required

## Java Code

```
import java.util.*;
class TUF{
static boolean ArmstrongNumber(int n)
{
int originalno = n;
int count = 0;
int temp = n;
while (temp != 0)
{
count++;
temp = temp / 10;
}
int sumofpower = 0;
while (n != 0)
{
int digit = n % 10;
sumofpower += Math.pow(digit,count);
n /= 10;
}
return (sumofpower == originalno);
}
public static void main(String args[])
{
int n1 = 153;
if (ArmstrongNumber(n1))
{
System.out.println("Yes, it is an Armstrong Number\n");
}
else
{
System.out.println("No, it is not an Armstrong Number\n");
}
}
}
```

**Output:**

Yes, it is an Armstrong Number

**Time Complexity: **O(n) where n is the number of digits since we need to traverse every digit and add digits raised to power no. of digits to sum.

**Space Complexity:** O(1) since no extra space is required

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