so the past couple of hours, i have been trying to understand how the filter function works in kotlin and if it has any correlation with that of Java.
basically, i have a code that's written in java and i would love to have it transcribed to kotlin
private List<Order> getFilteredOrders(Courier courier) {
String[] glovoBoxKeywords = glovoBoxWords.toLowerCase().split(",");
List<Vehicle> allowedVehicles = Arrays.asList(MOTORCYCLE, ELECTRIC_SCOOTER);
return orders.stream()
.filter(order -> {
String description = order.getDescription().toLowerCase();
if (!courier.getBox()) {
return Arrays.stream(glovoBoxKeywords).noneMatch(description::contains);
}
return true;
})
.filter(order -> {
Location pickupLocation = order.getPickup();
Location deliveryLocation = order.getDelivery();
Double distance = calculateDistance(pickupLocation, deliveryLocation);
if (distance > longDeliveryDistance) {
return allowedVehicles.contains(courier.getVehicle());
}
return true;
})
.collect(Collectors.toList());
}
i tried this but i got at this, and was literally stuck :(
private fun findFilteredOrder(courier: Courier) : List<Order> {
val glovoBoxKeyWords = glovoBoxWords.toLowerCase().split(",")
val allowedVehicles = listOf(Vehicle.ELECTRIC_SCOOTER, Vehicle.MOTORCYCLE)
orderList.filter { order ->
val description = order.getDescription().toLowerCase()
if(!courier.getBox()) {
}
true
}.filter {
val pickupLocation = it.getPickup()
val deliveryLocation = it.getDelivery()
val distance = calculateDistance(deliveryLocation, pickupLocation)
if(distance > longDeliveryDistance) {
courier.getVehicle() in allowedVehicles
}
true
}
}
Please this is my first attempt and doing something with kotlin, so please go easy guys. thanks, also i'd be appreciative if anyone could help me with informative stuff as to how to understand these kotlin functions better. let, apply, associateBy... etc.. THANKS
The filter function in Kotlin Collections has the same principle as other frameworks/libraries, including Java Streams. Given a predicate (a function from the type of the collection to Boolean) it will return a new collection with the elements matching the predicate. You can find more information and examples of other functions and operators in the official documentation and here.
Your code was almost there, I translate the Java Stream operation to Kotlin List and rewrite the return statements to remove the redundant if
private fun findFilteredOrder(courier: Courier) : List<Order> {
val glovoBoxKeyWords = glovoBoxWords.toLowerCase().split(",")
val allowedVehicles = listOf(Vehicle.ELECTRIC_SCOOTER, Vehicle.MOTORCYCLE)
orderList.filter { order ->
val description = order.getDescription().toLowerCase()
courier.getBox() || glovoBoxKeywords.none { it in description }
}.filter { order ->
val pickupLocation = order.getPickup()
val deliveryLocation = order.getDelivery()
val distance = calculateDistance(deliveryLocation, pickupLocation)
distance <= longDeliveryDistance || courier.getVehicle() in allowedVehicles
}
}
I don't know why no one mentioned the use of labels: https://kotlinlang.org/docs/returns.html#break-and-continue-labels.
Since this question has a nice google ranking, I'll add what I was originally searching for.
The OP probably was aware that filter needs a predicate that returns a Boolean and that the filter will return a list with the items that pass the predicate (the items which the predicate returned true).
What he was not aware is that we can "emulate" Java returns through Kotlin labels:
private fun findFilteredOrder(courier: Courier) : List<Order> {
val glovoBoxKeyWords = glovoBoxWords.toLowerCase().split(",")
val allowedVehicles = listOf(Vehicle.ELECTRIC_SCOOTER, Vehicle.MOTORCYCLE)
orderList.filter shouldSkip#{ order ->
val description = order.getDescription().toLowerCase()
if (courier.getBox()) {
return#shouldSkip true
}
if (glovoBoxKeywords.none { it in description }) {
return#shouldSkip true
}
return#shouldSkip false
}.filter shouldSkip# { order ->
val pickupLocation = order.getPickup()
val deliveryLocation = order.getDelivery()
val distance = calculateDistance(deliveryLocation, pickupLocation)
if (distance <= longDeliveryDistance) {
return#shouldSkip true
}
if (courier.getVehicle() in allowedVehicles) {
return#shouldSkip true
}
return#shouldSkip false
}
}
Since Kotlin allows us to return in the last block line and the return keyword returns to the outer scope, it is pretty easy to:
filter {
startPutting >= someMagic && andComplex ||
verificationsThat.is { hardToUnderstand }.because {
weNeedToReturnHere
}
}
The labels allow us to be more verbose but also more clear.
