Near the Parenthesis is Tim Arndt, a San Francisco based electronic musicia… Read Full Bio ↴Near the Parenthesis is Tim Arndt, a San Francisco based electronic musician. While playing piano and guitar in several bands through the years, Tim has continually been grounded in electronic and experimental music. His current approach focuses on full compositions, gentle and restrained, yet deeply layered and evolving. Passages never stray too far major or minor, centering themselves in between, and grow unexpectedly from simple rhythms and chords to swirling combinations of sound intensified with dissonance before resolving itself and quietly exiting

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Sarah Jarosz Get me back to where I started I'm folded over in…

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01modernisme7:22

02cerda's plan5:54

03a brief walk in the sea7:04

04guell5:50

05smdm5:11

06the second nave7:29

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Most interesting comments from YouTube:

Aakash Mudigonda

4y↑4

if anyone requires the "C" code for the above program ::

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#define MAX 100

struct Node{
int data;
struct Node* next;
};
typedef struct Node node;

node* top = NULL;

// PUSH()
void Push(char c)
{
node* temp = (node*)malloc(sizeof(node));
temp->data = c;
temp->next = top;
top = temp;
}

// POP()
void Pop()
{
node* temp;
if (top == NULL)
{
return;
}
temp = top;
top = top->next;
free(temp);
}

int isMatching(char c){
if ( top->data == '(' && c == ')' ){
return 1;
}
else if ( top->data == '[' && c == ']' ){
return 1;
}
else if ( top->data == '{' && c == '}' ){
return 1;
}

else
return 0;
}

void isBalanced(char exp[], unsigned long len){
for (int i = 0; i<len; i++) {
if ( (exp[i] == '(') || (exp[i] == '{') || (exp[i] == '[') ) {
Push(exp[i]);
}
else if ( (exp[i] == ')') || (exp[i] == ']') || (exp[i] == '}') ){
if (top == NULL){
return ;
}
if (isMatching(exp[i])) {
Pop();
}
else
return ;
}
}
return ;
}

int main(void){
char exp[MAX];
unsigned long len;
printf("Enter expression: ");
scanf("%s",exp);
len = strlen(exp);
isBalanced(exp, len);
if (top == NULL){
printf("Balanced\n");
}
else
printf("Not Balanced\n");

return 0;
}

Manfredo Weber

for anyone whos interested
i did it with an implicit stack/recursion but only for 1 kind of parentheses so far and overall not done yet, but an idea how it could look like

#include<iostream>

using std::cout;

bool equal_parentheses_r(char * str, char c=' '){
if(*str=='\0' && c==' ')return 1;
if(*str=='\0' && c!=' ')return 0;
if(*str=='('){
return equal_parentheses_r(str+1, ')');
}
else if(c==')' && *str==')'){
return equal_parentheses_r(str+1, ' ' );
}else return equal_parentheses_r(str+1, c );
}

int main(){
char str[51]{"asd(a)sd"};
cout <<"equal_parentheses_R? " << equal_parentheses_r(str) <<'\n';
return 0;
}

Raghul G

Much smaller code in C:
// Check for paranthesis using stack and the stack here is implemented using Linked lists

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int flag = 0;

struct Node{
int data;
struct Node *link;
};
struct Node *head;

void push(int x)
{
struct Node *nod=(struct Node*)malloc(sizeof(struct Node));
nod->data=x;
nod->link=head;
head=nod;
}
char pop()
{
char n;
struct Node *temp=head;
if(head==NULL) return;
n=temp->data;
head=head->link;
free(temp);
return n;
}
void checkforparanthesis(char *A,int x)
{
char c;
for(int i=0;i<x;i++)
{
if(A[i]=='{' || A[i]=='[' || A[i]=='(')
push(A[i]);
if(A[i]=='}' || A[i]== ']' || A[i]==')')
{
c=pop();
if((c=='{' && A[i]=='}')||(c=='(' && A[i]==')')|| (c=='{' && A[i]=='}'))
flag=1;
}
}
if(flag==1)
printf("Balanced");
else
printf("Not balanced");
}

int main()
{
head=NULL;
char A[100000];
gets(A);
checkforparanthesis(A,strlen(A));
}

Comment for any queries...

