We say a sequence of characters is a palindrome if it is the same written forwards and backwards. For exam- ple, ‘racecar’ is a palindrome, but ‘fastcar’ is not. A partition of a sequence of char- acters is a list of one or more disjoint non-empty groups of consecutive char- acters whose concatenation yields the initial sequence. For example, (‘race’, ‘car’) is a partition of ‘racecar’ into two groups. Given a sequence of characters, we can always create a partition of these characters such that each group in the partition is a palindrome! Given this observation it is natural to ask: what is the minimum number of groups needed for a given string such that ev- ery group is a palindrome? For example: • ‘racecar’ is already a palin- drome, therefore it can be par- titioned into one group. • ‘fastcar’ does not contain any non-trivial palindromes, so it must be partitioned as (‘f’, ‘a’, ‘s’, ‘t’, ‘c’, ‘a’, ‘r’). • ‘aaadbccb’ can be partitioned as (‘aaa’, ‘d’, ‘bccb’). Input Can you read upside-down? Input begins with the number n of test cases. Each test case consists of a single line of between 1 and 1000 lowercase letters, with no whitespace within. Output For each test case, output a line containing the minimum number of groups required to partition the input into groups of palindromes.
2/2 Sample Input 3 racecar fastcar aaadbccb Sample Output 1 7 3