In my Android app, I am trying to sort Bus route tags in order 1, 2, 3..etc.
For that I am using this
Collections.sort(directions, Comparator { lhs, rhs ->
var obj1 = lhs.short_names.firstOrNull() ?: ""
var obj2 = rhs.short_names.firstOrNull() ?: ""
if (obj1 === obj2) {
obj1 = lhs.headsigns.firstOrNull() ?: ""
obj2 = rhs.headsigns.firstOrNull() ?: ""
if (obj1 === obj2) {
return#Comparator 0
}
obj1.compareTo(obj2)
} else {
obj1.compareTo(obj2)
}
The issue I am having is this sorts them, but will run into the issue of
1, 2, 3, 30, 31, 4, 5
How should I change this to get the correct ordering.
If you need just a simple number comparison you can do it like that.
directions.sortWith(Comparator { lhs, rhs ->
val i1 = lhs.toInt()
val i2 = rhs.toInt()
when {
i1 < i2 -> -1
i1 > i2 -> 1
else -> 0
}
})
As hotkey pointed out the code above can be replaced with almost identical implementation that looks much simplier.
directions.sortBy { it.toInt() }
The general version of this algorithm is called alphanum sorting and described in details here. I made a Kotlin port of this algorithm, which you can use. It's more complicated than what you need, but it will solve your problem.
class AlphanumComparator : Comparator<String> {
override fun compare(s1: String, s2: String): Int {
var thisMarker = 0
var thatMarker = 0
val s1Length = s1.length
val s2Length = s2.length
while (thisMarker < s1Length && thatMarker < s2Length) {
val thisChunk = getChunk(s1, s1Length, thisMarker)
thisMarker += thisChunk.length
val thatChunk = getChunk(s2, s2Length, thatMarker)
thatMarker += thatChunk.length
// If both chunks contain numeric characters, sort them numerically.
var result: Int
if (isDigit(thisChunk[0]) && isDigit(thatChunk[0])) {
// Simple chunk comparison by length.
val thisChunkLength = thisChunk.length
result = thisChunkLength - thatChunk.length
// If equal, the first different number counts.
if (result == 0) {
for (i in 0..thisChunkLength - 1) {
result = thisChunk[i] - thatChunk[i]
if (result != 0) {
return result
}
}
}
} else {
result = thisChunk.compareTo(thatChunk)
}
if (result != 0) {
return result
}
}
return s1Length - s2Length
}
private fun getChunk(string: String, length: Int, marker: Int): String {
var current = marker
val chunk = StringBuilder()
var c = string[current]
chunk.append(c)
current++
if (isDigit(c)) {
while (current < length) {
c = string[current]
if (!isDigit(c)) {
break
}
chunk.append(c)
current++
}
} else {
while (current < length) {
c = string[current]
if (isDigit(c)) {
break
}
chunk.append(c)
current++
}
}
return chunk.toString()
}
private fun isDigit(ch: Char): Boolean {
return '0' <= ch && ch <= '9'
}
}
To use this Comparator just call
directions.sortWith(AlphanumComparator())
If you don't need it to be coded in Kotlin you can just take an original Java version on Dave Koelle's page. And the Kotlin version of the algorithm can be also found on GitHub.
