Merge pull request #4564 from manas-2003/problem1566B

Java Solution Added to cf-1566B

Author: SansPapyrus683

solutions/orphaned/cf-1566B.mdx

@@ -7,8 +7,7 @@ author: Ryan Chou
 
 [Official Editorial](https://codeforces.com/blog/entry/94803)
 
-
-# Explanation
+## Explanation
 
 Since the MEX can be at most 2 when there is a substring that contains a 0 and 1, there's only 3 possible cases. If there are no zeros, then the MEX is 0, since all substrings would be missing a 0.  If all zeros are adjacent to one another, we can cut the entire sequence out, leaving a MEX of 1. The other groups, all comprised of ones would have a MEX of 0. Leaving the answer to be $max(0, 1) = 1$. Lastly, if there are zeros and they're not adjacent to one another, then a substring like ${0, 1}$ must exist, resulting in a MEX of 2.
 
@@ -33,36 +32,67 @@ int main() {
 		// find the number of zeros
 		int zeros = count(s.begin(), s.end(), '0');
 
-		// no zeros, the answer is 0
 		if (zeros == 0) {
-			cout << 0 << endl;
+			cout << 0 << endl;  // no zeros, the answer is 0
 		} else {
 			// first and last occurences of zero
 			int l = s.find('0');
 			int r = s.rfind('0');
-
-			// if they're all adjacent to one another.
+			// if they're all adjacent to one another
 			cout << ((r - l + 1 == zeros) ? 1 : 2) << endl;
 		}
 	}
 }
 ```
+
 </CPPSection>
+<JavaSection>
+
+```java
+import java.io.*;
+
+public class MinMexCut {
+	public static void main(String[] args) throws IOException {
+		BufferedReader read =
+		    new BufferedReader(new InputStreamReader(System.in));
+		int t = Integer.parseInt(read.readLine());
+		for (int i = 0; i < t; i++) {
+			String s = read.readLine();
+			// find the number of zeros
+			int zeros = s.length() - s.replace("0", "").length();
+
+			if (zeros == 0) {
+				System.out.println(0);  // no zeros, the answer is 0
+			} else {
+				// first and last occurences of zero
+				int l = s.indexOf('0');
+				int r = s.lastIndexOf('0');
+				// if they're all adjacent to one another
+				System.out.println((r - l + 1 == zeros) ? 1 : 2);
+			}
+		}
+	}
+}
+```
 
+</JavaSection>
 <PySection>
 
 ```py
 for _ in range(int(input())):
 	s = input()
+	# find the number of zeros
 	zeros = s.count("0")
 
 	if zeros == 0:
-		print(0)
+		print(0)  # no zeros, the answer is 0
 	else:
+		# first and last occurences of zero
 		l = s.find("0")
 		r = s.rfind("0")
-
+		# if they're all adjacent to one another
 		print(1 if r - l + 1 == zeros else 2)
 ```
+
 </PySection>
 </LanguageSection>