#2544

Easy

auto_awesomeMath-driven solution strategy

LeetCode Problem Workspace

Alternating Digit Sum

Compute the alternating digit sum by applying a sign to each digit and summing efficiently using a math-driven approach.

math

Problem Statement

You are given a positive integer n. Each digit in n alternates in sign starting with positive for the most significant digit, then negative for the next, and so on. Your task is to calculate the sum of all digits with their corresponding sign.

Return the alternating sum of digits as an integer. For example, given n = 521, the sum is calculated as (+5) + (-2) + (+1) = 4. Constraints: 1 <= n <= 10^9.

Examples

Example 1

Input: n = 521

Output: 4

(+5) + (-2) + (+1) = 4.

Example 2

Input: n = 111

Output: 1

(+1) + (-1) + (+1) = 1.

Example 3

Input: n = 886996

Output: 0

(+8) + (-8) + (+6) + (-9) + (+9) + (-6) = 0.

Constraints

1 <= n <= 109

Solution Approach

Convert integer to digit array

Transform n into a string or array of digits so you can access each digit in order. This avoids mistakes with positional math and makes sign alternation straightforward.

Iterate with alternating sign

Use a loop over the digits, starting with a positive sign for the first digit and flip the sign at each step. Sum the digits according to the current sign during iteration.

Optimize without extra storage

Instead of storing digits, repeatedly extract the last digit using modulo and integer division, while keeping a sign flag. This reduces space complexity to O(1) while maintaining linear time relative to the number of digits.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(d) where d is the number of digits in n because each digit is processed once. Space complexity is O(d) if using an array of digits, or O(1) if processing digits in place without extra storage.

What Interviewers Usually Probe

Mentions looping through digits explicitly.
Asks about sign alternation and edge cases like single-digit numbers.
Checks if you can optimize space without losing clarity.

Common Pitfalls or Variants

Common pitfalls

Forgetting to start with positive for the first digit.
Incorrectly flipping signs when iterating over digits.
Assuming integer math alone without converting to accessible digit form.

Follow-up variants

Compute alternating sum starting from the least significant digit.
Apply different sign patterns such as ++-- repeating for every four digits.
Calculate modulo result of alternating digit sum for very large n.

FAQ

What is the best approach to solve Alternating Digit Sum efficiently?

Convert the integer to digits and iterate while flipping the sign, or process digits in-place using modulo/division for O(1) space.

Why does the first digit start positive in Alternating Digit Sum?

The problem defines the pattern starting with positive for the most significant digit; this ensures consistent summation for all n.

Can I use recursion for Alternating Digit Sum?

Yes, but recursion adds call stack overhead. Iterative digit processing is usually simpler and more efficient.

What happens if n has only one digit?

The sum is simply that digit since the first digit is always positive, so no alternation occurs.

How does the math-driven solution pattern help prevent errors?

It clarifies digit handling, ensures correct sign alternation, and allows optimizations like in-place processing without losing correctness.

terminal

Solution

Solution 1: Simulation

We can directly simulate the process as described in the problem.

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class Solution:
    def alternateDigitSum(self, n: int) -> int:
        return sum((-1) ** i * int(x) for i, x in enumerate(str(n)))

Solution 2

#### Python3

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class Solution:
    def alternateDigitSum(self, n: int) -> int:
        return sum((-1) ** i * int(x) for i, x in enumerate(str(n)))

Continue Practicing

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