I want to get int parameter and save it to playerprefs and then get it and change it to string and display as string, but when I use ToString(); there is error: Cannot implicitly convert type string' toint' for a=b;
C# unity
public int a;
public string b;
{
car = GameObject.Find ("mastergame").GetComponent<master> ().carr;
if (car == 1) {
busted = GameObject.Find ("1").GetComponent<ruchauta1> ().busted;
gameover = GameObject.Find ("1").GetComponent<ruchauta1> ().gameover;
if (gameover == false && busted == false) {
scoree.text = "" + score;
}
if (gameover == true || busted == true) {
if(score>bestscore){
bestscore=score;
PlayerPrefs.SetInt("bestscore",bestscore);}
score=0;
PlayerPrefs.SetInt("score",score);
}
} else {
busted = GameObject.Find ("Audi").GetComponent<ruchauta> ().busted;
gameover = GameObject.Find ("Audi").GetComponent<ruchauta> ().gameover;
if (gameover == false && busted == false) {
scoree.text = "" + score;
}
if (gameover == true || busted == true) {
if(score>bestscore){
bestscore=score;
PlayerPrefs.SetInt("bestscore",bestscore);
a=PlayerPrefs.GetInt("bestscore",bestscore);}
score=0;
PlayerPrefs.SetInt("score",score);
}
}
a.ToString ();
a = b;
bestscoree.text = b;
}
void Score(){
car = GameObject.Find ("mastergame").GetComponent<master> ().carr;
if (car == 1) {
busted = GameObject.Find ("1").GetComponent<ruchauta1> ().busted;
gameover = GameObject.Find ("1").GetComponent<ruchauta1> ().gameover;
if (gameover == false && busted == false) {
score+=5;
PlayerPrefs.SetInt("score",score);
}
} else {
busted = GameObject.Find ("Audi").GetComponent<ruchauta> ().busted;
gameover = GameObject.Find ("Audi").GetComponent<ruchauta> ().gameover;
if (gameover == false && busted == false) {
score+=5;
PlayerPrefs.SetInt("score",score);
}
}
toString itself does not convert a's type, what you should do is:
String newString = a.toString()
after that, you can play with your newString.
ToString() creates a string representation of an object and returns that value. It does not (or should not) change the value/contents of the object it is operating on and it certainly does not change the type of the object. So the line with a.ToString() likely results in no change.
Since b is a string (based upon the error you gave) and presumably contains a string representation of an integer, if you want to set the value of a with the value of b, you must first convert b to an integer. I'm not sure what language this is but that is typically done with one of several methods. Some examples:
a = Convert.ToInt(b);
a = int.Parse(b);
a = Int32.Parse(b);
Looking at the Scala 2.10.0's implementation of LinearSeqOptimized#find in LinearSeqOptimized.scala, why is it necessary to call var these = this?
Why couldn't this simply be used?
override /*IterableLike*/
def find(p: A => Boolean): Option[A] = {
var these = this
while (!these.isEmpty) {
if (p(these.head)) return Some(these.head)
these = these.tail
}
None
}
Because you would have to have the same condition and operation out of the loop for this and then start using these.
It's much simpler to just put everyone in the same basket and do it all in the loop itself. Example:
def find(p: A => Boolean): Option[A] = {
if (!this.isEmpty && p(this.head)) {
return Some(this.head)
}
var these = this.tail
while (!these.isEmpty) {
if (p(these.head)) return Some(these.head)
these = these.tail
}
None
}
Not very smart, as you can see.
You could also easily implement this as a #tailrec operation:
#tailrec final def find[A](p : A => Boolean) : Option[A] = {
if ( this.isEmpty ) {
None
} else {
if ( p(this.head) ) {
Some(this.head)
} else {
this.tail.find(p)
}
}
}
And it isn't done like this in Scala because tailrec calls have to be final or private.
I've written software in the past that uses a stack to check for balanced equations, but now I'm asked to write a similar algorithm recursively to check for properly nested brackets and parenthesis.
Good examples: () [] ()
([]()[])
Bad examples: ( (] ([)]
Suppose my function is called: isBalanced.
Should each pass evaluate a smaller substring (until reaching a base case of 2 left)? Or, should I always evaluate the full string and move indices inward?
First, to your original question, just be aware that if you're working with very long strings, you don't want to be making exact copies minus a single letter each time you make a function call. So you should favor using indexes or verify that your language of choice isn't making copies behind the scenes.
Second, I have an issue with all the answers here that are using a stack data structure. I think the point of your assignment is for you to understand that with recursion your function calls create a stack. You don't need to use a stack data structure to hold your parentheses because each recursive call is a new entry on an implicit stack.
I'll demonstrate with a C program that matches ( and ). Adding the other types like [ and ] is an exercise for the reader. All I maintain in the function is my position in the string (passed as a pointer) because the recursion is my stack.
/* Search a string for matching parentheses. If the parentheses match, returns a
* pointer that addresses the nul terminator at the end of the string. If they
* don't match, the pointer addresses the first character that doesn't match.
