Is Alternating Digit Sum easy or hard?

Alternating Digit Sum is categorized as an Easy problem. It focuses on digit extraction, iteration, and sign management rather than advanced algorithms or data structures.

Alternating Digit Sum Python/Java solution

In Python or Java, convert the number to a string and iterate through each character. Parse the digit, add it when the index is even, subtract it when the index is odd, and keep updating a running total. This implementation runs in O(d) time with constant auxiliary space.

How to solve Alternating Digit Sum in O(n)?

Treat n as a sequence of digits and iterate through them once. Maintain a running sum and a sign variable that flips between +1 and -1 after each digit. Adding digit * sign during the pass computes the alternating sum in linear time relative to the number of digits.

What is the best approach for Alternating Digit Sum?

The iterative digit processing approach is the most practical solution. Traverse the digits from left to right, add digits at even positions and subtract digits at odd positions. This runs in O(d) time where d is the number of digits and uses O(1) extra space.

Is Alternating Digit Sum asked at Google/Amazon/Meta?

Alternating Digit Sum is an easy-level problem commonly used in coding screens to test basic number manipulation and iteration logic. While not a classic FAANG interview staple, problems with similar digit-processing patterns appear in interviews at companies like Amazon and Google.

What data structure is used in Alternating Digit Sum?

No specialized data structure is required. The problem is solved using simple arithmetic operations or by iterating through a string representation of the number, along with a running accumulator and sign toggle.

What is the time complexity of Alternating Digit Sum?

The time complexity is O(d), where d is the number of digits in the integer. Each digit is processed exactly once while computing the alternating sum, and no additional nested operations are required.

Description Hints 3 Solution Video 4:18

2544. Alternating Digit Sum

Easy69.7% acceptance4 companies

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

Asked at 4 companies

eBay Visa Capital One Amazon

Topics

Math

Hints open one at a time

Hint 1

The first step is to loop over the digits. We can convert the integer into a string, an array of digits, or just loop over its digits.

Hint 2

Keep a variable sign that initially equals 1 and a variable answer that initially equals 0.

Hint 3

Each time you loop over a digit i, add sign * i to answer, then multiply sign by -1.

4:18

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