Pattern counting in a digit sequence – Count how many times the triple appears with 3 in the middle and 2 and 7 on either side (i.e., 2-3-7 or 7-3-2) in the given sequence.

Introduction / Context:The task is a local pattern count in a numeric sequence. We must find all consecutive triplets where the center is 3 and the neighbors are {2,7} in either order, i.e., 2-3-7 or 7-3-2.

Given Data / Assumptions:

• Sequence (spaces denote separation): 1 1 1 1 3 3 2 2 2 3 7 7 3 3 3 6 6 4 7 3 2 4 9 8 7 7 3 2 3 3 3 4 4 2 3 7
• Triplet window size = 3, slide by one each time.

Concept / Approach:Slide a 3-digit window along the sequence and count windows equal to (2,3,7) or (7,3,2). Each index i contributes the triplet (a[i], a[i+1], a[i+2]).

Step-by-Step Solution:Scan and mark matches: ... (2,3,7) at positions starting 9; (7,3,2) at 19; (7,3,2) at 26; (2,3,7) at 34 (1-based start indices).Total matches = 4.

Verification / Alternative check:Manual recount confirms exactly four occurrences; no overlaps are missed because windows advance one step at a time and all positions are checked.

Why Other Options Are Wrong:1/2/3 undercount the found instances; counts above 4 would require extra matches that do not exist.

Common Pitfalls:Ignoring the reversed pattern 7-3-2, or double-counting overlapping windows that are not valid matches.

Final Answer:4