TheBadBunny

// C code Implementation:
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>

typedef struct stack{
char data;
struct stack *next;
} stack;

stack *push(stack *top, char chart){

stack *new = malloc(sizeof *new);
new->data = chart;
new->next = top;
return new;
}

stack *pop(stack *top){

if(!top) return NULL;

stack *temp = top;
top = top->next;
free(temp);
return top;
}

char Stack_top(stack *top){

if(!top) return '\0';
return top->data;

}

bool isEmpty(stack *top){

return !top ? true : false;
}

bool isPair(char op, char ed){

if(op == '(' && ed == ')') return true;
if(op == '[' && ed == ']') return true;
if(op == '{' && ed == '}') return true;

return false;
}

bool balanced_parantheses(char str[]){

stack *tmp = NULL;
for(size_t i = 0; i < strlen(str); i++)
{
if(str[i] == '(' || str[i] == '{' || str[i] == '[')
tmp = push(tmp, str[i]);
else if(str[i] == ')' || str[i] == '}' || str[i] == ']')
{
if(!isEmpty(tmp) && isPair(Stack_top(tmp), str[i]))
tmp = pop(tmp);
else return false;
}
}
return isEmpty(tmp) ? true : false;
}

int main(void){

char str[] = "{()(((((";
balanced_parantheses(str) ? printf("Balanced") : printf("NOT Balanced");

}

omkar kulkarni

JAVA CODE

public class CharStack {

public Node head;

public CharStack(char item) {
Node newNode = new Node(item);
head=newNode;
}

public CharStack() {

}

public void push(char item) {
Node newNode = new Node(item);
newNode.setNext(head);
head=newNode;

}

public char pop() {
if(!isEmpty()) {
char value =head.getValue();
head= head.getNext();
return value;
}
else {
System.out.println("Stack is empty!!!");
return ' ';
}

}

public boolean isEmpty() {

if(head==null) {
return true;
}
else {
return false;
}

}

public char top() {
if(!isEmpty()) {
return head.getValue();
}
else
return ' ';
}

public void show() {
Node pointer = head;
while(pointer!=null) {
System.out.print(pointer.getValue()+"|");
pointer=pointer.getNext();
}
System.out.println();
}

}

Main class:

public class App2 {

public static void main(String[] args) {
boolean flag;
String str1 = "{(a+b)*(c+d)}";
flag=matchPair(str1);
System.out.println("Balanced Paranthesis "+ flag);

}
public static boolean matchPair(String str1) {
CharStack stack1 = new CharStack();
for(int i=0;i<str1.length();i++) {
char ch = str1.charAt(i);
//Opening Brackets are pushed to the stack
if(ch=='(' || ch=='{' || ch== '[') {
stack1.push(ch);
}
/**For closing brackets
* a) if the stack is empty we have extra closing bracket
* b) check if it is matching closing bracket for the stack top**/
else if(ch==')' || ch=='}' || ch== ']') {

if(stack1.isEmpty()||!checkPair(stack1.top(),ch)) {
return false;
}
else {
stack1.pop();
}

}

}
if(stack1.isEmpty())
return true;
else
return false;

}

public static boolean checkPair(char ch1, char ch2) {

if(ch1=='(' && ch2 ==')')
return true;
else if(ch1=='[' && ch2 ==']')
return true;
else if(ch1=='{' && ch2 =='}')
return true;
else return false;
}

}

Gautham Pai

8y↑1

For Objective-C Folks
Paste below in main.m on Command Line Mac Project Template of Xcode.

