How do you solve Alternating Digit Sum on LeetCode?

Alternating Digit Sum (LeetCode #2544) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding how to manipulate digits directly can lead to more efficient solutions.

What is the brute force solution for Alternating Digit Sum?

The brute force approach for Alternating Digit Sum is Brute Force, with O(n) time complexity and O(1) space complexity. The brute-force approach involves converting the number into a string to easily access each digit. We then iterate through the digits, applying the alternating sign based on the position of each digit. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Alternating Digit Sum?

Alternating Digit Sum uses the following concepts: Math. The recommended patterns to study are: Math, Array.

What are the interview tips for Alternating Digit Sum?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as the smallest and largest possible values for n. Explain your thought process while coding to demonstrate your understanding.

What are common mistakes when solving Alternating Digit Sum?

Confusing the order of operations when applying signs. Not handling single-digit numbers correctly.

#2544

Alternating Digit Sum

Easy

MathMathArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages the same logic as the brute force but avoids the overhead of string manipulation. Instead, we directly work with the number by extracting digits using modulus and division.

⚙️

Algorithm

6 steps

1Step 1: Initialize total_sum to 0 and sign to 1.
2Step 2: While n is greater than 0, extract the last digit using n % 10.
3Step 3: Add the digit multiplied by sign to total_sum.
4Step 4: Alternate the sign by multiplying by -1.
5Step 5: Remove the last digit from n by performing integer division n //= 10.
6Step 6: Return total_sum after processing all digits.

solution.py11 lines

1# Full working Python code
2
3def alternating_digit_sum(n):
4    total_sum = 0
5    sign = 1
6    while n > 0:
7        digit = n % 10
8        total_sum += sign * digit
9        sign *= -1  # Alternate sign
10        n //= 10  # Remove last digit
11    return total_sum

ℹ

Complexity note: The time complexity remains O(n) as we still process each digit once. The space complexity is O(1) since we only use a fixed number of variables.

1Understanding how to manipulate digits directly can lead to more efficient solutions.
2Recognizing patterns in alternating sums can simplify the implementation.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.