What feature new to Windows Server 2012 provides the ability to find identical sets of data on a SAN based storage array and red
1 answer:
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Answer:
Explanation:
The following code is written in Java. It uses for loops to loop through the array of characters, and uses IF statements to check if the symbols in the expression are balanced. If so it returns true, otherwise it breaks the function and returns false. Two test cases have been provided in the main method and the output can be seen in the attached image below.
import java.util.ArrayList;
class Brainly {
public static void main(String[] args) {
char[] exp = { '[', 'A', '*', '{', 'B', '(', 'C', 'D', ')', '}', ']'};
char[] exp2 = {'[', 'A', '*', '(', 'C', '}', ']'};
System.out.println(checkBalance(exp));
System.out.println(checkBalance(exp2));
}
public static boolean checkBalance(char exp[]) {
ArrayList<Character> symbols = new ArrayList<>();
for (char x: exp) {
if ((x =='[') || (x == '{') || (x == '(')) {
symbols.add(x);
}
if (x ==']') {
if (symbols.get(symbols.size()-1) == '[') {
symbols.remove(symbols.size()-1);
} else {
return false;
}
}
if (x =='}') {
if (symbols.get(symbols.size()-1) == '{') {
symbols.remove(symbols.size()-1);
} else {
return false;
}
}
if (x ==')') {
if (symbols.get(symbols.size()-1) == '(') {
symbols.remove(symbols.size()-1);
} else {
return false;
}
}
}
return true;
}
}
