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Alternating Digit Sum

Easy

Asked by:

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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

Alternating Digit Sum

Easy

Asked by:

5 views

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

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum possible value of the input integer n?
Should I handle the input as a string or an integer, and is there a preferred method?
Is n guaranteed to be a positive integer, or should I handle cases where n is zero or negative?
Could you provide an example of the expected output for a larger value of n, such as 12345?
Should I validate the input n before processing?

Brute Force Solution

Approach

To find the alternating digit sum using a brute force approach, we will explore every possible way to apply pluses and minuses between the digits. We will calculate the result for each possibility and then choose the correct answer based on those results.

Here's how the algorithm would work step-by-step:

Consider all the possible ways to put either a plus or a minus sign (or no sign at the beginning) between each pair of digits.
For each of these combinations of pluses and minuses, calculate the final sum.
Once you have calculated the sums for all the possible combinations, compare the sums.
If the first digit is implicitly positive, start with a plus. if the number is negative, start with a minus.
Starting from left to right, alternate the operations, plus then minus, until you reach the end of the number.
Add all of these alternating sums together to get the final result.

Code Implementation

def alternating_digit_sum(number):    number_string = str(number)

    number_length = len(number_string)

    total_sum = 0

    # Iterate through all possible combinations of plus and minus signs
    for i in range(2**number_length):
        current_sum = 0

        sign = 1

        for j in range(number_length):

            digit = int(number_string[j])

            # Apply the correct sign based on the iteration.
            current_sum += sign * digit

            sign *= -1

        # Add the current sum to the total sum
        total_sum = current_sum

        return total_sum

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm considers all possible combinations of plus and minus signs between the digits of the input number, which has n digits. For each of the n-1 spaces between digits, there are two choices: either a plus or a minus. Therefore, there are 2^(n-1) possible combinations to evaluate. For each combination, calculating the sum takes O(n) time. This results in a total time complexity of O(n * 2^(n-1)), which simplifies to O(2^n) as the exponential term dominates.

Space Complexity

O(2^N) – The algorithm explores all possible combinations of plus and minus signs between the digits of the input number. Since there are N-1 positions between N digits where we can place a plus or a minus, there are 2^(N-1) possible combinations. For each combination, the algorithm calculates a sum. While the sum itself takes constant space, a naive brute force implementation might store all the 2^(N-1) sums before comparing them. Therefore, the space complexity is O(2^N), where N is the number of digits in the input number.

Optimal Solution

Approach

To find the alternating digit sum, we treat the digits as positive or negative based on their position. The trick is to avoid calculating powers and instead focus on identifying a pattern to reduce calculations. We can determine the sign of each digit based on whether we start with positive or negative and proceed from left to right.

Here's how the algorithm would work step-by-step:

Start by assuming the first digit is positive.
Keep track of a 'current sign' which starts as positive.
Go through the digits of the number one by one, from left to right.
If the 'current sign' is positive, add the digit to a running total. If it's negative, subtract the digit from the running total.
After processing a digit, flip the 'current sign' to its opposite (from positive to negative, or negative to positive).
Once all the digits are processed, the running total will be the alternating digit sum.

Code Implementation

def alternating_digit_sum(number: int) -> int:
    string_representation = str(number)
    alternating_sum = 0
    current_sign = 1

    # Iterate through the digits
    for digit_char in string_representation:
        digit = int(digit_char)

        # Apply the appropriate sign to the digit
        alternating_sum += current_sign * digit

        # Flip the sign for the next digit
        current_sign *= -1

    return alternating_sum

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the digits of the input number once. Let n be the number of digits. In each iteration, it performs a constant-time operation: adding or subtracting the digit from the running sum and flipping the sign. Therefore, the time complexity is directly proportional to the number of digits, n, resulting in a linear time complexity of O(n).

