# The Alphabet Score > Every letter has a secret numeric value - what's your total? Canonical URL: Domain: Python · Difficulty: easy · Seniority: L3 ## Problem Given a string, return the sum of alphabetic character values where a=1..z=26, case-insensitive. Non-alphabetic characters contribute 0. ## Worked solution and explanation ### Why this problem exists in real interviews Character-to-value scoring tests whether you know the ord/chr ASCII trick or whether you'd build a 26-entry dict by hand. Interviewers grade this on how cleanly you state the alphabetic-only filter, how you handle case, and whether your final expression is a single sum() over a generator (Pythonic) or a manual accumulator (junior). --- ### Break down the requirements #### Step 1: Lowercase once, filter alphabetic Lowercase the entire string up front so you only need one ord-anchor (ord('a')). Use str.isalpha() to skip non-letters; this also correctly handles digits, punctuation, and whitespace as zero contributions. #### Step 2: Convert each letter with ord arithmetic For a lowercase letter c, the value is ord(c) - ord('a') + 1. This maps 'a' to 1 and 'z' to 26 in one expression with no lookup table. Memorizing this idiom saves you from constructing a 26-item dict, which is the most common over-engineering. #### Step 3: Sum in one expression Wrap the conversion in a generator expression inside sum(). It runs in C, allocates no intermediate list, and reads as the spec sentence: 'sum of alphabetic character values.' Empty input returns 0 with no special case. --- ### The solution **Lowercase, filter, ord-arithmetic, sum** ```python def convert_string_mapping(s: str) -> int: s = s.lower() return sum(ord(c) - ord('a') + 1 for c in s if c.isalpha()) ``` > **Cost Analysis** > > Time is O(n) for one lowercase pass and one summation pass. Space is O(n) for the lowercased copy (Python strings are immutable). The generator expression avoids materializing a second list, so peak memory stays at the lowercased string. > **Interviewers Watch For** > > Whether you reach for ord arithmetic instead of building a 26-entry dict, whether you handle case explicitly with .lower(), and whether you use isalpha() rather than a manual range check. Strong candidates also mention that isalpha() on Python strings is Unicode-aware and would accept letters outside a-z, which may or may not be desired. > **Common Pitfall** > > Using c.isalpha() without lowercasing and then computing ord(c) - ord('a') + 1 for an uppercase letter, which gives negative or wrong values (uppercase A is ord 65, lowercase a is 97). Either lowercase first or branch on case. Another miss is filtering with c.isascii() and c.isalpha() together for safety, but then forgetting that isalpha() on Greek or Cyrillic letters returns True; clarify scope with the interviewer. --- ## Common follow-up questions - How would the solution change for case-sensitive scoring (uppercase contributes 27..52)? _(branch on isupper() or build a small dict; show both.)_ - What if the alphabet weights are arbitrary (e.g., Scrabble values)? _(replace ord-arithmetic with a dict lookup; mention str.translate for very hot paths.)_ - How would you score a 10 GB text file without loading it all? _(stream chunk by chunk, accumulate the sum; note that lowercasing on chunk boundaries is safe for ASCII but not for some Unicode case-folding edge cases.)_ ## Related - [All practice problems](https://datadriven.io/problems) - [Mock interview mode](https://datadriven.io/interview/the_alphabet_score) - [Python Interview Questions](https://datadriven.io/python-interview-questions) - [Data Engineering Interview Prep Guide](https://datadriven.io/data-engineer-interview-prep) - [Daily Challenge](https://datadriven.io/daily) --- Source: DataDriven (https://datadriven.io). 100% free data engineering interview prep. Live code execution against Postgres 16, Python 3.11, and Spark sandboxes. No paywall, no premium tier, no signup gate.