*/
const char *match(const char *str)
{
if( *str == '\0' || *str == ')' ) { return str; }
if( *str == '(' )
{
const char *closer = match(++str);
if( *closer == ')' )
{
return match(++closer);
}
return str - 1;
}
return match(++str);
}
Tested with this code:
const char *test[] = {
"()", "(", ")", "", "(()))", "(((())))", "()()(()())",
"(() ( hi))) (())()(((( ))))", "abcd"
};
for( index = 0; index < sizeof(test) / sizeof(test[0]); ++index ) {
const char *result = match(test[index]);
printf("%s:\t", test[index]);
*result == '\0' ? printf("Good!\n") :
printf("Bad # char %d\n", result - test[index] + 1);
}
Output:
(): Good!
(: Bad # char 1
): Bad # char 1
: Good!
(())): Bad # char 5
(((()))): Good!
()()(()()): Good!
(() ( hi))) (())()(((( )))): Bad # char 11
abcd: Good!
There are many ways to do this, but the simplest algorithm is to simply process forward left to right, passing the stack as a parameter
FUNCTION isBalanced(String input, String stack) : boolean
IF isEmpty(input)
RETURN isEmpty(stack)
ELSE IF isOpen(firstChar(input))
RETURN isBalanced(allButFirst(input), stack + firstChar(input))
ELSE IF isClose(firstChar(input))
RETURN NOT isEmpty(stack) AND isMatching(firstChar(input), lastChar(stack))
AND isBalanced(allButFirst(input), allButLast(stack))
ELSE
ERROR "Invalid character"
Here it is implemented in Java. Note that I've switched it now so that the stack pushes in front instead of at the back of the string, for convenience. I've also modified it so that it just skips non-parenthesis symbols instead of reporting it as an error.
static String open = "([<{";
static String close = ")]>}";
static boolean isOpen(char ch) {
return open.indexOf(ch) != -1;
}
static boolean isClose(char ch) {
return close.indexOf(ch) != -1;
}
static boolean isMatching(char chOpen, char chClose) {
return open.indexOf(chOpen) == close.indexOf(chClose);
}
static boolean isBalanced(String input, String stack) {
return
input.isEmpty() ?
stack.isEmpty()
: isOpen(input.charAt(0)) ?
isBalanced(input.substring(1), input.charAt(0) + stack)
: isClose(input.charAt(0)) ?
!stack.isEmpty() && isMatching(stack.charAt(0), input.charAt(0))
&& isBalanced(input.substring(1), stack.substring(1))
: isBalanced(input.substring(1), stack);
}
Test harness:
String[] tests = {
"()[]<>{}",
"(<",
"]}",
"()<",
"(][)",
"{(X)[XY]}",
};
for (String s : tests) {
System.out.println(s + " = " + isBalanced(s, ""));
}
Output:
()[]<>{} = true
(< = false
]} = false
()< = false
(][) = false
{(X)[XY]} = true
The idea is to keep a list of the opened brackets, and if you find a closing brackt, check if it closes the last opened:
If those brackets match, then remove the last opened from the list of openedBrackets and continue to check recursively on the rest of the string
Else you have found a brackets that close a nerver opened once, so it is not balanced.
