algoadvance

Given an integer num, repeatedly add all its digits until the result has only one digit, and return it.

Example:

• Input: num = 38
• Output: 2
  • Explanation: The process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.

Clarifying Questions

1. Is the input always a non-negative integer?
   • Yes, the problem statement implies that the input integer is non-negative.
2. What should be returned if the input is 0?
   • The process would indicate that 0 is returned since no further addition is required.
3. Is there a limit to the size of the integer num?
   • The problem does not specify a limit, so we should assume num can be a large non-negative integer.

Strategy

The task essentially requires iteratively summing the digits of the number until we get a single digit. There are two common approaches to solve this problem:

1. Iterative Approach:
   • Convert the number to a string to easily access each digit.
   • Sum the digits repeatedly until a single digit is obtained.
2. Mathematical Approach (Digital Root):
   • Utilize the properties of numbers, specifically using the digital root congruence formula to find the result in constant time. The digital root for a number is given by 1 + (num - 1) % 9.

Code

We’ll use the mathematical approach for its efficiency.

def addDigits(num: int) -> int:
    if num == 0:
        return 0
    else:
        return 1 + (num - 1) % 9

# Example usage
print(addDigits(38))  # Output: 2

Time Complexity

• Iterative Approach: O(log(num))
  • Each iteration reduces the number of digits, hence the overall time complexity is proportional to the number of digits in num.
• Mathematical Approach: O(1)
  • The formula 1 + (num - 1) % 9 directly computes the result in constant time.

In this implementation, we’ve chosen the mathematical approach due to its straightforward and efficient constant time complexity.

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