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问题:
For example if the parenthesis/brackets is matching in the following:
({})
(()){}()
()
and so on but if the parenthesis/brackets is not matching it should return false, eg:
{}
({}(
){})
(()
and so on. Can you please check this code? Thanks in advance.
public static boolean isParenthesisMatch(String str) {
Stack<Character> stack = new Stack<Character>();
char c;
for(int i=0; i < str.length(); i++) {
c = str.charAt(i);
if(c == '{')
return false;
if(c == '(')
stack.push(c);
if(c == '{') {
stack.push(c);
if(c == '}')
if(stack.empty())
return false;
else if(stack.peek() == '{')
stack.pop();
}
else if(c == ')')
if(stack.empty())
return false;
else if(stack.peek() == '(')
stack.pop();
else
return false;
}
return stack.empty();
}
public static void main(String[] args) {
String str = "({})";
System.out.println(Weekly12.parenthesisOtherMatching(str));
}
回答1:
Your code has some confusion in its handling of the '{' and '}' characters. It should be entirely parallel to how you handle '(' and ')'.
This code, modified slightly from yours, seems to work properly:
public static boolean isParenthesisMatch(String str) {
if (str.charAt(0) == '{')
return false;
Stack<Character> stack = new Stack<Character>();
char c;
for(int i=0; i < str.length(); i++) {
c = str.charAt(i);
if(c == '(')
stack.push(c);
else if(c == '{')
stack.push(c);
else if(c == ')')
if(stack.empty())
return false;
else if(stack.peek() == '(')
stack.pop();
else
return false;
else if(c == '}')
if(stack.empty())
return false;
else if(stack.peek() == '{')
stack.pop();
else
return false;
}
return stack.empty();
}
回答2:
This code is easier to understand:
public static boolean CheckParentesis(String str)
{
if (str.isEmpty())
return true;
Stack<Character> stack = new Stack<Character>();
for (int i = 0; i < str.length(); i++)
{
char current = str.charAt(i);
if (current == '{' || current == '(' || current == '[')
{
stack.push(current);
}
if (current == '}' || current == ')' || current == ']')
{
if (stack.isEmpty())
return false;
char last = stack.peek();
if (current == '}' && last == '{' || current == ')' && last == '(' || current == ']' && last == '[')
stack.pop();
else
return false;
}
}
return stack.isEmpty();
}
回答3:
The algorithm:
- scan the string,pushing to a stack for every '(' found in the string
- if char ')' scanned, pop one '(' from the stack
Now, parentheses are balanced for two conditions:
- '(' can be popped from the stack for every ')' found in the string, and
- stack is empty at the end (when the entire string is processed)
回答4:
public static boolean isValidExpression(String expression) {
Map<Character, Character> openClosePair = new HashMap<Character, Character>();
openClosePair.put(')', '(');
openClosePair.put('}', '{');
openClosePair.put(']', '[');
Stack<Character> stack = new Stack<Character>();
for(char ch : expression.toCharArray()) {
if(openClosePair.containsKey(ch)) {
if(stack.pop() != openClosePair.get(ch)) {
return false;
}
} else if(openClosePair.values().contains(ch)) {
stack.push(ch);
}
}
return stack.isEmpty();
}
回答5:
Actually, there is no need to check any cases "manually". You can just run the following algorithm:
Iterate over the given sequence. Start with an empty stack.
If the current char is an opening bracket, just push it to the stack.
If it's a closing bracket, check that the stack is not empty and the top element of the step is an appropriate opening bracket(that it is, matches this one). If it is not, report an error. Otherwise, pop the top element from the stack.
In the end, the sequence is correct iff the stack is empty.
Why is it correct? Here is a sketch of a proof: if this algorithm reported that the sequence is corrected, it had found a matching pair of all brackets. Thus, the sequence is indeed correct by definition. If it has reported an error:
If the stack was not empty in the end, the balance of opening and closing brackets is not zero. Thus, it is not a correct sequence.
If the stack was empty when we had to pop an element, the balance is off again.
If there was a wrong element on top of the stack, a pair of "wrong" brackets should match each other. It means that the sequence is not correct.
I have shown that:
If the algorithm has reported that the sequence is correct, it is correct.
If the algorithm has reported that the sequence is not correct, it is incorrect(note that I do not use the fact that there are no other cases except those that are mentioned in your question).