When the string is finally empty, if the list of brackes is empty too (so all the brackes has been closed) return true, else false
ALGORITHM (in Java):
public static boolean isBalanced(final String str1, final LinkedList<Character> openedBrackets, final Map<Character, Character> closeToOpen) {
if ((str1 == null) || str1.isEmpty()) {
return openedBrackets.isEmpty();
} else if (closeToOpen.containsValue(str1.charAt(0))) {
openedBrackets.add(str1.charAt(0));
return isBalanced(str1.substring(1), openedBrackets, closeToOpen);
} else if (closeToOpen.containsKey(str1.charAt(0))) {
if (openedBrackets.getLast() == closeToOpen.get(str1.charAt(0))) {
openedBrackets.removeLast();
return isBalanced(str1.substring(1), openedBrackets, closeToOpen);
} else {
return false;
}
} else {
return isBalanced(str1.substring(1), openedBrackets, closeToOpen);
}
}
TEST:
public static void main(final String[] args) {
final Map<Character, Character> closeToOpen = new HashMap<Character, Character>();
closeToOpen.put('}', '{');
closeToOpen.put(']', '[');
closeToOpen.put(')', '(');
closeToOpen.put('>', '<');
final String[] testSet = new String[] { "abcdefksdhgs", "[{aaa<bb>dd}]<232>", "[ff{<gg}]<ttt>", "{<}>" };
for (final String test : testSet) {
System.out.println(test + " -> " + isBalanced(test, new LinkedList<Character>(), closeToOpen));
}
}
OUTPUT:
abcdefksdhgs -> true
[{aaa<bb>dd}]<232> -> true
[ff{<gg}]<ttt> -> false
{<}> -> false
Note that i have imported the following classes:
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Map;
public static boolean isBalanced(String str) {
if (str.length() == 0) {
return true;
}
if (str.contains("()")) {
return isBalanced(str.replaceFirst("\\(\\)", ""));
}
if (str.contains("[]")) {
return isBalanced(str.replaceFirst("\\[\\]", ""));
}
if (str.contains("{}")) {
return isBalanced(str.replaceFirst("\\{\\}", ""));
} else {
return false;
}
}
Balanced Parenthesis (JS)
The more intuitive solution is to use stack like so:
function isBalanced(str) {
const parentesis = {
'(': ')',
'[': ']',
'{': '}',
};
const closing = Object.values(parentesis);
const stack = [];
for (let char of str) {
if (parentesis[char]) {
stack.push(parentesis[char]);
} else if (closing.includes(char) && char !== stack.pop()) {
return false;
}
}
return !stack.length;
}
console.log(isBalanced('{[()]}')); // true
console.log(isBalanced('{[(]]}')); // false
console.log(isBalanced('([()]')); // false
And using recursive function (without using stack), might look something like so:
function isBalanced(str) {
const parenthesis = {
'(': ')',
'[': ']',
'{': '}',
};
if (!str.length) {
return true;
}
for (let i = 0; i < str.length; i++) {
const char = str[i];
if (parenthesis[char]) {
for (let j = str.length - 1; j >= i; j--) {
const _char = str[j];
if (parenthesis[_char]) {
return false;
} else if (_char === parenthesis[char]) {
return isBalanced(str.substring(i + 1, j));
}
}
} else if (Object.values(parenthesis).includes(char)) {
return false;
}
}
return true;
}
console.log(isBalanced('{[()]}')); // true
console.log(isBalanced('{[(]]}')); // false
console.log(isBalanced('([()]')); // false
* As #Adrian mention, you can also use stack in the recursive function without the need of looking backwards
It doesn't really matter from a logical point of view -- if you keep a stack of all currently un-balanced parens that you pass to each step of the recursion, you'll never need to look backwards, so it doesn't matter if you cut up the string on each recursive call, or just increment an index and only look at the current first character.
In most programming languages, which have non-mutable strings, it's probably more expensive (performance-wise) to shorten the string than it is to pass a slightly larger string on the stack. On the other hand, in a language like C, you could just increment a pointer within the char array. I guess it's pretty language-dependent which of these two approaches is more 'efficient'. They're both equivalent from a conceptual point of view.
In the Scala programming language, I would do it like this:
def balance(chars: List[Char]): Boolean = {
def process(chars: List[Char], myStack: Stack[Char]): Boolean =
if (chars.isEmpty) myStack.isEmpty
else {
chars.head match {
case '(' => process(chars.tail, myStack.push(chars.head))
case ')' => if (myStack.contains('(')) process(chars.tail, myStack.pop)
else false
case '[' => process(chars.tail, myStack.push(chars.head))
case ']' => {
if (myStack.contains('[')) process(chars.tail, myStack.pop) else false
}
case _ => process(chars.tail, myStack)
}
}
val balancingAuxStack = new Stack[Char]
process(chars, balancingAuxStack)
}
Please edit to make it perfect.
I was only suggesting a conversion in Scala.
