Delete-One-Character Positions

Given strings `s1` and `s2`, where `s2` can be obtained by deleting one character from `s1`, return all deletion positions that make the transformation valid. Requirements Input: two strings s1 and s2. s2 is obtainable from s1 by deleting one character. Return every index in s1 whose deletion produces s2. If repeated characters make multiple deletions equivalent, include all valid positions. Notes First verify len(s1) == len(s2) + 1 unless the prompt guarantees it. A simple O(n²) check deletes each index and compares strings; acceptable only for small constraints. The O(n) approach finds the first mismatch, then expands across the run of identical characters in s1 that could all be deleted to produce the same s2. Duplicate-character runs are the main trap: s1 = "aaab", s2 = "aab" should return multiple positions, not just the first mismatch. Preparation Implement the brute-force version as a correctness oracle, then the O(n) run-expansion version. Test leading deletion, trailing deletion, middle deletion, and a repeated-character block with several valid answers.