#import <Foundation/Foundation.h>

@interface NSStack : NSObject

@property (nonatomic, strong)NSMutableArray *elements;

@property (nonatomic, readonly)BOOL isEmpty;

- (void)push:(id)number;
- (void)pop;
- (id)top;

@end

@implementation NSStack

- (instancetype)init
{
self = [super init];
if (self) {
_elements = [[NSMutableArray alloc] init];
}
return self;
}

- (void)push:(id)number{
[_elements addObject:number];
}

- (void)pop{
return [self.elements removeLastObject];
}

- (id)top{
return [self.elements lastObject];
}

-(BOOL)isEmpty{
return self.elements.count == 0 ? YES : NO;
}

- (void)printElements{
NSLog(@"%@",self.elements);
}

@end

int main(int argc, const char * argv[]) {

@autoreleasepool {

NSString *expression = @"[asd(asd)as12( 980)fas]";

NSMutableCharacterSet *charSetToRemove = [NSMutableCharacterSet whitespaceCharacterSet];
NSCharacterSet *alphaNumericCharSet = [NSCharacterSet alphanumericCharacterSet];
[charSetToRemove formUnionWithCharacterSet:alphaNumericCharSet];

expression = [[expression componentsSeparatedByCharactersInSet:[NSCharacterSet whitespaceCharacterSet]] componentsJoinedByString:@""];

NSMutableArray *letterArray = [NSMutableArray array];
for (int i = 0; i < [expression length]; i++) {
[letterArray addObject:[NSString stringWithFormat:@"%C", [expression characterAtIndex:i]]];
}

NSArray *open = @[@"(",@"[",@"{"];
NSArray *close = @[@")",@"]",@"}"];

NSStack *stack = [[NSStack alloc] init];

for (int i=0; i<letterArray.count; i++) {

NSString *letter = letterArray[i];
if ([open containsObject:letter]) {
[stack push:letter];
}else if([close containsObject:letter]){
if (stack.isEmpty) {

[stack push:letter];
break;
}else{
if ([open containsObject:[stack top]]) {
[stack pop];
}
}
}
}

if (stack.isEmpty) {
NSLog(@"Expression is Balanced");
}else{
NSLog(@"Expression is Not Balanced");
}
}
return 0;
}

bolu

Python implementation of solution

#Node Class for Stack implementation
class Node:
def __init__(self,data=None):
self.data = data
self.next = None

#Stack implementation with Linked Lists
class stack():
def __init__(self):
self.headNode = None
def push(self,val):
NN = Node(val)
NN.next = self.headNode
self.headNode = NN
def pop(self):
if(not self.headNode):
return None
CNode = self.headNode
self.headNode = self.headNode.next
return CNode
def not_emp(self):
if(self.headNode!=None):
return True
return False
def top(self):
return self.headNode

#function
def balanced(string):
#create stack instance
Lbrackets = stack()
#create dictionaries of open- and close- brackets
OpenB = {"(":1,"{":2,"[":3}
CloseB = {")":1,"}":2,"]":3}
#loop through each character in string
for i in string:
#if open-character, push value from dictionary with character as key
if i in OpenB:
Lbrackets.push(OpenB[i])
#if close-character...
if i in CloseB:
if(not Lbrackets.not_emp()):
return "Unbalanced"
#if value in dictionary not equal to top item in stack
if (not (CloseB[i] == Lbrackets.top().data)):
return "Unbalanced"
Lbrackets.pop()
if(Lbrackets.not_emp()):
return "Unbalanced"
return "Balanced"

Learn about linked lists: https://www.youtube.com/watch?v=NobHlGUjV3g&list=PL2_aWCzGMAwI3W_JlcBbtYTwiQSsOTa6P&index=3
Learn about Stacks: https://www.youtube.com/watch?v=F1F2imiOJfk&list=PL2_aWCzGMAwI3W_JlcBbtYTwiQSsOTa6P&index=14

Nikhil Goyal

5y↑1

Simple Java Solution

import java.util.Stack;

class Test {

private static String s = "[{}]";
// private static String s = "[]{}";
// private static String s = "[()]{}{[()()]()}";

public static void main(String[] args) {
Stack<Character> stack = new Stack<>();
stack.push(s.charAt(0));
for (int i = 1; i < s.length(); i++) {
char incomingCharacter = s.charAt(i);
// if stack becomes empty in between insert the character example - []{}
if (stack.isEmpty()) {
stack.push(incomingCharacter);
}
// incoming character is a closing pair of the top character of stack
else if (isAPairOfStackTopCharacter(stack.peek(), incomingCharacter)) {
stack.pop();
}
// push the incoming character if it is not a closing pair of the top character of stack
else {
stack.push(incomingCharacter);
}
}
// if stack is empty after passing through the string then brackets are balanced
System.out.println(stack.isEmpty());
}

private static boolean isAPairOfStackTopCharacter(char topCharacterOfStack,
char incomingCharacter) {
return (topCharacterOfStack == '[' && incomingCharacter == ']')
|| (topCharacterOfStack == '{' && incomingCharacter == '}')
|| (topCharacterOfStack == '(' && incomingCharacter == ')');
}
}

All comments from YouTube:

Myron Baggett

5y↑134

I clearly understand this subject now. I saw this for free and am in debt from school for something they couldn't teach. The irony. Thanks for the lovely video!!

Higlights

Pòseìdòn

4y↑44

This guy is one enough to make learn coding the whole student community

Tejas Rawat

8y↑11

your classes are very interactive as programming perspective. I could say this is the best data structure tutorial till now we have in web.

David Sumich

8y↑3

Your classes are all fantastic. Thanks so much!!!

William Ryan

8y↑50

Thank you! Literally went from "no clue" to fair idea about stacks from watching this! Great Vid!

Suvan Singh

5y↑3

The best explanation of this concept , Last Un-Closed , First Un-Closed , makes it clear why a stack must be used. This detail has been left out it most explanations. Great work. @9:27

RebelutionaryGaming

7y↑12

I've learned more abuout data structures from your videos, than I have from school. Thank you.

aboci dejofa

10y↑25

No one can explain any clearer that this!!! thank you

Raed Khader

8y↑2

I cant thank you enough, so simple and clear ;)

Lam Le

You make it sooo easy to understand. Thanks a lot!

Prashant Bajpai

11y↑2

good work sir thnk u so much fr make a awesome creation of videos u r the real teacher thnks a lot sir plz upload more videos related to Tree,Graphs,Constructor,exceptional handling,file handling etc

Selenay Yalçın

the best data structures video on the internet i've ever seen. thanks for the clarity and understandability even 9 years later. hope you're okey

Dinesh Kumar

Words are short to express my gratitude. I am able to code it in C# with your advised alsogrithm, it works in all conditions. Thank you once again.

KRITIKA SHUKLA

So smoothly explained. Thank you!

Jang.b

7y↑2

Really awesome and helpful explaination. Thanks.

Davide Chiappetta

2y↑1

using a stack seems a bit excessive to me, since for each character to match it must make a call to the C ++ library and this consumes cpu and memory, you can simply use a byte array of fixed length, as large as the number of characters or symbols to compare, in which in each element you can save a number corresponding to the character to be balanced, e.g. 1 corresponds to square brackets and 2 corresponds to round brackets and a global pointer which moves to the right (g_ptr ++) to insert char x,y,z and to move to the left (g_ptr--) to compare and remove char z,y,x , (if the characters do not match or if the pointer goes below zero then there is an unbalance) a method that can also be used to keep track of the states of C directives (for example excluding everything that descends from #else (including other #if #else #endif and resuming the block at the exit of the first #endif) where it is not worth using a stack or a parser e.g. LALR

Vedang Kavathiya

4y↑1

Sir your explanation is too good thanks for explaining me this topic of stack... 😃

Aymar Deo OLATOUNDE

Hi Sir,
I think after set variable n with expression length, we can check if n is even. If it’s not, n is odd and return false (Expression will contain only brackets).

Rinja Kundu

Nicely explained!
Really helpful

Aakash Mudigonda

4y↑4

Dharshanaa Piskala

TQ so much !!!!

maycodes

Thank you , what a great contribution.

indavarapu aneesh

Excellent marvellous extraordinary mind blowing fantastic awesome.no words to describe your work.thanks guruji

Manfredo Weber

brando

excellent video! i understand this concept much better now.

Andrew Ma

10y↑16

Wow, exactly the same as what I did in an Amazon phone interview. I forgot to check s.isEmpty while peeking the stack until it threw exception in one testing, but after adding it back, it is exactly the same as what it shows here. Cool!

Muhammad Bilal Malik

@nikunj chapagain yes he was refrenced by me !!

nikunj chapagain

7y↑19

did you get the job tho?

PSVgaming - CLASH ROYALE & MORE

Thanks for the help! really help me alot in my studies! !

Geraldo Macias

8y↑10

Thank you for this explanation. I have to do this exact problem for a job interview coming in the fall.

Geraldo Macias

3y↑5

@Rohan Bangash Nailed this question on every interview. Funny enough I work for a company that did not ask this question. It's a great algorithm to help think about the benefits of a stack data structure.

Rohan Bangash

3y↑3

4 years later, how did it go lol

Raghul G

Sant Gupta

Nice videos.
To the point and easy to understand.

Where can I get code if I want ?

Mayur Satbhai

Sant Gupta open "see more" below the video and click on "see source code here.

Xin He

10y

Very clear instruction!

Daniel Senik

thanks, this video helped a lot to understand the general idea of the algorithm

Your Average Software Developer

hats off bro... thank you so much for this amazing series

싱 니키타 Nikita 🎶

3y↑1

Crystal clear, thanks😊😊

Hyperflow

3y↑1

thank you for this input method, made it way easier for me to understand

Md Solaiman Chowdhury

It was really helpful !

Foodie Food sagar indore bangalore

I am really grateful to the person for this awesome compilation of Data Structures Thanks You!

Chandra Shekhar

Amazing Video..please upload some more videos on Interview Questions and MultiThreading and Design Pattern

kriti kaushal

Great explanation !

@rpit Patel

Amazing explanation Thank you sir

TheBadBunny

Bhawna Kukreja

great explanation sir...thanks a lot

hana

4y↑1

thank u very much, very well explained! helped me a lot.

memo Fahood

10y

Amazing explanation :)

Rawan

10y

so clear , thanks alot

p 16

4y↑2

Thank you this really helped! c:

md. kaif

You have the best way of teaching..🔥🔥🔥

hemant yadav

We can maintain two data structure stack and queue. stack for the opening and queue for the closing parentheses.

The we can compare these two if all the parentheses are in the order.

eyescream

Nicely explained :)

Aswin Pranav

8y↑63

You're a bloody genius... Thanks!

Ashish Menon

4y↑2

if length of the string is "odd" its invalid , only we should check for the even ones.

Brahm Karan Singh

If the length is odd you should simply return false as there is no chance of matching of brackets

faizan nisar

7y↑1

amazing lectures as well as brilliant voice

Yoga and Living with Chai

Can you solve for String Anagram and word ladder?

Viraj Doshi

Why would you use stack to 'park' the parantheses ? how is the possibility of any bracket finding a match within the 'parked' list/array ? Dosen;t the pop method give you always the topmost (last) instance ?

Diksha

Thank you. It was very helpful

Pachava Manasa

thanku you so much sir
you are healping us a lot

omkar kulkarni

Tharu Geethz

Thank u so much! It was really helpful !

Maidul

would this o(n^2) time complexity? because you have for loop and inside you append items which take O(n)?

Ömer Kurttekin

Append, Pop, Top, IsEmpty operations are O(1) in Stack. Even if we pop or append all the chars to stack, we would do that n times.

Aymen Sekhri

Thanks a lot, very helpful

Shahriar badhon

Just awesome

byron perez

please, don't get me wrong i am study really hard this topics, but what are the benefits of implement your own stack even if c++ have their own??
i spend all of my day studying data structures, but the way...

tech savvy

There is no difference between implementing your own stack and using the inbuilt c++ stack. But just to make us understand concisely
he chose the former.

byron perez

you are a great comunicator, it is easy for me to understand you, even when my english is very bad

Chayan Kanungo

Thank you so much Sir!

