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Thousand Separator

Difficulty: Easy

Problem Description

Given an integer n, add a dot (".") as the thousands separator and return it in string format.

Key Insights

  • The problem requires formatting an integer to include thousands separators.
  • The dot should be added every three digits from the right.
  • Edge cases include handling small numbers (less than 1000) that do not require any separators.

Space and Time Complexity

Time Complexity: O(log n) - The number of digits in n determines the number of iterations needed to format the string. Space Complexity: O(k) - Where k is the number of characters in the resulting string.

Solution

To solve the problem, the approach involves:

  1. Converting the integer to a string to manipulate its characters.
  2. Iterating through the string representation of the number from the end towards the beginning.
  3. Inserting a dot (".") after every three digits.
  4. Joining the processed characters back into a single string and returning it.

This approach uses a simple loop and string operations, ensuring that the algorithm remains efficient and easy to understand.

Code Solutions

def thousandSeparator(n: int) -> str:
    # Convert the integer to string
    s = str(n)
    result = []
    # Iterate over the string in reverse order
    for i in range(len(s)):
        # Append the current digit
        result.append(s[-1 - i])
        # If we are at every third position, add a dot unless it's the last digit
        if (i + 1) % 3 == 0 and (i + 1) != len(s):
            result.append('.')
    # The result is currently reversed, reverse it back and join to form the final string
    return ''.join(reversed(result))
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