LeetCode 2544: Alternating Digit Sum Solution

Problem Explanation

Explanation

To solve this problem, we can iterate over the digits of the given number from left to right. We will keep track of the current sign, starting with positive. For each digit, we will calculate its value with the corresponding sign and update the sign for the next digit based on the sign of the current digit.

Algorithm

Initialize a variable sign to 1 (positive).
Initialize a variable result to 0.
Iterate over the digits of the number from left to right.
For each digit:
- Calculate the value of the digit by multiplying it with the current sign.
- Add this value to the result.
- Update the sign for the next digit by multiplying the current sign with -1.
Return the final result.

Time Complexity

The time complexity of this algorithm is O(log n), where n is the given number.

Space Complexity

The space complexity of this algorithm is O(1).

Solution Code

class Solution {
    public int getSum(int n) {
        int sign = 1;
        int result = 0;
        
        while (n > 0) {
            int digit = n % 10;
            result += sign * digit;
            sign *= -1;
            n /= 10;
        }
        
        return result;
    }
}

Frequently Asked Questions

How to solve LeetCode 2544 (Alternating Digit Sum)?

This page provides optimized solutions for LeetCode problem 2544 (Alternating Digit Sum) in Java, C++, and Python, along with a detailed explanation and an interactive code editor to test your code.

What is the time complexity of LeetCode 2544 (Alternating Digit Sum)?

The time complexity for LeetCode 2544 (Alternating Digit Sum) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

Can I run code for LeetCode 2544 on DevExCode?

Yes, DevExCode provides an interactive code editor where you can write, test, and run your code for LeetCode 2544 in Java, C++, or Python.

2544. Alternating Digit Sum