## Algorithm

Problem Name: Java Stack

In this HackerRank Functions in Java programming problem solution,

``````In computer science, a stack or LIFO (last in, first out) is an abstract data type that serves as a collection of elements, with two principal operations: push, which adds an element to the collection, and pop, which removes the last element that was added.(Wikipedia)
``````

A string containing only parentheses is balanced if the following is true: 1. if it is an empty string 2. if A and B are correct, AB is correct, 3. if A is correct, (A) and {A} and [A] are also correct.

Examples of some correctly balanced strings are: "{}()", "[{()}]", "({()})"

Examples of some unbalanced strings are: "{}(", "({)}", "[[", "}{" etc.

Given a string, determine if it is balanced or not.

Input Format

There will be multiple lines in the input file, each having a single non-empty string. You should read input till end-of-file.

The part of the code that handles input operation is already provided in the editor.

Output Format

For each case, print 'true' if the string is balanced, 'false' otherwise.

Sample Input

``````{}()
({()})
{}(
[]
``````

Sample Output

``````true
true
false
true
``````

## Code Examples

### #1 Code Example with Java Programming

```Code - Java Programming```

``````
import java.util.Scanner;
import java.util.HashMap;
import java.util.ArrayDeque;

// ArrayDeque is "likely to be faster than Stack when used as a stack" - Java documentation

class Solution {
public static void main(String[] args) {
/* Create HashMap to match opening brackets with closing brackets */
HashMap map = new HashMap();
map.put('(', ')');
map.put('[', ']');
map.put('{', '}');

/* Test each expression for validity */
Scanner scan = new Scanner(System.in);
while (scan.hasNext()) {
String expression = scan.next();
System.out.println(isBalanced(expression, map) ? "true" : "false" );
}
scan.close();
}

private static boolean isBalanced(String expression, HashMap map) {
if ((expression.length() % 2) != 0) {
return false; // odd length Strings are not balanced
}
ArrayDeque deque = new ArrayDeque(); // use deque as a stack
for (int i = 0; i < expression.length(); i++) {
Character ch = expression.charAt(i);
if (map.containsKey(ch)) {
deque.push(ch);
} else if (deque.isEmpty() || ch != map.get(deque.pop())) {
return false;
}
}
return deque.isEmpty();
}
}
``````
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