Symmetric Triplets

Count the number of length-3 contiguous substrings whose first and last characters match (case-insensitive). The classic Q1 of recent Capital One OAs — simple, but a quick win that frees time for the harder later problems. Requirements Input: a single string s. For every contiguous triplet s[i:i+3], increment a counter if s[i].lower() == s[i+2].lower(). Return the total count. Strings of length < 3 return 0. Examples s = "axA" -> 1 (a-x-A) s = "cxcbdb" -> 2 (c-x-c at i=0, b-d-b at i=3) Notes One pass over the string with an index loop; constant extra memory. The only non-trivial bit is the case-insensitive compare — normalise either both characters or the whole string up front. The middle character is irrelevant; do not be tempted to add it to the predicate. This is almost always Q1 on the OA. Finishing it in 3-4 minutes leaves a 65-minute budget for the heavier problems; spending more than 8 minutes here is a signal that something is wrong with the read. Preparation Write it without any standard-library help (no zip, no slicing) so the index arithmetic is automatic under pressure. Add a quick sanity test for non-ASCII input — interviewers occasionally throw a UTF-8 string at the same template; str.lower() is the safe normalisation.