Given a string 'AAAVGXFHHFSGFGGLK', find the non-repeating letters.

To find non-repeating letters in a string, the most efficient approach involves a two-pass scan: first, count the frequency of each character, then iterate through the string again to identify characters with a count of one. Python's collections.Counter simplifies the frequency counting step.

Mechanics and Why

This problem leverages character frequency analysis. The first pass builds a frequency map (or hash table) where keys are characters and values are their counts. This takes O(n) time, where n is the length of the string, as each character is visited once. The second pass iterates through the original string, checking the count of each character in the pre-computed frequency map. If a character's count is 1, it's a non-repeating letter. This second pass also takes O(n) time. The overall time complexity is therefore O(n) because constant factors are dropped. Space complexity is O(k) where k is the number of unique characters (e.g., at most 26 for English alphabet, or 256 for ASCII).

Example and Production Considerations

The definition of "non-repeating" is crucial in a production context. Does it mean characters that appear exactly once in the entire string, or the first occurrence of a character that hasn't been seen yet? The provided solution addresses the former. For the latter (e.g., finding the first unique character), the approach would need modification, potentially using an ordered dictionary or tracking insertion order.

from collections import Counter
def find_non_repeating_letters(s: str) -> list[str]:
    """
    Finds all characters that appear exactly once in the input string.
    Example: 'AAAVGXFHHFSGFGGLK' -> ['V', 'X', 'L', 'K']
    """
    if not s:
        return []
char_counts = Counter(s)
    non_repeating = [char for char in s if char_counts[char] == 1]
    return non_repeating

This pattern of counting frequencies is fundamental and scales to large datasets. For instance, in distributed processing frameworks like Apache Spark, counting word frequencies (a similar problem) would involve a flatMap to emit individual words, followed by a groupByKey or reduceByKey operation to aggregate counts across partitions, demonstrating how this core algorithmic concept applies to massive-scale data engineering tasks.

In the interview, also mention…

Discuss edge cases like empty strings, strings with only repeating characters, or strings with only unique characters. Also, consider the impact of character set (ASCII, Unicode) on space complexity.