Any base addition

  1. You are given a base b

2. You are given 2 numbers of base b

3. You are required to add two numbers and print their values in base b.

Input Format:

- A base b

- A number n1

- A number n2

Output:

sum of n1 & n2 in base b

Constraints:

2 <= b <= 10

0 <= n1 <= 256

0 <= n2 <= 256

package jan19;

public class AnyBaseAddition {
    public static void main(String[] args) {
        int n1_b_8 = 236;
        int n2_b_8 = 754;
        int base = 8;

        sum(n1_b_8, n2_b_8, base);
    }

    /**
     * Adds two numbers given in a specific base and prints the result in that base.
     *
     * Time Complexity: O(max(d1, d2))
     * - Where d1 and d2 are the number of digits in n1 and n2.
     * - In each iteration, one digit from each number is processed.
     *
     * Space Complexity: O(1)
     * - Uses a fixed amount of space regardless of input size.
     */
    private static void sum(int n1B8, int n2B8, int base) {
        int carry = 0, ans = 0, power = 0;

        while (n1B8 > 0 || n2B8 > 0 || carry > 0) {
            int a = n1B8 % 10;
            int b = n2B8 % 10;
            n1B8 = n1B8 / 10;
            n2B8 = n2B8 / 10;

            int sum = a + b + carry;
            carry = sum / base;

            ans += (sum % base) * (int) Math.pow(10, power++);
        }

        System.out.println(ans);
    }
}

Case	Time Complexity	Space Complexity	Explanation
Best Case	O(1)	O(1)	When both numbers are 0, only one loop iteration runs.
Average Case	O(d)	O(1)	Where d is the number of digits in the longer of the two numbers.
Worst Case	O(d)	O(1)	Carries continue through all digits, requiring processing every digit.

💡 Note: The number of digits d in a number n is O(log₁₀(n)), so technically it's O(log₁₀(n)), but in competitive programming and base conversion contexts, we usually denote it as O(d) to reflect digit-based iterations.