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Java Stack hackerrank solution
Problem:-
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
trueSolution:-
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static boolean isBalanced(String s) {
Stack<Character> stack = new Stack<Character>();
for (int i=0; i<s.length();++i){
if (s.charAt(i) == '(') stack.push('(');
else if (s.charAt(i) == '{') stack.push('{');
else if (s.charAt(i) == '[') stack.push('[');
else if (s.charAt(i) == ')') {
if (stack.isEmpty()) return false;
if (stack.pop() != '(') return false;
}
else if (s.charAt(i) == '}') {
if (stack.isEmpty()) return false;
if (stack.pop() != '{') return false;
}
else if (s.charAt(i) == ']') {
if (stack.isEmpty()) return false;
if (stack.pop() != '[') return false;
}
}
return stack.isEmpty();
}
public static void main(String[] args) {
Stack<Character> stack = new Stack<Character>();
Scanner sc = new Scanner(System.in);
String line;
while (sc.hasNextLine()){
line = sc.nextLine();
if (isBalanced(line)) System.out.println("true");
else System.out.println("false");
}
}
}
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