package com.thealgorithms.stacks;
import java.util.Stack;
/**
* The nested brackets problem is a problem that determines if a sequence of
* brackets are properly nested. A sequence of brackets s is considered properly
* nested if any of the following conditions are true: - s is empty - s has the
* form (U) or [U] or {U} where U is a properly nested string - s has the form
* VW where V and W are properly nested strings For example, the string
* "()()[()]" is properly nested but "[(()]" is not. The function called
* is_balanced takes as input a string S which is a sequence of brackets and
* returns true if S is nested and false otherwise.
*
* @author akshay sharma
* @author <a href="https://github.com/khalil2535">khalil2535<a>
* @author shellhub
*/
final class BalancedBrackets {
private BalancedBrackets() {
}
/**
* Check if {@code leftBracket} and {@code rightBracket} is paired or not
*
* @param leftBracket left bracket
* @param rightBracket right bracket
* @return {@code true} if {@code leftBracket} and {@code rightBracket} is
* paired, otherwise {@code false}
*/
public static boolean isPaired(char leftBracket, char rightBracket) {
char[][] pairedBrackets = {
{'(', ')'},
{'[', ']'},
{'{', '}'},
{'<', '>'},
};
for (char[] pairedBracket : pairedBrackets) {
if (pairedBracket[0] == leftBracket && pairedBracket[1] == rightBracket) {
return true;
}
}
return false;
}
/**
* Check if {@code brackets} is balanced
*
* @param brackets the brackets
* @return {@code true} if {@code brackets} is balanced, otherwise
* {@code false}
*/
public static boolean isBalanced(String brackets) {
if (brackets == null) {
throw new IllegalArgumentException("brackets is null");
}
Stack<Character> bracketsStack = new Stack<>();
for (char bracket : brackets.toCharArray()) {
switch (bracket) {
case '(':
case '[':
case '<':
case '{':
bracketsStack.push(bracket);
break;
case ')':
case ']':
case '>':
case '}':
if (bracketsStack.isEmpty() || !isPaired(bracketsStack.pop(), bracket)) {
return false;
}
break;
default:
return false;
}
}
return bracketsStack.isEmpty();
}
}