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Hi, everyone.

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So today we are going to solve this problem, evaluate expression.

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OK, so the problem statement is very simple.

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So this is basically that prop string.

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So this is a very interesting so this will be our input and we have to evaluate our postfix notation.

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OK, so there are basically two types of annotations.

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So we have three notations.

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So these notations are infix.

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We have postfix.

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And we have prefix notation.

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OK, so generally we are only aware of infix notation, so infix means basically two plus one.

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Multiple-entry, so this is an example of infix notation.

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OK, but there are postfix and prefix notation.

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So what happens?

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So if you will give this input to the calculator, so, for example, if you will give this input to

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the calculator, then how the calculator will evaluate our answer.

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So first of all, Calculator will convert this infix expression into postfix and then postfix expression

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is evaluated and then we will get our answer.

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OK, so that's how the calculator works.

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So Calculator, the working of calculator is very simple.

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First of all, the first step is to convert the infix expression into the postfix.

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And then evaluate the postfix expression.

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OK, so evaluate postfix expression so they'll tell the calculator works.

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So basically we are given postfix expression.

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So this is a postfix.

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Expression and we have to evaluate this, OK, so we have to evaluate this.

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So the other name of this question can be evaluated, postfix expression.

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The other name of postfix expression is also called reverse Polish notation.

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OK, so reverse Polish notation.

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This is the name of Postfix.

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OK, so basically how we can convert infix to postfix.

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So this is very simple.

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OK, so this is given postfix expression and this is basically its corresponding infix.

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So this is infix and this we already know.

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Similarly, this is infix and we are very familiar with the infix views and fixing our day to day life

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so how we can convert infix into postfix.

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OK, so for example, let us consider infix expression, so let us consider this one only.

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So let's take it today.

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OK, so this is a very good example.

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So this is infix.

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Now, how we can convert infix to postfix, so the idea is very simple, what we will do.

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So first of all, move the opponent.

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So we have operators.

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And we have operands.

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So operators means plus, multiply, minus and divide.

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OK, plus or minus divide and so on.

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OK, and operators are basically indigence on which we have to perform the opera operations.

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OK, so what will do, move the operators to the closing brackets.

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So similarly.

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Fort Bliss, this is the closing record.

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OK, so remove, multiply, remove, divide from here and put it here similarly, remove place from

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here and put glass here.

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Now read this from left to right.

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So first of all, we have four.

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So we will write for then.

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Plus is not here because it is here, then we will not read the opening record.

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So we will read 13.

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So I will read 13.

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Then the divide is not there.

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We have five.

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So I will write five.

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Then we have this multiply.

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So remove this.

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So we have divide.

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Sorry.

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So I will write divide then.

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At this point I have plus so I will write plus.

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So this is basically.

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The postfix notation.

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OK, so this is a postfix notation.

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Let us take one more example.

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OK, so suppose I have let's say.

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Two plus one.

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And two, three.

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OK, so this is basically infix and we have to convert this into postfix so how we can do this.

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So the idea is very simple, but we have to do we have to move the operators.

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So for plus it's closing bracket.

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Is this so right.

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Plus here.

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Again, remove it from here.

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So for multiply, it's closing record.

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Is this OK?

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So remove, multiply from here and right, multiply here and then read from left to right.

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So first I have this to you.

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Do not consider this opening and closing brackets.

86
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OK, so first I have to.

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So I will write to then Plus's naughtier then I have one so I will write one.

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Then I have Pless so I will write plus then I have three so I will write three.

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Then I have multiply so I will write multiply.

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OK, so this is the corresponding postfix notation you can see in the above example.

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Also saw two one plus three multiply.

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OK, so do one blessed three multiply.

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OK, so the input is given a postfix expression and we have to evaluate.

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So basically if the expression was infix, we can see our answer will be nine.

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We can easily calculate.

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OK, so in this case, our answer will be six.

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So this is an integer division.

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So whenever integer upon integer we will get integer only.

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So this is basically two.

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We will get two and four plus two is six.

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OK, so now let's see how to evaluate postfix expression.

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So basically there are some rules on how to evaluate for expression and we will simply follow the rules.

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OK, if you don't know the rules, you cannot solve this question.

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OK, so there are basically defined rules.

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To solve the postfix expression.

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OK, and you have to remember this rules, OK, if you don't know the rules, you can resolve this question.

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OK, so the rules are very simple, so let us consider an example first.

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So the first two and one plus three and multiply.

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So this is a postfix notation.

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So the rules are very, very simple.

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OK, so what are the rules?

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So obviously, I will have operators.

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And I will have opening's.

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So operators will be like plus minus multiply and divide.

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OK, so for operators to work, I should have two options.

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I'll get to should be able to be for example, you can only use plus when I have to options E plus B

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minus B, E multiplier baby and a divided baby.

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OK, so we need to operands.

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To apply the operators.

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OK, so the idea is very simple, maintain a stake so you have a stake in the stake is empty.

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So you are standing at this point.

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So if it is an open end, so if it is to then push, this is open push.

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Now I have an operator, so I have operator.

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So I have to perform operation.

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OK, so we're performing the operation.

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What I will do, I will pop one, I will pop two and I will do the operation.

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So I will do two plus one.

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OK, I am writing it here.

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So at this point what will happen?

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I will pop the top two element of the stack and I will do the operation.

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So I will write two plus one.

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OK, so at this point, what I am doing pop the two elements.

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So that we can apply the plus operating between them, so pop, two elements apply plus operator, so

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after playing the plus operators, so we will get the result three and then put Bush result three inside

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the stick Bush result.

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So the result is three.

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So what we will do, we will push the result.

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So three will come inside the stack.

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OK.

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Now, again, so I have an option, if it is an option, you can basically push, OK, now I have an

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operator, so if it is an operator, what we have to do, we have to pop two elements.

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So simply follow the step up, two elements.

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So two elements are three and three.

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So the two elements are three and three.

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Then what I have to do, I have to apply the operator.

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So in this case, our operators multiply.

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So we will apply the operator.

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We will get the result and result will be nine and then we have to push the result.

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OK, so basically these two elements are public.

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I have to push the result.

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So I will push the result nine and then I will reach at the end.

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OK, so what will be our answer?

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So finally, our answer will be the element present at this Dec. 11 percent detail of the strike.

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So stack as dot, dot, dot, dot, dot will be our answer.

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So our answer is nine.

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OK, let's see so we can see.

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Our answer is nine.

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OK, now let us.

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Used the same approach on this for three, four, 13, five.

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Divide and class, OK, 48 in five divided plus.

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So I have four, 13, five divide plus.

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So we know the rules are very simple.

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If it is an integer, push inside the stick.

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OK, so at this point, what will happen?

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So this is an integer, which we have integer, which we have integer.

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Bush, we have an op ed, so if it is an operator, obviously we need to eliminate, so we need to eliminate.

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So how we will get to elements, we will get the two elements with the help of Steck.

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So with the help of Steg, I will Bob Fife and I will pop 13.

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OK, so remove five and 13.

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So I have to eliminate 13 and five.

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And then obviously we have to apply the operator.

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So applied the operator after applying the operator.

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Our result will be 13 by five, which is two.

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OK, so this is our result.

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And now what we have to do, we have to push that result inside the stack.

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So too will come here.

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OK, now, again, move ahead, so I have basically plus so plus that means we need two elements.

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OK, and how we will get there to eliminate simply we will take the two elements from the top of the

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stack, so remove two and remove four.

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So I have four and I have two.

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And then what we have to do, we have to apply the operator operators.

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So we will apply the operator.

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Our result will come out to be six and then we would have to do we have to push the result inside the

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stack.

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OK, and then I will reach at the end.

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So after reaching the end, what will be our answer?

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Our answer is basically the top element of the stack, which is six.

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OK, so for this case, our output is six and we can see here.

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Our output was six.

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OK, so basically our goal is working, our logic is 100 percent correct.

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OK, now you can assume here.

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OK, so you can assume that.

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What we can assume, so the input given to us, they input postfix expression given to us is basically

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a valid expression.

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OK, this is a valid expression that we can assume.

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OK, so now let us write the code.

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So basically the rules are very simple.

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So no one, if it is an open.

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If operand, then simply push.

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Second guess it is operator.

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So if it is operator, then it is very simple, you need two elements to perform the operation and how

202
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you will get to elements simply the top of this track.

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OK.

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Perform the operation, we will perform plus, minus multiply, divide, so plus minus, multiply and

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divide and then you will get the result.

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So Bush that is held inside this deck.

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And finally, what will happen is finally our answer basically will be the element, there will be only

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one element at the end.

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OK, so at the end there will be only one element inside the stack and that element will be our answer.

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OK, there will be only one element at the end and it will be our answer.

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So that is basically the last remaining element of the stack.

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And we can get that element with the help of the top.

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OK, now let us read the gold.

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So if you have understood the solution, if you understood the logic, writing code is very easy.

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So first of all, what we need, we need a stack and stack of integers, basically, OK, we are storing

216
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and beta inside the stack.

217
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So the name of the stack is empty.

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Now, if we have to do, we are simply iterating.

219
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So let us separate.

220
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So equal zero I listin.

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Tokens, Dot says.

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And I placeless, so we are trading OK, now what we have to do so.

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If the top element is an apprentice, we will push and if it is appropriate, so if it is an operator.

224
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So basically if tokens of a equal equal plus or tokens of a

225
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minus or tokens of a equal equal multiply or tokens of a ecological divide.

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So if this is the scenario, then what we have to do, so if it is the operator, we have to push,

227
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we have to pop two elements, OK, so we need two elements here to perform the operation.

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We need two elements.

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OK, so now let's get both the elements and we will get the elements from the stack, so let's name

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it value one, which is a start top.

231
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And astronaut Bob.

232
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Now, let us get our second value, our second element, so we do again our top, an astronaut, Bob

233
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Stegner.

234
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Double standard.

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Double standard, Bob.

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OK, so we have our board, the elements now, let us perform the operation, so if.

237
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Tokens of equal, equal, plus, then what we have to do, we will add we want and we do OK so as to

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not so first find the result and then push.

239
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So I have to do with two plus one.

240
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OK, I am finding the result.

241
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So Bush result inside the deck.

242
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OK, so push the result inside stick.

243
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Now, similarly.

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So elusive.

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If tokens of A is minus, we have separate operations, so I will do we do minus Reven.

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Similarly.

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And if I have multiply operation.

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I will do over to multiply one, and if I have the divide operation.

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So I will do we do divided by one.

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OK, and that's all now.

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00:16:30,080 --> 00:16:33,590
Now, in the Elzbieta, what we have, so we have options here.

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Now, we discussed if we have an option, we can basically push inside the stack.

253
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So what I will write is to not push and we can basically push Dawkins off a.

254
00:16:46,850 --> 00:16:49,370
But there's a problem here and what is the problem?

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So first of all, this will be Bush.

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So what is the problem?

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So basically this bag is often details.

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OK, so I have created a stack of often teachers and tokens of I is basically a string.

259
00:17:09,410 --> 00:17:15,920
OK, so it is basically a string, for example, tartine, for example, to five, for example, to

260
00:17:16,250 --> 00:17:22,069
OK, so this is basically a string and our stack, our stack is of integers.

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So what I have to do, I have to convert the string into integers.

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OK, so this is our problem.

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I have to convert the string into integers so that I can push integer inside the deck.

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00:17:36,010 --> 00:17:39,620
OK, so basically we do not have to write code for this.

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We have inbuilt function and the name of the function is to say, OK, so what it does, it will take

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a string as input.

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And it will give me the integer as output.

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OK, so if I will give it away basically a string, let's say 45, then it will give me integer 45.

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OK, so that is the functionality of it to wavefunction.

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OK, so this function is present inside the inside some library of C++.

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OK, so this function is already implemented.

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We do not have to implement this, but there is a problem with this, a two way function and what is

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the problem.

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So the problem is basically very simple.

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So I'm explaining it here.

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So first thing you understand that why we have to convert string into integer, because our input,

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the input that is given to us is basically a string and the stack that we have created is basically

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often tedious.

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So that's why I have to convert a string in 2010.

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OK, now the second thing, how to convert into integer.

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So I'm telling you, we have a function in the way I will give a string as input and it will give me

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the integer value.

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Simple, but there's again a problem with a two way function and the problem with it is so in C++ there

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is basically two types of string.

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OK, so there are two types of strings, so first one is the C strange string C style.

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So this is basically the character.

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OK.

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And the second one is basically the string class.

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00:19:22,520 --> 00:19:24,170
OK, so this is.

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This is basically distinctness.

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And it wavefunction takes in Putsy style strength and not this C++ type string.

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00:19:34,960 --> 00:19:40,550
OK, so what I have to do, so I have to convert this string in to see Strela string.

293
00:19:41,050 --> 00:19:45,900
OK, so basically I am repeating myself in C++ we have two types of string.

294
00:19:46,180 --> 00:19:50,200
So first one is this e-mail and the second one is basically the string class.

295
00:19:50,580 --> 00:19:53,050
OK, so here I will write string as.

296
00:19:55,100 --> 00:20:02,260
And this is the character, Mary, so it wavefunction takes input, see style string, basically correct

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type string.