This two points imply that this algorithm works for all possible inputs.
回答6:
public static boolean isBalanced(String s) {
Map<Character, Character> openClosePair = new HashMap<Character, Character>();
openClosePair.put('(', ')');
openClosePair.put('{', '}');
openClosePair.put('[', ']');
Stack<Character> stack = new Stack<Character>();
for (int i = 0; i < s.length(); i++) {
if (openClosePair.containsKey(s.charAt(i))) {
stack.push(s.charAt(i));
} else if ( openClosePair.containsValue(s.charAt(i))) {
if (stack.isEmpty())
return false;
if (openClosePair.get(stack.pop()) != s.charAt(i))
return false;
}
// ignore all other characters
}
return stack.isEmpty();
}
回答7:
Algorithm is:
1)Create a stack
2)while(end of input is not reached)
i)if the character read is not a sysmbol to be balanced ,ignore it.
ii)if the character is {,[,( then push it to stack
iii)If it is a },),] then if
a)the stack is empty report an error(catch it) i.e not balanced
b)else pop the stack
iv)if element popped is not corresponding to opening sysmbol,then report error.
3) In the end,if stack is not empty report error else expression is balanced.
In Java code:
public class StackDemo {
public static void main(String[] args) throws Exception {
System.out.println("--Bracket checker--");
CharStackArray stack = new CharStackArray(10);
stack.balanceSymbol("[a+b{c+(e-f[p-q])}]") ;
stack.display();
}
}
class CharStackArray {
private char[] array;
private int top;
private int capacity;
public CharStackArray(int cap) {
capacity = cap;
array = new char[capacity];
top = -1;
}
public void push(char data) {
array[++top] = data;
}
public char pop() {
return array[top--];
}
public void display() {
for (int i = 0; i <= top; i++) {
System.out.print(array[i] + "->");
}
}
public char peek() throws Exception {
return array[top];
}
/*Call this method by passing a string expression*/
public void balanceSymbol(String str) {
try {
char[] arr = str.toCharArray();
for (int i = 0; i < arr.length; i++) {
if (arr[i] == '[' || arr[i] == '{' || arr[i] == '(')
push(arr[i]);
else if (arr[i] == '}' && peek() == '{')
pop();
else if (arr[i] == ']' && peek() == '[')
pop();
else if (arr[i] == ')' && peek() == '(')
pop();
}
if (isEmpty()) {
System.out.println("String is balanced");
} else {
System.out.println("String is not balanced");
}
} catch (Exception e) {
System.out.println("String not balanced");
}
}
public boolean isEmpty() {
return (top == -1);
}
}
Output:
--Bracket checker--
String is balanced
回答8:
import java.util.*;
class StackDemo {
public static void main(String[] argh) {
boolean flag = true;
String str = "(()){}()";
int l = str.length();
flag = true;
Stack<String> st = new Stack<String>();
for (int i = 0; i < l; i++) {
String test = str.substring(i, i + 1);
if (test.equals("(")) {
st.push(test);
} else if (test.equals("{")) {
st.push(test);
} else if (test.equals("[")) {
st.push(test);
} else if (test.equals(")")) {
if (st.empty()) {
flag = false;
break;
}
if (st.peek().equals("(")) {
st.pop();
} else {
flag = false;
break;
}
} else if (test.equals("}")) {
if (st.empty()) {
flag = false;
break;
}
if (st.peek().equals("{")) {
st.pop();
} else {
flag = false;
break;
}
} else if (test.equals("]")) {
if (st.empty()) {
flag = false;
break;
}
if (st.peek().equals("[")) {
st.pop();
} else {
flag = false;
break;
}
}
}
if (flag && st.empty())
System.out.println("true");
else
System.out.println("false");
}
}
回答9:
Ganesan's answer above is not correct and StackOverflow is not letting me comment or Edit his post. So below is the correct answer. Ganesan has an incorrect facing "[" and is missing the stack isEmpty() check.
The below code will return true if the braces are properly matching.
public static boolean isValidExpression(String expression) {
Map<Character, Character> openClosePair = new HashMap<Character, Character>();
openClosePair.put(')', '(');
openClosePair.put('}', '{');
openClosePair.put(']', '[');
Stack<Character> stack = new Stack<Character>();
for(char ch : expression.toCharArray()) {
if(openClosePair.containsKey(ch)) {
if(stack.isEmpty() || stack.pop() != openClosePair.get(ch)) {
return false;
}
} else if(openClosePair.values().contains(ch)) {
stack.push(ch);
}
}
return stack.isEmpty();
}
回答10:
Optimized implementation using Stacks and Switch statement:
public class JavaStack {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
Stack<Character> s = new Stack<Character>();
while (sc.hasNext()) {
String input = sc.next();
for (int i = 0; i < input.length(); i++) {
char c = input.charAt(i);
switch (c) {
case '(':
s.push(c); break;
case '[':
s.push(c); break;
case '{':
s.push(c); break;
case ')':
if (!s.isEmpty() && s.peek().equals('(')) {
s.pop();
} else {
s.push(c);
} break;
case ']':
if (!s.isEmpty() && s.peek().equals('[')) {
s.pop();
} else {
s.push(c);
} break;
case '}':
if (!s.isEmpty() && s.peek().equals('{')) {
s.pop();
} else {
s.push(c);
} break;
default:
s.push('x'); break;
}
}
if (s.empty()) {
System.out.println("true");
} else {
System.out.println("false");
s.clear();
}
}
} }
Cheers !
回答11:
//basic code non strack algorithm just started learning java ignore space and time.
/// {[()]}[][]{}
// {[( -a -> }]) -b -> replace a(]}) -> reverse a( }]))->
//Split string to substring {[()]}, next [], next [], next{}
public class testbrackets {
static String stringfirst;
static String stringsecond;
static int open = 0;
public static void main(String[] args) {
splitstring("(()){}()");
}
static void splitstring(String str){
int len = str.length();
for(int i=0;i<=len-1;i++){
stringfirst="";
stringsecond="";
System.out.println("loop starttttttt");
char a = str.charAt(i);
if(a=='{'||a=='['||a=='(')
{
open = open+1;
continue;
}
if(a=='}'||a==']'||a==')'){
if(open==0){
System.out.println(open+"started with closing brace");
return;
}
String stringfirst=str.substring(i-open, i);
System.out.println("stringfirst"+stringfirst);
String stringsecond=str.substring(i, i+open);
System.out.println("stringsecond"+stringsecond);
replace(stringfirst, stringsecond);
}
i=(i+open)-1;
open=0;
System.out.println(i);
}
}
static void replace(String stringfirst, String stringsecond){
stringfirst = stringfirst.replace('{', '}');
stringfirst = stringfirst.replace('(', ')');
stringfirst = stringfirst.replace('[', ']');
StringBuilder stringfirst1 = new StringBuilder(stringfirst);
stringfirst = stringfirst1.reverse().toString();
System.out.println("stringfirst"+stringfirst);
System.out.println("stringsecond"+stringsecond);
if(stringfirst.equals(stringsecond)){
System.out.println("pass");
}
else{
System.out.println("fail");
System.exit(0);
}
}
}
回答12:
import java.util.Stack;
class Demo
{
char c;
public boolean checkParan(String word)
{
Stack<Character> sta = new Stack<Character>();
for(int i=0;i<word.length();i++)
{
c=word.charAt(i);
if(c=='(')
{
sta.push(c);
System.out.println("( Pushed into the stack");
}
else if(c=='{')
{
sta.push(c);
System.out.println("( Pushed into the stack");
}
else if(c==')')
{
if(sta.empty())
{
System.out.println("Stack is Empty");
return false;
}
else if(sta.peek()=='(')
{
sta.pop();
System.out.println(" ) is poped from the Stack");
}
else if(sta.peek()=='(' && sta.empty())
{
System.out.println("Stack is Empty");
return false;
}
}
else if(c=='}')
{
if(sta.empty())
{
System.out.println("Stack is Empty");
return false;
}
else if(sta.peek()=='{')
{
sta.pop();
System.out.println(" } is poped from the Stack");
}
}
else if(c=='(')
{
if(sta.empty())
{
System.out.println("Stack is empty only ( parenthesis in Stack ");
}
}
}
// System.out.print("The top element is : "+sta.peek());
return sta.empty();
}
}
public class ParaenthesisChehck {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
Demo d1= new Demo();
// d1.checkParan(" ");
// d1.checkParan("{}");
//d1.checkParan("()");
//d1.checkParan("{()}");
// d1.checkParan("{123}");
d1.checkParan("{{{}}");
}
}
回答13:
I tried this using javascript below is the result.
