Check whether the given parenthesis filled string is balanced? (Java)
By Priyam SurIn this Java article, you will learn how to check if the parenthesis filled string is balanced or not in JAVA. Here we have provided a Java program on check parenthesis filled string is balanced or not in JAVA
Check if the given parenthesis containing string is balanced or not
STACK using JAVA
STACK or LIFO (last in, first out), according to Wikipedia, is an abstract data type that serves as a collection of elements in computer science, with two principal operations:
- push: adds an element to the collection
- pop: removes the last element that was added
When the following conditions are true, then we might call the given parenthesis containing string as balanced.
- If it’s an empty string
- If B and A are correct, BA is correct,
- If B is correct, (B) and {B} and [B] are also correct.
Examples are:
- Balanced strings : “(){}”, “[({})]”, “({[])})”,{()}
- Unbalanced strings : “{}([“, “([)}”, “[[{“, “}({” etc.
Now, we will see how to determine if it is balanced or not using Java.
Format of Input
The input file must have multiple lines, each having a single non-empty string. You should read input till end-of-file.
Format of Output
For the string in each input line, print corresponding output as ‘true’ if the string is balanced, else print ‘false’.
Program: Check parenthesis filled string is balanced or not in JAVA
//JAVA8
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.util.Vector;
import java.io.OutputStream;
import java.util.Stack;
import java.io.PrintWriter;
import java.io.IOException;
import java.util.StringTokenizer;
public class Solution {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
Task solver = new Task();
solver.solution(1, in, out);
out.close();
}
}
class Task {
public void solution(int testNumber, InputReader in, PrintWriter out) {
String str;
try {
while ((str = in.reader.readLine()) != null) {
Stack<Character> s = new Stack<>();
boolean ans = true;
for (int i = 0; i < str.length(); i++) {
Character my_char = str.charAt(i);
if (my_char == '(' || my_char == '{' || my_char == '[') {
s.push(my_char );
} else if (my_char == ')') {
if (!s.isEmpty() && s.peek() == '(') {
s.pop();
} else {
ans = false;
}
} else if (my_char == '}') {
if (!s.isEmpty() && s.peek() == '{') {
s.pop();
} else {
ans = false;
}
} else if (my_char == ']') {
if (!s.isEmpty() && s.peek() == '[') {
s.pop();
} else {
ans = false;
}
}
}
out.println(ans && s.isEmpty());
}
} catch (IOException e) {
}
}
}
class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
}Input
{}()
({()})
{}(
[]Output
true true false true
Hope this tutorial helped you to clear your doubts and solve your problem.
Also Read:
public class CountParens
{
public static void main(String args[])
{
for(String arg : args) {
int lp = arg.replaceAll(“[^(]”, “”).length();
int rp = arg.replaceAll(“[^)]”, “”).length();
String result = String.format(
“String \”%s\” : lp = %d rp = %d rp-lp = %d”
, arg, lp, rp, rp-lp);
System.out.println(result);
}
}
}
Example:
$ java CountParens “(((( try this out ))))” “(())())”
String “(((( try this out ))))” : lp = 4 rp = 4 rp-lp = 0
String “(())())” : lp = 3 rp = 4 rp-lp = 1
I think this would be much better. Using stack will just make it more complicated
String “)(“: lp = 1 rp = 1 rp-lp = 0