Last Two Digit Sum in C++

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

Example:

Input: n = 2379

Output: 16

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

Understanding the Problem

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 checksum calculations, digital root computations, and more. A common pitfall is to overcomplicate the extraction of the last two digits, but it can be efficiently done using the modulo operator.

Approach

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

Use the modulo operator with 100 to get the last two digits of the number.
Extract the last digit using modulo 10.
Extract the second last digit by performing integer division by 10.
Sum the two extracted digits.

Naive Solution

A naive solution might involve converting the number to a string, extracting the last two characters, converting them back to integers, and summing them. However, this approach is not optimal due to the overhead of string operations.

Optimized Solution

The optimized solution leverages the modulo operator to directly extract the last two digits and compute their sum. This approach is efficient with a constant time complexity of O(1).

Algorithm

Here is a step-by-step breakdown of the optimized algorithm:

Compute last_two_digits = n % 100 to get the last two digits.
Compute last_digit = last_two_digits % 10 to get the last digit.
Compute second_last_digit = last_two_digits / 10 to get the second last digit.
Return the sum of last_digit and second_last_digit.

Code Implementation


#include <iostream>

int sumOfLastTwoDigits(int n) {
    // Get the last two digits
    int last_two_digits = n % 100;
    // Get the last digit
    int last_digit = last_two_digits % 10;
    // Get the second last digit
    int second_last_digit = last_two_digits / 10;
    // Return the sum of the last two digits
    return last_digit + second_last_digit;
}

int main() {
    int n = 2379;
    std::cout << "Sum of last two digits: " << sumOfLastTwoDigits(n) << std::endl;
    return 0;
}

Complexity Analysis

The time complexity of the optimized solution is O(1) because the operations involved (modulo and division) are constant time operations. The space complexity is also O(1) as we are using a fixed amount of extra space.

Edge Cases

Potential edge cases include:

Numbers with exactly two digits (e.g., 10, 99).
Numbers with trailing zeros (e.g., 100, 2300).

Each of these cases is handled correctly by the algorithm as it directly extracts and sums the last two digits.

Testing

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

n = 2379 (Expected output: 16)
n = 10 (Expected output: 1)
n = 99 (Expected output: 18)
n = 100 (Expected output: 1)
n = 2300 (Expected output: 0)

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Break down the problem into smaller, manageable parts.
Leverage mathematical operations like modulo and division for efficient solutions.
Think about edge cases and how your solution handles them.
Practice similar problems to improve your problem-solving skills.

Conclusion

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 in programming.

Additional Resources

For further reading and practice, consider the following resources: