3.7 Infix to Postfix conversion using Stack | Data structures and algorithms

In this video, I have explained the conversion of INFIX expression to POSTFIX expression using STACK with the help of examples. Keeping in mind the priority of operators(precedence) as well as associativity of operators. Associativity comes into picture when two operators of the same precedence arrive like addition and subtraction, multiplication and division etc.
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[@AmanatAli-gm2xq]

honestly, you explain this in 8 minuts more clearly than my prof who took two lectures thank you

[@gueshteklu8624]

Mrs Jenny( My On Line Teacher) Has Higher Precedece than My Class Teacher Of Data Structure No Need Associativity Rule🙏🙏🙏I Always Fellow You.... Thanks More!!!!!!!!!

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EXCELLENT teaching mam! Can you please upload videos about 'Monotonic stack' and 'Monotonic queues'?

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[@priyanshgoswami2201]

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[@thecloudterminal]

Thank you for great explanation and putting so much effort mam 🙏

[@imrohityadav23]

Thank you ma'am ❤

[@priyankashaw1238]

When we were supposed to push '+' on the top of ' -' , you said associativity is left to right. When '-' was already there in stack and we were supposed to push '+' , then also you said left to right associativity. When does right to left associativity rule applies?

[@editprousama]

same qs here bruh

[@khushwantmaan791]

when you have ^ that operator at top of the stack and the incoming operator is also ^ that operator, then right to left associativity rule will be applied and put that operator in stack directly.

[@manishgaurav5129]

@@khushwantmaan791 right brother

[@OptimisticForce]

In future you will cover this topic again , This current video will be in recommendation. I will click it

[@jatinranpariya1510]

Implementation in C++ class expression { private: string s; public: expression(string t) { s = t; } bool isOperator(char ch) { return ch == '+' or ch == '-' or ch == '*' or ch == '/' or ch == '^'; } int precedence(char ch) { if (ch == '^') return 3; if (ch == '*' or ch == '/') return 2; if (ch == '+' or ch == '-') return 1; return 0; } string postfix() { string ans; stack st; for (auto i : s) { if (i == '(') st.push(i); else if (i == ')') { while (!st.empty() and st.top() != '(') { ans.push_back(st.top()); st.pop(); } st.pop(); } else if (isOperator(i)) { if (st.empty()) st.push(i); else if (precedence(i) > precedence(st.top())) st.push(i); else if (precedence(i)

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Very useful information

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Mam really appreciate your hard work !!! It is helping me a lot !! Please keep up the good work........ Baaki mam please agar time mile to algorithms ke lectures bhi add kijiyega !!

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Thanks mam

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[@someonesoumya]

thanks mam. good explanation

[@parikshitsharma3393]

I have a doubt. Ma'am, I think instead of checking the higher precedence condition two times only, we have to check until an operator of lower precedence is found. Technically what I want to say is we should use while(higherprecedence(s.top)) instead of using two ifs i.e if(higherprecedence(s.top)) -> if(higherprecedence(s.top)). If I am wrong please correct me.

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Great Respect form PAKISTAN and thanks .

[@Shivang-rj9lq]

How much has changed in the first video and now in the last wow mam

[@bachinaganesh1891]

mam make a video on this example "a+b*(c^d-e)^(f+g*h)-i"

[@sushmasa7817]

abcd^e-fgh*+*+i-

[@kailashuv3028]

Thank you very much didi😊😊😊

[@baskaranbaluchamy3667]

very good explanation mam.thanks.... can u explain follow this . (a+b)*(c*d-e)* f/g ?

[@baskaranbaluchamy3667]

Mam. I need the post order conversion for this. (a+b) * ( c*d-e)*f/g. My answer is ab+cd*e-*f*g/ ... Is it correct ? If else pls give me the step by step procedure for this mam....

[@JennyslecturesCSIT]

yes your answer is correct..

[@baskaranbaluchamy3667]

(a*b+d) --> in stack ( *,+) in this case, how we pop out operators in stack. Right to left or we rearrange the operators based on their precedence ? abc+* Or abc*+ ? Which one is correct madam?

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[@sushantarora5566]

crystal clear🙂

[@sowqatali4377]

Madam,please bataiye, aagar ye situation aaye ,,, (+*) ,to ye dono pop ho kar postfix me jayega ,,to postfix me kya lokhungi (+*/*+) matlab ,, first+then* ya fir first *then +

[@atharvakatre3256]

first * then + coz stack me top pr * tha toh woh pehle pop hoga

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Thank You so much!

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[@kevalshah5594]

mam associativity of ^ these operator is left to right i have searched on google and its showing left to right

[@shahriarkhan7749]

Thanku so much

[@imtiyazhussain2901]

maam In this case if (/( next operator is - then what should we do.

[@ritikasharma2513]

Plz make the video on conversion of infix to prefix expression but di pls table bnakr btana ache se smjh a jayga abi thoda sa confusion h isliy table Vali plz

[@JennyslecturesCSIT]

okay dear.. will upload by Friday..

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thnkuuu dii

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Mam if ur not here, I would've failed my DS exam

[@chiragjoshi6118]

I am still confused with associativity rule it should be more detail about right to left or left to right

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a huge thanks

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mam thanks for lovely explanation .u are so cute i understood everything😍👌

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thnkss alott mam..

[@JennyslecturesCSIT]

Your gratitude is humbly appreciated

[@gurupreetsingh8347]

plz help with infix to prefix also with the help of stack............ waiting.......

[@JennyslecturesCSIT]

Okiee.

[@EBKCS_MOKSHASHARMA]

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thank you

[@kusumanandavarapu1073]

Madam...code kuda...cheppochu kada..infix to postfix ki...

[@ShriyaSawant2023]

What is the priority of power operator

[@saritakumari-oh8fs]

Nice

[@cheekatiramyaramya3056]

Hloo mam is + having the high priority than -

[@cheekatiramyaramya3056]

Plse reply fast mam there is an exam tommorow