## DancingSentence : TopCoder Problem 5950

[The problem appeared in TopCoder SRM 279 (Div-2, Level-1) and SRM 279 (Div-1, Level-1)]

Problem Statement:
A sentence is called dancing if its first letter is uppercase and the case of each subsequent letter is the opposite of the previous letter. Spaces should be ignored when determining the case of a letter. For example, “A b Cd” is a dancing sentence because the first letter (‘A’) is uppercase, the next letter (‘b’) is lowercase, the next letter (‘C’) is uppercase, and the next letter (‘d’) is lowercase.

You will be given a String sentence. Turn the sentence into a dancing sentence by changing the cases of the letters where necessary. All spaces in the original sentence must be preserved.

## PalindromeMaker : TopCoder Problem 5881

[The problem appeared in TopCoder SRM 274 (Div-2, Level-2) and SRM 274 (Div-1, Level-1)]

Problem Statement:
A palindrome is a string that is spelled the same forward and backward. We want to rearrange letters of the given string baseString so that it becomes a palindrome.

You will be given a String baseString. Return the palindrome that can be made from baseString. When more than one palindrome can be made, return the lexicographically earliest (i.e., the one that occurs first in alphabetical order). Return “” (the empty string) if no palindromes can be made from baseString.

## FewestFactors : TopCoder Problem 5886

[The problem appeared in TopCoder SRM 272 (Div-1, Level-1) and SRM 271 (Div-2, Level-2)]

Problem Statement:
You will be given some decimal digits in a int[] digits. Build an integer with the minimum possible number of factors, using each of the digits exactly once (be sure to count all factors, not only the prime factors). If more than one number has the same (minimum) number of factors, return the smallest one among them.

## CompletingBrackets : TopCoder Problem 5977

[The problem appeared in TopCoder SRM 280 (Div-1, Level-1) and TopCoder SRM 280 (Div-2, Level-2)]

Problem Statement:
A series of brackets is complete if we can pair off each left bracket ‘[‘ with a right bracket ‘]’ that occurs later in the series. Every bracket must participate in exactly one such pair.

Given a String text add the minimal number of brackets to the beginning and/or end of text to make it complete. Return the result.