Space Complexity

O(1) – The algorithm uses a constant amount of extra space. It only stores a 'current sign' variable and a 'running total' variable. The memory used by these variables does not depend on the number of digits N in the input number. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Input is 0	The alternating digit sum of 0 is 0, so the function should return 0.
Input is a single-digit number	The alternating digit sum is the number itself, so return n.
Large input number leading to potential integer overflow	The sum of the digits, even with alternating signs, will never overflow since each digit is at most 9, and the maximum number of digits is bounded by the integer representation.
Negative input	The problem states the input is a positive integer, so negative inputs are invalid and should be handled with an exception or appropriate error code.
Very large number of digits (close to integer limit)	The algorithm should still function correctly as it iterates through the digits of the number; performance would be linear.
Input resulting in alternating sum of 0	The algorithm will correctly calculate the alternating sum and return 0 in this case.
All digits are the same	The alternating sum will be either the digit itself or zero depending on the number of digits, which will be calculated correctly.
Number consisting only of 9's	This doesn't introduce any special behavior, and the normal process of alternating signs will work.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum possible value of the input integer n?
Should I handle the input as a string or an integer, and is there a preferred method?
Is n guaranteed to be a positive integer, or should I handle cases where n is zero or negative?
Could you provide an example of the expected output for a larger value of n, such as 12345?
Should I validate the input n before processing?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Consider all the possible ways to put either a plus or a minus sign (or no sign at the beginning) between each pair of digits.
For each of these combinations of pluses and minuses, calculate the final sum.
Once you have calculated the sums for all the possible combinations, compare the sums.
If the first digit is implicitly positive, start with a plus. if the number is negative, start with a minus.
Starting from left to right, alternate the operations, plus then minus, until you reach the end of the number.
Add all of these alternating sums together to get the final result.

Code Implementation

def alternating_digit_sum(number):    number_string = str(number)

    number_length = len(number_string)

    total_sum = 0

    # Iterate through all possible combinations of plus and minus signs
    for i in range(2**number_length):
        current_sum = 0

        sign = 1

        for j in range(number_length):

            digit = int(number_string[j])

            # Apply the correct sign based on the iteration.
            current_sum += sign * digit

            sign *= -1

        # Add the current sum to the total sum
        total_sum = current_sum

        return total_sum

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start by assuming the first digit is positive.
Keep track of a 'current sign' which starts as positive.
Go through the digits of the number one by one, from left to right.
If the 'current sign' is positive, add the digit to a running total. If it's negative, subtract the digit from the running total.
After processing a digit, flip the 'current sign' to its opposite (from positive to negative, or negative to positive).
Once all the digits are processed, the running total will be the alternating digit sum.

Code Implementation

def alternating_digit_sum(number: int) -> int:
    string_representation = str(number)
    alternating_sum = 0
    current_sign = 1

    # Iterate through the digits
    for digit_char in string_representation:
        digit = int(digit_char)

        # Apply the appropriate sign to the digit
        alternating_sum += current_sign * digit

        # Flip the sign for the next digit
        current_sign *= -1

    return alternating_sum

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input is 0	The alternating digit sum of 0 is 0, so the function should return 0.
Input is a single-digit number	The alternating digit sum is the number itself, so return n.
Large input number leading to potential integer overflow	The sum of the digits, even with alternating signs, will never overflow since each digit is at most 9, and the maximum number of digits is bounded by the integer representation.
Negative input	The problem states the input is a positive integer, so negative inputs are invalid and should be handled with an exception or appropriate error code.
Very large number of digits (close to integer limit)	The algorithm should still function correctly as it iterates through the digits of the number; performance would be linear.
Input resulting in alternating sum of 0	The algorithm will correctly calculate the alternating sum and return 0 in this case.
All digits are the same	The alternating sum will be either the digit itself or zero depending on the number of digits, which will be calculated correctly.
Number consisting only of 9's	This doesn't introduce any special behavior, and the normal process of alternating signs will work.