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The problem requires us to count the number of symmetric integers within a specified range [low, high]. A symmetric integer is defined as an integer that remains the same when its digits are reversed. For example, 121 and 1221 are symmetric integers, but 123 is not.

Clarifying Questions

  1. Range: Are both low and high inclusive in the range?
    • Yes, both low and high are inclusive.
  2. Range Limits: What are the possible values for low and high? (e.g., can they be negative?)
    • We assume low and high are non-negative integers since we’re talking about digits. For simplicity, assume we are only dealing with positive integers where low < high.
  3. Output: Is the output expected to be an integer indicating the count of symmetric integers?
    • Yes, the output is the count of symmetric integers within the range [low, high].

Strategy

To solve the problem, we will:

  1. Iterate through each integer in the range [low, high].
  2. For each integer, convert it to its string representation and check if it reads the same forwards and backwards.
  3. If it is symmetric, increment the count.
  4. Return the final count after checking all integers in the specified range.

Code

def count_symmetric_integers(low: int, high: int) -> int:
    def is_symmetric(n: int) -> bool:
        s = str(n)
        return s == s[::-1]
    
    count = 0
    for i in range(low, high + 1):
        if is_symmetric(i):
            count += 1
            
    return count

# Example usage
low, high = 10, 200
result = count_symmetric_integers(low, high)
print(f'The number of symmetric integers between {low} and {high} is {result}.')

Time Complexity

The time complexity for this solution is (O(n \cdot d)), where:

This should be efficient enough for reasonably sized ranges. If low and high can be larger, further optimizations might be needed, but this solution is straightforward and should work well for typical input sizes in an interview scenario.

Feel free to ask further clarifying questions or adjustments!

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