function bracesChecker(str) {
if(!str) {
return true;
}
var openingBraces = ['{', '[', '('];
var closingBraces = ['}', ']', ')'];
var stack = [];
var openIndex;
var closeIndex;
//check for opening Braces in the val
for (var i = 0, len = str.length; i < len; i++) {
openIndex = openingBraces.indexOf(str[i]);
closeIndex = closingBraces.indexOf(str[i]);
if(openIndex !== -1) {
stack.push(str[i]);
}
if(closeIndex !== -1) {
if(openingBraces[closeIndex] === stack[stack.length-1]) {
stack.pop();
} else {
return false;
}
}
}
if(stack.length === 0) {
return true;
} else {
return false;
}
}
var testStrings = [
'',
'test',
'{{[][]()()}()}[]()',
'{test{[test]}}',
'{test{[test]}',
'{test{(yo)[test]}}',
'test{[test]}}',
'te()s[]t{[test]}',
'te()s[]t{[test'
];
testStrings.forEach(val => console.log(`${val} => ${bracesChecker(val)}`));
回答14:
I have seen answers here and almost all did well. However, I have written my own version that utilizes a Dictionary for managing the bracket pairs and a stack to monitor the order of detected braces. I have also written a blog post for this.
Here is my class
public class FormulaValidator
{
// Question: Check if a string is balanced. Every opening bracket is matched by a closing bracket in a correct position.
// { [ ( } ] )
// Example: "()" is balanced
// Example: "{ ]" is not balanced.
// Examples: "()[]{}" is balanced.
// "{([])}" is balanced
// "{ ( [ ) ] }" is _not_ balanced
// Input: string, containing the bracket symbols only
// Output: true or false
public bool IsBalanced(string input)
{
var brackets = BuildBracketMap();
var openingBraces = new Stack<char>();
var inputCharacters = input.ToCharArray();
foreach (char character in inputCharacters)
{
if (brackets.ContainsKey(character))
{
openingBraces.Push(character);
}
if (brackets.ContainsValue(character))
{
var closingBracket = character;
var openingBracket = brackets.FirstOrDefault(x => x.Value == closingBracket).Key;
if (openingBraces.Peek() == openingBracket)
openingBraces.Pop();
else
return false;
}
}
return openingBraces.Count == 0;
}
private Dictionary<char, char> BuildBracketMap()
{
return new Dictionary<char, char>()
{
{'[', ']'},
{'(', ')'},
{'{', '}'}
};
}
}
回答15:
If you want to have a look at my code. Just for reference
public class Default {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int numOfString = Integer.parseInt(br.readLine());
String s;
String stringBalanced = "YES";
Stack<Character> exprStack = new Stack<Character>();
while ((s = br.readLine()) != null) {
stringBalanced = "YES";
int length = s.length() - 1;
for (int i = 0; i <= length; i++) {
char tmp = s.charAt(i);
if(tmp=='[' || tmp=='{' || tmp=='('){
exprStack.push(tmp);
}else if(tmp==']' || tmp=='}' || tmp==')'){
if(!exprStack.isEmpty()){
char peekElement = exprStack.peek();
exprStack.pop();
if(tmp==']' && peekElement!='['){
stringBalanced="NO";
}else if(tmp=='}' && peekElement!='{'){
stringBalanced="NO";
}else if(tmp==')' && peekElement!='('){
stringBalanced="NO";
}
}else{
stringBalanced="NO";
break;
}
}
}
if(!exprStack.isEmpty()){
stringBalanced = "NO";
}
exprStack.clear();
System.out.println(stringBalanced);
}
}
}
回答16:
public String checkString(String value) {
Stack<Character> stack = new Stack<>();
char topStackChar = 0;
for (int i = 0; i < value.length(); i++) {
if (!stack.isEmpty()) {
topStackChar = stack.peek();
}
stack.push(value.charAt(i));
if (!stack.isEmpty() && stack.size() > 1) {
if ((topStackChar == '[' && stack.peek() == ']') ||
(topStackChar == '{' && stack.peek() == '}') ||
(topStackChar == '(' && stack.peek() == ')')) {
stack.pop();
stack.pop();
}
}
}
return stack.isEmpty() ? "YES" : "NO";
}
回答17:
Here's a solution in Python.