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00:20:03,800 --> 00:20:08,390
And the input which is given to us is basically the second by posting.

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00:20:08,750 --> 00:20:14,960
So there should be a way to convert this string and do this character, Eddie, and for converting.

300
00:20:15,410 --> 00:20:19,580
So we have a function called Dot C and a score SDR.

301
00:20:20,790 --> 00:20:25,660
OK, so this is the function which is used to convert this type of string into the correct.

302
00:20:26,370 --> 00:20:29,100
So what I will write, I will write you to away.

303
00:20:30,080 --> 00:20:38,510
I have the input, which is this tape and what I will do, I will write, not see and the score SDR.

304
00:20:40,430 --> 00:20:42,460
OK, so that's how it will work.

305
00:20:44,010 --> 00:20:50,870
OK, so this string is basically this is a token of a token of fire is basically a string.

306
00:20:51,630 --> 00:20:53,040
Now, let us write the code.

307
00:20:57,260 --> 00:20:59,890
So I cannot write like this sort of.

308
00:21:00,170 --> 00:21:08,690
This is basically a string, so for converting string into integer, I have to use it to wavefunction.

309
00:21:10,430 --> 00:21:17,440
But this is not enough, not enough, because it's wavefunction will give us added that this dog that

310
00:21:17,780 --> 00:21:19,900
they did expect a sea style sting.

311
00:21:20,270 --> 00:21:26,810
So what I have to do, I have to use dot see in the square star so that ci a. could steer.

312
00:21:28,580 --> 00:21:30,560
So all these functions are in functions.

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00:21:30,590 --> 00:21:36,980
OK, you do not have to worry about this, and if your doubt is how I know about these functions is

314
00:21:36,980 --> 00:21:42,110
basically at this point when I first of all, this question, at this point, I get confused how to

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00:21:42,110 --> 00:21:42,490
solve.

316
00:21:42,500 --> 00:21:45,830
Then I search on the Internet and I came up with the solutions.

317
00:21:45,860 --> 00:21:47,900
OK, so that's how everyone learns.

318
00:21:49,820 --> 00:21:52,900
So finally, when I will come out of this loop.

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00:21:52,910 --> 00:21:53,990
So what will happen?

320
00:21:56,820 --> 00:21:58,170
So this is Ms.

321
00:22:03,400 --> 00:22:09,090
OK, so when I will come out of this loop, so basically I will have only one element inside the stick,

322
00:22:09,400 --> 00:22:13,200
so only one element inside the stick, and that will be our answer.

323
00:22:15,540 --> 00:22:20,840
So let us return our answer so we're done Steck, dot, dot.

324
00:22:21,630 --> 00:22:25,010
OK, you can see the return type of the function is in Pejar.

325
00:22:25,290 --> 00:22:27,630
That's why I have created this stack of integers.

326
00:22:27,640 --> 00:22:30,420
OK, so finally I can write like this.

327
00:22:30,420 --> 00:22:31,710
They don't stack up.

328
00:22:32,400 --> 00:22:35,220
So I think I have done everything correctly.

329
00:22:35,940 --> 00:22:37,560
Now let us try to run our code.

330
00:22:40,940 --> 00:22:43,460
OK, so there is some problem with the Blackwood's.

331
00:23:00,000 --> 00:23:02,060
OK, so there will be one more Beckert.

332
00:23:06,850 --> 00:23:08,200
Not at a summit, our called.

333
00:23:10,350 --> 00:23:15,000
OK, so our goal is working fine, so what is the time and the space complexity?

334
00:23:16,020 --> 00:23:22,350
OK, so basically the time complexity, we are just getting over the edge over the strings of time.

335
00:23:22,350 --> 00:23:25,860
Complexity is on and we are using we are creating a stack.

336
00:23:26,200 --> 00:23:32,040
So the space complexities on OK, so this is the time and the space complexity of a solution.

337
00:23:33,610 --> 00:23:36,010
OK, so I hope you understood this.

338
00:23:37,270 --> 00:23:39,950
I hope you understood this problem and also the solution.

339
00:23:40,180 --> 00:23:43,450
So basically you can see reverse policy notation.

340
00:23:43,480 --> 00:23:48,770
OK, so this is the another name for postfix rotation policy expression.

341
00:23:48,820 --> 00:23:55,360
OK, so if you have any doubt in this problem, it everyday approach or either with the or you can ask

342
00:23:55,360 --> 00:23:55,630
me.

343
00:23:55,730 --> 00:23:57,280
OK, thank you.