Given a non-negative integer **n** with at least **two digits**, compute and return the **sum** of the last **two digits**

**Example:**

Input:n= 2379Output:16Explanation:Last two digits of n are 7 and 9, so our answer is 7 + 9 = 16

The core challenge of this problem is to extract the last two digits of a given number and compute their sum. This problem is significant in various applications such as digital signal processing, checksum calculations, and more. A common pitfall is to overcomplicate the solution when a simple mathematical approach suffices.

To solve this problem, we can use the modulo operator to extract the last two digits and then sum them. Here's a step-by-step approach:

- Use modulo 100 to get the last two digits of the number.
- Extract the last digit using modulo 10.
- Extract the second last digit by dividing the two-digit number by 10.
- Sum the two digits and return the result.

A naive solution might involve converting the number to a string and then extracting the last two characters. However, this approach is not optimal due to the overhead of string manipulation.

The optimized solution uses simple arithmetic operations, which are more efficient:

```
public class LastTwoDigitSum {
public static int sumLastTwoDigits(int n) {
// Get the last two digits
int lastTwoDigits = n % 100;
// Extract the last digit
int lastDigit = lastTwoDigits % 10;
// Extract the second last digit
int secondLastDigit = lastTwoDigits / 10;
// Return the sum of the last two digits
return lastDigit + secondLastDigit;
}
public static void main(String[] args) {
int n = 2379;
System.out.println("Sum of last two digits: " + sumLastTwoDigits(n)); // Output: 16
}
}
```

Here's a detailed breakdown of the algorithm:

- Compute
`lastTwoDigits = n % 100`

to get the last two digits of the number. - Compute
`lastDigit = lastTwoDigits % 10`

to get the last digit. - Compute
`secondLastDigit = lastTwoDigits / 10`

to get the second last digit. - Return the sum of
`lastDigit`

and`secondLastDigit`

.

The time complexity of this solution is **O(1)** because it involves a constant number of arithmetic operations. The space complexity is also **O(1)** as no additional space is required.

Consider the following edge cases:

- When
`n`

is exactly two digits (e.g., 10, 99). - When the last two digits are zeros (e.g., 100, 200).

These cases are handled correctly by the algorithm as it directly extracts and sums the last two digits.

To test the solution comprehensively, consider the following test cases:

Test Case 1:Input: 2379 Output: 16Test Case 2:Input: 105 Output: 5Test Case 3:Input: 100 Output: 1Test Case 4:Input: 99 Output: 18

When approaching such problems, consider the following tips:

- Break down the problem into smaller, manageable parts.
- Use mathematical operations to simplify the solution.
- Consider edge cases and test your solution against them.

In this blog post, we discussed how to compute the sum of the last two digits of a given number using an efficient algorithm. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for developing strong problem-solving skills.

For further reading and practice, consider the following resources: