Multiple Parenthesis Matching Using Stack with C Code

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[@nextnotification9857]

I have written code myself after understanding the logic but it took me 1 hr to write it

[@aryanshrivastava2344]

To understand this problem : Just remember As we traverse the string ( expression ) from left to right for ( int i = 0; i < len; i++) { 1. if (we encounter a opening parentheses push it on to the stack) 2. Else if ( we encounter a closing parentheses ) if ( stack is empty || top element don't pair with closing parentheses ) return false; else{ pop() } } After the loop just check at last if ( stack is empty ) // means all opening parentheses found their closing counterparts return true; else // some opening parentheses not found their closing counterparts so they are not popped return false; For expression to be balanced every closure should be for the last unclosed [ ( ) ] ( -> last unclosed ) -> closure we encounter first // it will be for the last unclosed which is ) ( -> will be popped Updated list : [ ] [ -> last unclosed ] -> closure we encounter next Hope you understand !!

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[@dariofairhall2695]

We can use stacktop() If we check open and close parenthesis match befor poping element from stack. At that case, we need top element from stack which is need to be checked and then If matched, then pop and continue checking Else expression not balanced

[@parallax8916]

exactly!! i was also thinking about the same logic. we do not need to pop directly, instead we check first , match it with stackTop, if it matches, then only we pop, else we return 0.

[@Shorts-iv2zk]

int mulParenMatch(char* exp){ struct Stack*s; s->top=-1; s->size=50; s->arr=(char*)malloc(s->size*sizeof(char)); for(int i=0;exp[i]!='\0';i++){ if(exp[i]=='(' || exp[i]=='{' || exp[i]=='['){ push(s,exp[i]); } else if(exp[i]==')' && s->arr[s->top]=='('){ if(isEmpty(s)){ return 0; } else{ pop(s); } } else if(exp[i]=='}' && s->arr[s->top]=='{'){ if(isEmpty(s)){ return 0; } else{ pop(s); } } else if(exp[i]==']' && s->arr[s->top]=='['){ if(isEmpty(s)){ return 0; } else{ pop(s); } } } return isEmpty(s); } then what's wrong in this code??

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Harry bhai aap pls competitive programming ke bhee questions karwa diya karo from Codechef

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Sir, could you please make a video on character input and output, file copying , and character counting in c?

[@anjalibajaj8824]

Badiya Harry bhai very helpful

[@IG_Hunter]

To great explanation Love u bro You are great

[@shatviksmit3388]

0:41 Sir I suppose that in our previous code while traversing the list our condition should be if it is '(' , '[' '{' then push in the stack and if it is ')', ']' or '}' then pop from the stack Like if correct

[@Lila12386]

Nice Explanation Harry Bhai👍👍👍👍

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Harry bhai app to ds ke sath advance c bhi sikha rhe ho 👍🏻

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you are best teacher

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Excellent explanation ❤️

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Excellent explanation 👌

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superb bro

[@samuelfrank9947]

Thanks a lot bhaiya, Enjoyed this video

[@joyrajlongjam7121]

If code is not running then allocate memory dynamically for the stack in the parenthesisMatch function i.e struct stack*sp = (struct stack *)malloc(sizeof(struct stack));

[@kushal6065]

but it works on mine without allocating memory, how?

[@rohanprasad7601]

thanks..

[@AbhishekGupta-qh8xh]

thanks bro..but can you explain why was it was not working without dynamic memory allocation

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Yes

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Setup tour plzzz!! 👇

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Excellent

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thanks a lot sir this playlist is very helpful

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Awesome video 👍👍👍👍

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Thankyou 🙏

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Bhai JavaScript Framework ke upar Ek combine video bnao jismein aap angular, react aur vue ka tips and tricks do ki Ham kisko choose Kare or fir uske upar aap course banoo, Main aapka web development ka JavaScript tutorials in Hindi ka course complete kar liya hai aur Mujhe front end web developer banna hai to front end web developer ke liye JavaScript and CSS frameworks bhi Aane chahie aur mujhe bahut jyada confusion ho rahi hai ki main JavaScript Framework kaunsa Sikho to please bhai kam karo.AUR EK VIDEO BANAO. My humble request to you. You are such a good teacher to teach programming languages that's why i am asking to you.

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Very good video 👍😀😀 keep it up 👌

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Thank you sir🤩

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Thanks harry bhi

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thankew

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Appki puri video dekh rha hu

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Harry sir did it before 1:20

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Done done

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Mast video

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Thank you sir.

[@YASHSINGH-se7qu]

sir ek doubt hai , apne kahi pr stack ko heap me memory k liye request nhi kiya .....struct stack *sp=(struct stact*)malloc(sizeof(struct stack)) ....?

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Harry bahi issue hai (how to minimize main-thread work in blogger) please batiya Harry bhaiya

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[@utkarshsingh245]

#include using namespace std; struct stack{ int size; int top; char* arr; }; void push(struct stack* p, char element){ if(p->top == p->size-1){ coutarr = new char(p->size); for(int i = 0; itop == -1){ return 0; } else{ if(p->arr[p->top] == '(' && s[i] == ')'){ pop(p); } else if(p->arr[p->top] == '[' && s[i] == ']'){ pop(p); } else if(p->arr[p->top] == '{' && s[i] == '}'){ pop(p); } else{ return 0; } } } } if(p->top == -1){ return 1; } else{ return 0; } } int main() { string s = "([7-9}-7)"; if(parathesischeck(s)){ cout

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Please upload vedios on Graph data structure harry bhai

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Please make data structures and algorithms using python with notes in 1 video

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React JS 😍😍

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*sir aap Multiple Parenthesis Matching ko as an homework bhi de sakhte thay*

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Mere DS ke karta dharta aap hi ho🙏😁

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Aaj to java wala episode aane wala tha na sir ???

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10k to go

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1m Near.

[@pupjetofficial]

bhai why don't we just increase value of top when we find opening parentheses and decrease it when we find closing one.....and using if else we can solve this.... we don't need to make an array....or stack for this

[@ansarikashif4514]

18:52

[@adarshraj3208]

We can also this problem by creating three structure pointers for three different brackets as it is shown below int parenthesis(char *equation) { struct stack *s; s = (struct stack *)malloc(sizeof(struct stack)); s->size = strlen(equation); s->arr = (char *)malloc(s->size * sizeof(char)); s->top = -1; struct stack *t; t = (struct stack *)malloc(sizeof(struct stack)); t->size = strlen(equation); t->arr = (char *)malloc(s->size * sizeof(char)); t->top = -1; struct stack *u; u = (struct stack *)malloc(sizeof(struct stack)); u->size = strlen(equation); u->arr = (char *)malloc(s->size * sizeof(char)); u->top = -1; for (int i = 0; i < s->size; i++) { if (equation[i] == '(') { push(s, '('); } else if (equation[i] == ')') { if (isempty(s)) { return 0; break; } else { pop(s); } } else if (equation[i] == '{') { push(t, '{'); } else if (equation[i] == '}') { if (isempty(t)) { return 0; break; } else { pop(t); } } else if (equation[i] == '[') { push(u, '['); } else if (equation[i] == ']') { if (isempty(u)) { return 0; break; } else { pop(u); } } } if (isempty(s)==isempty(t)==isempty(u)==1) { return 1; } else { return 0; } }

[@Pranauv]

bro i dont think this will work for wrong brackets expression like -- [(34*21])

[@himanshudash-c4c]

no problem

[@rayangamedev1912]

bhai c# ka course kab laho ge

[@elon_chacha]

Harry sir, ye data structures ka course kitne moth ka h? Plz ek bar bta dijiye bdi kripa hogi....🙏

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👍

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👍👍👍

[@SachinKumar-xt8ny]

Bss ek dil chahiye Bhai.... 7th comment

[@abhhishekabs]

Why do we need a different match function ? Can't we match the popped character and the exp[I] directly ? Will it give errors ?

[@manjitsharma889]

First of all, the popped character will always be one of these three ' ( ' or ' { ' or ' [ ' , now according to you, popped_char == exp[i] will be ' ( ' == ' ) ' which is false. ' ( ' is a different character than ' ) '. Matching the popped character and exp[i] will be matching ' ( ' with ' ) ' or ' ( ' with ' ] ' both of these are always false. suppose you've this expression " [ ( a + b ] ) " then first we'll push the character '[', then '(' will be pushed, when ']' is encountered, the top element in the stack is popped, that is '(' is popped, now we use the match function to match the following set '(' with ' ) ' and ' [ ' with ']'. here above, the popped element is ' ( ', but the current character is ' ] ', hence the match function will return false, since the following pair - ' ( ' , ' ] ' didn't match.

[@amalasebastian9968]

@@manjitsharma889 thank you so much for the explanation

[@yashwanthpasaladi2714]

same doubt

[@MEAmitKourav]

Sir, Whenever you add new part to program Please tell b/w which to part you pasted it.

[@lakhveer0362]

😍😘🙏🙏🙏✌✌✌legend 😍

[@noi375]

Sir I am using online compiler and your code was giving same error as in the previous video i.e. segmentation fault please reply my question 🥺🥺🥺🥺

[@RohitSharma-rk6xd]

run those code in code blocks!! bro! i am also try to run those code in online compiler and it throws TLE error!

[@NitinKumar-xm7hu]

You've to use "struct Parenthesis * p= (struct Parenthesis *)malloc(sizeof(struct Parenthesis))" instead of "struct Parenthesis * p" only .

[@ShivamKumar-yp4vv]

Sir big fan heart plz

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Please bahi apple kay lia appstore may app publish karnay kay liay video banaw or hum 1 app ko kitnay jaga publish karsaktay hai

[@arindamdutta7369]

Sir ji DS ALGO ka kitna % Padhaye AAP??..AUR kitna baaki hai???....I am excited to know

[@shazwatching5697]

7:40

[@jinaykothari4511]

Harry bhai please reply fast .... I am not able to use jupyter notebook

[@anmol_pandey14]

We want dsa videos fast!

[@sb_infi]

8:30

[@GovtExamsPrep1]

Alternative code: #include #include #include #define max 100 char stack[max],expr[max]; int top =-1; int match(char); void check(); int isEmpty(); void push(char); char pop(); void push(char x){ if(top == max-1){ printf("Stack_overflow "); return; } else{ top = top + 1; stack[top] = x; } } char pop(){ if(top == -1){ printf("Stack_underflow "); printf(" Unbalanced expression !! "); exit(0); } else{ char c = stack[top]; top = top - 1; return c; } } int isEmpty(){ if(top == -1) return 1; else return 0; } int match(char x){ switch(x){ case '(': case ')': return 3; case '{': case '}': return 2; case '[': case ']': return 1; } } void check(){ int i = 0; char element,x; while(i