2544. Alternating Digit Sum | Weekly Contest 329 | LeetCode 2544
Alternating Digit Sum - Video Solution
Watch 10 video solutions for Alternating Digit Sum, a easy level problem involving Math. This walkthrough by Bro Coders has 1,216 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
Problem Statement
You are given a positive integer n. Each digit of n has a sign according to the following rules:
- The most significant digit is assigned a positive sign.
- Each other digit has an opposite sign to its adjacent digits.
Return the sum of all digits with their corresponding sign.
Example 1:
Input: n = 521 Output: 4 Explanation: (+5) + (-2) + (+1) = 4.
Example 2:
Input: n = 111 Output: 1 Explanation: (+1) + (-1) + (+1) = 1.
Example 3:
Input: n = 886996 Output: 0 Explanation: (+8) + (-8) + (+6) + (-9) + (+9) + (-6) = 0.
Constraints:
1 <= n <= 109
Approach Overview
Problem Overview: Given an integer n, compute the alternating sum of its digits. The leftmost digit is added, the next digit is subtracted, the next added again, and the pattern continues until the last digit.
Approach 1: Iterative Digit Processing (O(d) time, O(1) space)
This approach walks through the digits and flips the sign after every step. The easiest implementation converts the number to a string and iterates from left to right. Add the digit when the index is even and subtract it when the index is odd. The key insight is that the problem is purely positional; you only need a running sum and a sign toggle.
You can also process digits mathematically using n % 10 and n / 10, but then you must account for the fact that digits are processed from right to left. Converting to a string keeps the implementation simple and avoids parity adjustments. This solution performs a single pass over d digits, giving O(d) time complexity with constant auxiliary space.
The approach is essentially a small math simulation problem: iterate, apply sign, flip sign, repeat.
Approach 2: Recursive Digit Evaluation (O(d) time, O(d) space)
The recursive version models the alternating pattern directly with function calls. Convert the number to a string and define a helper function that processes the digit at index i. Each call contributes either +digit or -digit depending on whether i is even or odd, then recursively evaluates the next index.
The recursion naturally mirrors the alternating structure of the problem: process current digit, flip the sign implicitly via index parity, and move forward. The recursion depth equals the number of digits, so the stack uses O(d) space. Time complexity remains O(d) because each digit is processed once.
This version is mainly educational and highlights how simple numeric problems can map to recursion. In production or interviews, the iterative solution is usually preferred because it avoids call stack overhead.
Recommended for interviews: The iterative approach is the expected solution. It runs in O(d) time with O(1) extra space and clearly shows you understand digit extraction and sign toggling. Mentioning the recursive variant demonstrates deeper understanding of problem decomposition, but the loop-based method is cleaner and more practical.
Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Iterative Digit Processing | O(d) | O(1) | Best general solution. Simple loop with sign toggling and minimal memory usage. |
| Recursive Digit Evaluation | O(d) | O(d) | Useful for demonstrating recursion concepts or when modeling position-based digit processing. |