TCHS-10-1000

Problem Statement
	You have a string s containing only ones and zeroes. Your goal is to make all the characters in s equal using a minimal number of operations. A single operation consists of inverting all the characters in a contiguous substring of s. Inverting a character means transforming a one into a zero, and vice versa. For example, consider the string "0001100" (quotes for clarity). We can transform this string into all ones using 2 operations: Invert the entire string: 0001100 -> 1110011. Invert the substring from the 4th character to the 5th character, inclusive: 1110011 -> 1111111. We can transform the same string into all zeros using just 1 operation: Invert the substring from the 4th character to the 5th character, inclusive: 0001100 -> 0000000. Thus, the answer for s="0001100" is 1. You are given s as a String[]. Concatenate the elements of s to obtain the actual string, and return the minimal number of operations necessary to transform s into all ones or all zeroes.
Definition
	Class: SwitchingBits Method: minSwitches Parameters: String[] Returns: int Method signature: int minSwitches(String[] s) (be sure your method is public)
Constraints
-	s will contain between 1 to 50 elements, inclusive.
-	Each element of s will contain between 1 and 50 characters, inclusive.
-	Each element of s will contain only ones ('1') and zeroes ('0').
Examples
0)
	{"0001100"} Returns: 1 Invert the two ones in the middle.
1)
	{"11111"} Returns: 0 Already all ones.
2)
	{"00000001"} Returns: 1 There are two possibilities here: invert the one, or invert all the zeroes. Either way, only 1 operation is needed.
3)
	{"1100110011001100","000","1"} Returns: 4
4)
	{"11","101","101"} Returns: 2 There are several ways. For example: 11101101 -> 11110011 -> 11111111 or 11101101 -> 11111101 -> 11111111

public class SwitchingBits {

	public int minSwitches(String[] ss) {
		String str = "";
		for (String s : ss)
			str += s;
		int a = str.replaceAll("1+", "1").replaceAll("0", "").length();
		int b = str.replaceAll("0+", "0").replaceAll("1", "").length();
		return Math.min(a, b);
	}

}