**Problem Statement:** Given an Octal Number, convert it into a Decimal Number.

**Examples:**

Example 1:Input:345Output:229Explanation:Decimal equivalent of given Octal expressionis 229Example 2:Input:170Output:121Explanation:Decimal equivalent of given Octal expression is121

**Solution**

** Disclaimer**:

*Don’t jump directly to the solution, try it out yourself first.*

**Approach**:

In Octal to Decimal Conversion, We will take every digit of the number and multiply it with 8 raised to power i which will increase by 1 when we move to the next digit and then add it to sum. This task is repeated until n becomes 0.

For Eg, 345

**Step 1**: Take digit 5, multiply 8 raised to power 0 with it so we get 5 here and now n becomes 34

**Step 2:** Take digit 4, multiply 8 raised to power 1 with it so we get 32 and add it to the sum which now becomes 37. Also, n becomes 3.

**Step 3**: Take digit 3 and multiply 8 raised to power 2 with it so we get 192 and add it to sum which now becomes 229. Now n becomes zero.

**Code:**

## C++ Code

```
#include <iostream>
#include <math.h>
using namespace std;
int OctaltoDecimal(int Octal)
{
int Decimal = 0;
int i = 0;
while (Octal != 0)
{
int rem = Octal % 10;
Decimal += rem * pow(8, i);
i++;
Octal /= 10;
}
return Decimal;
}
int main()
{
int Octal = 345;
cout <<"The decimal equivalent of the given octal number is "<<OctaltoDecimal(Octal) << endl;
return 0;
}
```

**Output:**

The decimal equivalent of the given octal number is 229

**Time Complexity: **O(n) where n is the number of digits in the Octal Number.

**Space Complexity: **O(1)

## Java Code

```
import java.util.*;
public class Main {
public static int OctaltoDecimal(int Octal) {
int Decimal = 0;
int i = 0;
while (Octal != 0) {
int rem = Octal % 10;
Decimal += rem * Math.pow(8, i);
i++;
Octal /= 10;
}
return Decimal;
}
public static void main(String[] args) {
int Octal = 345;
System.out.println("The decimal equivalent of the given octal number is
"+OctaltoDecimal(Octal));
}
}
```

**Output:**

The decimal equivalent of the given octal number is 229

**Time Complexity: **O(n) where n is the number of digits in the Octal Number.

**Space Complexity: **O(1)

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