Please refer to the palindromic pangram problem on this page. AnalysisLet \(A_{p}\) represent a word and let \(A_{pr}\) represent its reverse. Let \(A_{1}\), \(A_{2}\), ..., \(A_{n1}\), \(A_{n}\) be the subsequences of \(A_{pr}\) such that all of \(A_{1}\) \(A_{2}\) ..., \(A_{n1}\) are valid words while \(A_{n}\) could be a valid word, but must be a palindrome. Then we can say that the following is a pangram for word \(A_{p}\): $$Pangram_{A_{p}} \leftarrow \sum_{i = 1}^{i = n1} A + A_{p}$$ Let us verify if this holds true with the following two examples cited in the problem statement: $$Pangram_{daffodil} \leftarrow lid + off + a + daffodil$$ where $A_{n}$ = 'd' which is a palindrome, but not a valid word! and, $$Pangram_{ayatollahs} \leftarrow shallot + ayatollahs$$ where \(A_{n}\) = 'aya' which is a palindraome, but not a valid word! It can be easily observed here that \(A_{1}\), \(A_{2}\), ...,\(A_{n}\) are all prefixes of \(A_{pr}\). Or we can think of them as substring q of string \(A_{pr}\). We also know then that a suffix tree is a good data structure to do string matching! Algorithm
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MeI am a polyglot software engineer specializing in shipping iOS and 3d scientific visualization applications. Archives
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