Mohit

7y↑3

you r the god of data structures!!! thanks a lot! BTW where are u from in India??

Amrita Basu

Sid I see.

Karmyog Sewa Sansthan (KYSS)

Amrita Basu both.. But the voice is Animesh's

Amrita Basu

6y↑1

Sid So Harsha created these videos ? I was under the impression that it was Animesh.

Karmyog Sewa Sansthan (KYSS)

Amrita Basu i was talking abt Harsha

Amrita Basu

6y↑1

Sid He died in 2015? You mean him or the person he founded My Code School with, Harsha Suryanarayana? He's the one who unfortunately passed away in 2014 or so right? This person is Animesh Nayan and he is pretty much alive?

2 More Replies...

Samuel K

excellent explanation

Rownita Khanam

By mercy of Allah,you 're saving careers of a lot of people.

Eklavya Prasad

3y↑10

Remembering Lord Harsha ❤️

k mahesh

the given code in the git hub will fail for the test case when we include the space in the user input because the taking of string input will not execute after a blank space I'm still having doubt on the concepts of strings and characters of user input oriented.. i will also do silly mistakes like this code which was uploaded on git hub

Pramod Bhat

The best explanation.

Gautham Pai

8y↑1

csanchezcuba

can someone tell me how to aproach this problem using recursion instead of stacks? the signature of my method should be
public boolean isBalanced(String in)

Sagar Mittal

Nice explanation👍👍

SAURAV

great job sir....

Cecilia Quintana

pseudo code rocks! Thanks!

MB007

It was Helpful :)

Shashwat Singh

Great video sir

SUB FOR SUB COMMUNITY CENTER

Wow u have all the thing i searching for

aarb17

10y

Great, thank you

IMAAD NAEEM ANSARI

thank you sir!

Ibrahim Shaikh

Many many thanks,keep it up

Krishnachytanya Ayyagari

9y↑1

superb!!

Braden Mills

what would the Big0 of this be?

SlumDog

u r legend sir

Paras Deshbhratar

3y↑3

I’ve been asked this question in coding interview.

Trisha Chander

what's the algorithm to check that the top doesn't pair with i?

Ralph Mente

Thank youuu!!!

Dr Constantinos

Thanks so much!

bolu

Xahid Sakib

HELPFUL

Khaltouma Ch12

What does POP () function do

keerthi chandu

9y↑22

I would add this piece of code,
if(expression.length() % 2 != 0){
return false;
}
in the beginning, anyway good video :)

Rohan Mehto

6y↑4

when in use, we dont check for expressions which only contain paranthesis (thats stupid) , so the expression with odd length can have balanced pararenthesis.

Kanishk Jain

nice logic
it will reduce the time and space complexity for odd cases to a great extend

keerthi chandu

8y↑6

+Prem Kumar Hi Prem,

Thanks for the answer.
That piece of code I wanted to add would only fit if the expression contains no other characters except parentheses (could be any closing ones "'<>'" too). Since the video was only mentioning a string with parentheses, I considered that would fit in there. Yes, if the string can include all sorts of characters, as you said that won't work. Thanks for taking your time anyway :)

SeriousFox

How do I deal with the brackets inside the quote, or comments?

Nands

7y↑7

What are some more popular programming interview questions like this one?

Nands

7y↑3

Christian Jusino that too is outdated but

Christian Jusino

7y↑5

In the about section of his channel he has a link to an article about programing interviews.

Manas Srivastava

This reminds me of Push Down Automata(PDA)

Noor Eldali

my exam is in two hours you are lifesaver

Rajdeep khandelwal

thank you 🙌

kasyap dharanikota

best explanation

Nikhil Goyal

5y↑1

Arun dev

Thank you.

Muhammad Ali

how to check pair??

daswarudo

thanks!

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