#!/usr/bin/env python
def brackets_match(brackets):
stack = []
for char in brackets:
if char == "{" or char == "(" or char == "[":
stack.append(char)
if char == "}":
if stack[-1] == "{":
stack.pop()
else:
return False
elif char == "]":
if stack[-1] == "[":
stack.pop()
else:
return False
elif char == ")":
if stack[-1] == "(":
stack.pop()
else:
return False
if len(stack) == 0:
return True
else:
return False
if __name__ == "__main__":
print(brackets_match("This is testing {([])} if brackets have match."))
回答18:
Was asked to implement this algorithm at live coding interview, here's my refactored solution in C#:
Git Tests
回答19:
package com.balance.braces;
import java.util.Arrays;
import java.util.Stack;
public class BalanceBraces {
public static void main(String[] args) {
String[] values = { "()]", "[()]" };
String[] rsult = match(values);
Arrays.stream(rsult).forEach(str -> System.out.println(str));
}
static String[] match(String[] values) {
String[] returnString = new String[values.length];
for (int i = 0; i < values.length; i++) {
String value = values[i];
if (value.length() % 2 != 0) {
returnString[i] = "NO";
continue;
} else {
Stack<Character> buffer = new Stack<Character>();
for (char ch : value.toCharArray()) {
if (buffer.isEmpty()) {
buffer.add(ch);
} else {
if (isMatchedBrace(buffer.peek(), ch)) {
buffer.pop();
} else {
buffer.push(ch);
}
}
if (buffer.isEmpty()) {
returnString[i] = "YES";
} else {
returnString[i] = "FALSE";
}
}
}
}
return returnString;
}
static boolean isMatchedBrace(char start, char endmatch) {
if (start == '{')
return endmatch == '}';
if (start == '(')
return endmatch == ')';
if (start == '[')
return endmatch == ']';
return false;
}
}
回答20:
Check balanced parenthesis or brackets with stack--
var excp = "{{()}[{a+b+b}][{(c+d){}}][]}";
var stk = [];
function bracket_balance(){
for(var i=0;i<excp.length;i++){
if(excp[i]=='[' || excp[i]=='(' || excp[i]=='{'){
stk.push(excp[i]);
}else if(excp[i]== ']' && stk.pop() != '['){
return false;
}else if(excp[i]== '}' && stk.pop() != '{'){
return false;
}else if(excp[i]== ')' && stk.pop() != '('){
return false;
}
}
return true;
}
console.log(bracket_balance());
//Parenthesis are balance then return true else false
回答21:
You're doing some extra checks that aren't needed. Doesn't make any diff to functionality, but a cleaner way to write your code would be:
public static boolean isParenthesisMatch(String str) {
Stack<Character> stack = new Stack<Character>();
char c;
for (int i = 0; i < str.length(); i++) {
c = str.charAt(i);
if (c == '(' || c == '{')
stack.push(c);
else if (stack.empty())
return false;
else if (c == ')') {
if (stack.pop() != '(')
return false;
} else if (c == '}') {
if (stack.pop() != '{')
return false;
}
}
return stack.empty();
}
There is no reason to peek at a paranthesis before removing it from the stack. I'd also consider wrapping instruction blocks in parantheses to improve readability.
