Java Program For Decimal to Octal Conversion
Last Updated :
14 Apr, 2025
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Given a decimal number N, convert N into an equivalent octal number, i.e., convert the number with base value 10 to base value 8. The decimal number system uses 10 digits from 0-9, and the octal number system uses 8 digits from 0-7 to represent any numeric value.
Illustration: Basic input/output for decimal to octal conversion.
Input : 33
Output: 41Input : 10
Output: 12
Approach 1:
- Store the remainder when the number is divided by 8 in an array. Divide the number by 8 now.
- Repeat the above two steps until the number is not equal to 0.
- Print the array in reverse order now.
Example:
// Java program to convert a Decimal Number to Octal Number
// Importing input output classes
import java.io.*;
// Main class
class Geeks
{
// Method
// To convert decimal to octal
static void decToOctal(int n)
{
// Creating an Integer array of
// array to store octal number
int[] octalNum = new int[100];
// counter for octal number array
int i = 0;
while (n != 0) {
// Storing remainder in octal array
octalNum[i] = n % 8;
n = n / 8;
i++;
}
// Printing octal number array in reverse order
for (int j = i - 1; j >= 0; j--)
System.out.print(octalNum[j]);
}
// Method 2
// Main driver method
public static void main(String[] args)
{
// Custom input Integer number
int n = 33;
// Calling the above method to convert
// Decimal to Octal number
decToOctal(n);
}
}
Output
41
Complexity Analysis:
- Time complexity: O(log N)
- Auxiliary space: O(1) it is using constant space for variable and array OctalNum.
Approach 2:
- Initialize ocatalNum with 0 and countVal with 1 and the decimal number as n
- Calculate the remainder when the decimal number is divided by 8
- Update octal number with octalNum + (remainder * countval)
- Increment the countval by countval*10
- Divide the decimal number by 8
- Repeat from step 2 until the decimal number is zero
Example:
// Java Program to Convert Decimal Number to Octal Number
// Importing input output classes
import java.io.*;
// Main class
class Geeks
{
// Method 1
// To calculate the octal value of the given
// decimal number
static void decimaltooctal(int deciNum)
{
// Initially declaring and initializing the
// octal number with zero
int octalNum = 0, countval = 1;
int dNo = deciNum;
// Condition check
while (deciNum != 0) {
// Decimals remainder is calculated
int remainder = deciNum % 8;
// Storing the octalvalue
octalNum += remainder * countval;
// Storing exponential value
countval = countval * 10;
deciNum /= 8;
}
// Print and display the octal number
System.out.println(octalNum);
}
// Method 2
// Main driver method
public static void main(String[] args)
{
// Custom input decimal number
int n = 33;
// Calling the Method1 to convert above number
// to Octal number system
decimaltooctal(n);
}
}
Output
41
Complexity Analysis:
- Time complexity : O(log N)
- Auxiliary space : O(1)