I would say this depends on your design. You could either use two counters or stack with two different symbols or you can handle it using recursion, the difference is in design approach.
func evalExpression(inStringArray:[String])-> Bool{
var status = false
var inStringArray = inStringArray
if inStringArray.count == 0 {
return true
}
// determine the complimentary bracket.
var complimentaryChar = ""
if (inStringArray.first == "(" || inStringArray.first == "[" || inStringArray.first == "{"){
switch inStringArray.first! {
case "(":
complimentaryChar = ")"
break
case "[":
complimentaryChar = "]"
break
case "{":
complimentaryChar = "}"
break
default:
break
}
}else{
return false
}
// find the complimentary character index in the input array.
var index = 0
var subArray = [String]()
for i in 0..<inStringArray.count{
if inStringArray[i] == complimentaryChar {
index = i
}
}
// if no complimetary bracket is found,so return false.
if index == 0{
return false
}
// create a new sub array for evaluating the brackets.
for i in 0...index{
subArray.append(inStringArray[i])
}
subArray.removeFirst()
subArray.removeLast()
if evalExpression(inStringArray: subArray){
// if part of the expression evaluates to true continue with the rest.
for _ in 0...index{
inStringArray.removeFirst()
}
status = evalExpression(inStringArray: inStringArray)
}
return status
}
PHP Solution to check balanced parentheses
<?php
/**
* #param string $inputString
*/
function isBalanced($inputString)
{
if (0 == strlen($inputString)) {
echo 'String length should be greater than 0';
exit;
}
$stack = array();
for ($i = 0; $i < strlen($inputString); $i++) {
$char = $inputString[$i];
if ($char === '(' || $char === '{' || $char === '[') {
array_push($stack, $char);
}
if ($char === ')' || $char === '}' || $char === ']') {
$matchablePairBraces = array_pop($stack);
$isMatchingPair = isMatchingPair($char, $matchablePairBraces);
if (!$isMatchingPair) {
echo "$inputString is NOT Balanced." . PHP_EOL;
exit;
}
}
}
echo "$inputString is Balanced." . PHP_EOL;
}
/**
* #param string $char1
* #param string $char2
* #return bool
*/
function isMatchingPair($char1, $char2)
{
if ($char1 === ')' && $char2 === '(') {
return true;
}
if ($char1 === '}' && $char2 === '{') {
return true;
}
if ($char1 === ']' && $char2 === '[') {
return true;
}
return false;
}
$inputString = '{ Swatantra (() {} ()) Kumar }';
isBalanced($inputString);
?>
It should be a simple use of stack ..
private string tokens = "{([<})]>";
Stack<char> stack = new Stack<char>();
public bool IsExpressionVaild(string exp)
{
int mid = (tokens.Length / 2) ;
for (int i = 0; i < exp.Length; i++)
{
int index = tokens.IndexOf(exp[i]);
if (-1 == index) { continue; }
if(index<mid ) stack .Push(exp[i]);
else
{
if (stack.Pop() != tokens[index - mid]) { return false; }
}
}
return true;
}
#indiv's answer is nice and enough to solve the parentheses grammar problems. If you want to use stack or do not want to use recursive method you can look at the python script on github. It is simple and fast.
BRACKET_ROUND_OPEN = '('
BRACKET_ROUND__CLOSE = ')'
BRACKET_CURLY_OPEN = '{'
BRACKET_CURLY_CLOSE = '}'
BRACKET_SQUARE_OPEN = '['
BRACKET_SQUARE_CLOSE = ']'
TUPLE_OPEN_CLOSE = [(BRACKET_ROUND_OPEN,BRACKET_ROUND__CLOSE),
(BRACKET_CURLY_OPEN,BRACKET_CURLY_CLOSE),
(BRACKET_SQUARE_OPEN,BRACKET_SQUARE_CLOSE)]
BRACKET_LIST = [BRACKET_ROUND_OPEN,BRACKET_ROUND__CLOSE,BRACKET_CURLY_OPEN,BRACKET_CURLY_CLOSE,BRACKET_SQUARE_OPEN,BRACKET_SQUARE_CLOSE]
def balanced_parentheses(expression):
stack = list()
left = expression[0]
for exp in expression:
if exp not in BRACKET_LIST:
continue
skip = False
for bracket_couple in TUPLE_OPEN_CLOSE:
if exp != bracket_couple[0] and exp != bracket_couple[1]:
continue
if left == bracket_couple[0] and exp == bracket_couple[1]:
if len(stack) == 0:
return False
stack.pop()
skip = True
left = ''
if len(stack) > 0:
left = stack[len(stack) - 1]
if not skip:
left = exp
stack.append(exp)
return len(stack) == 0
if __name__ == '__main__':
print(balanced_parentheses('(()())({})[]'))#True
print(balanced_parentheses('((balanced)(parentheses))({})[]'))#True
print(balanced_parentheses('(()())())'))#False