回答22:
import java.util.*;
public class Parenthesis
{
public static void main(String...okok)
{
Scanner sc= new Scanner(System.in);
String str=sc.next();
System.out.println(isValid(str));
}
public static int isValid(String a) {
if(a.length()%2!=0)
{
return 0;
}
else if(a.length()==0)
{
return 1;
}
else
{
char c[]=a.toCharArray();
Stack<Character> stk = new Stack<Character>();
for(int i=0;i<c.length;i++)
{
if(c[i]=='(' || c[i]=='[' || c[i]=='{')
{
stk.push(c[i]);
}
else
{
if(stk.isEmpty())
{
return 0;
//break;
}
else
{
char cc=c[i];
if(cc==')' && stk.peek()=='(' )
{
stk.pop();
}
else if(cc==']' && stk.peek()=='[' )
{
stk.pop();
}
else if(cc=='}' && stk.peek()=='{' )
{
stk.pop();
}
}
}
}
if(stk.isEmpty())
{
return 1;
}else
{
return 0;
}
}
}
}
回答23:
import java.util.*;
public class MatchBrackets {
public static void main(String[] argh) {
String input = "[]{[]()}";
System.out.println (input);
char [] openChars = {'[','{','('};
char [] closeChars = {']','}',')'};
Stack<Character> stack = new Stack<Character>();
for (int i = 0; i < input.length(); i++) {
String x = "" +input.charAt(i);
if (String.valueOf(openChars).indexOf(x) != -1)
{
stack.push(input.charAt(i));
}
else
{
Character lastOpener = stack.peek();
int idx1 = String.valueOf(openChars).indexOf(lastOpener.toString());
int idx2 = String.valueOf(closeChars).indexOf(x);
if (idx1 != idx2)
{
System.out.println("false");
return;
}
else
{
stack.pop();
}
}
}
if (stack.size() == 0)
System.out.println("true");
else
System.out.println("false");
}
}
回答24:
public static bool IsBalanced(string input)
{
Dictionary<char, char> bracketPairs = new Dictionary<char, char>() {
{ '(', ')' },
{ '{', '}' },
{ '[', ']' },
{ '<', '>' }
};
Stack<char> brackets = new Stack<char>();
try
{
// Iterate through each character in the input string
foreach (char c in input)
{
// check if the character is one of the 'opening' brackets
if (bracketPairs.Keys.Contains(c))
{
// if yes, push to stack
brackets.Push(c);
}
else
// check if the character is one of the 'closing' brackets
if (bracketPairs.Values.Contains(c))
{
// check if the closing bracket matches the 'latest' 'opening' bracket
if (c == bracketPairs[brackets.First()])
{
brackets.Pop();
}
else
// if not, its an unbalanced string
return false;
}
else
// continue looking
continue;
}
}
catch
{
// an exception will be caught in case a closing bracket is found,
// before any opening bracket.
// that implies, the string is not balanced. Return false
return false;
}
// Ensure all brackets are closed
return brackets.Count() == 0 ? true : false;
}
回答25:
in java you don't want to compare the string or char by == signs. you would use equals method. equalsIgnoreCase or something of the like. if you use == it must point to the same memory location. In the method below I attempted to use ints to get around this. using ints here from the string index since every opening brace has a closing brace. I wanted to use location match instead of a comparison match. But i think with this you have to be intentional in where you place the characters of the string. Lets also consider that Yes = true and No = false for simplicity. This answer assumes that you passed an array of strings to inspect and required an array of if yes (they matched) or No (they didn't)
import java.util.Stack;
public static void main(String[] args) {
//String[] arrayOfBraces = new String[]{"{[]}","([{}])","{}{()}","{}","}]{}","{[)]()}"};
// Example: "()" is balanced
// Example: "{ ]" is not balanced.
// Examples: "()[]{}" is balanced.
// "{([])}" is balanced
// "{([)]}" is _not_ balanced
String[] arrayOfBraces = new String[]{"{[]}","([{}])","{}{()}","()[]{}","}]{}","{[)]()}","{[)]()}","{([)]}"};
String[] answers = new String[arrayOfBraces.length];
String openers = "([{";
String closers = ")]}";
String stringToInspect = "";
Stack<String> stack = new Stack<String>();
for (int i = 0; i < arrayOfBraces.length; i++) {
stringToInspect = arrayOfBraces[i];
for (int j = 0; j < stringToInspect.length(); j++) {
if(stack.isEmpty()){
if (openers.indexOf(stringToInspect.charAt(j))>=0) {
stack.push(""+stringToInspect.charAt(j));
}
else{
answers[i]= "NO";
j=stringToInspect.length();
}
}
else if(openers.indexOf(stringToInspect.charAt(j))>=0){
stack.push(""+stringToInspect.charAt(j));
}
else{
String comparator = stack.pop();
int compLoc = openers.indexOf(comparator);
int thisLoc = closers.indexOf(stringToInspect.charAt(j));
if (compLoc != thisLoc) {
answers[i]= "NO";
j=stringToInspect.length();
}
else{
if(stack.empty() && (j== stringToInspect.length()-1)){
answers[i]= "YES";
}
}
}
}
}
System.out.println(answers.length);
for (int j = 0; j < answers.length; j++) {
System.out.println(answers[j]);
}
}
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