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Hello, everyone, so in the last video, we discussed the algorithm for implementing in place observed,

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and I told you guys to just give it a shot, try to write the code yourself.

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So what is the time, complexities or the time?

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Complexity is actually in Logan and the space complexity.

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So as the name suggests in place.

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So the meaning of in place is that it will actually constrain time so that space is going to be big

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of one.

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Now as discussed.

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So what we will also consider this area.

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So let's say the array elements are five, eight, three, one and six.

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And what was our approach?

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So I will divide the array into two parts.

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So the first part is actually containing the hip element and the second part is actually containing

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our unsorted area.

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And we discussed it.

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But we will also I will start reversing the area from the next one.

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So this is an this is an X zero, so I will start traversing the area from the next one.

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And our approach is very simple.

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So one by one, I will push the elements into my hip.

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So what will happen to the hip range will increase and the unsorted range will decrease.

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So you can insert elements and obviously the next zero is one.

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So this is my child index.

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And again, you have to find out the pain index, which is simply index minus one by two.

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And this loop is actually it is containing the child makes such an index value when then its value will

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become three.

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It will become two.

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And then it's three.

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And then it will become four.

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So simply you have to compare it in five and nothing will happen.

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Then you will take elementary inside your heap.

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Then heap sizes increase and assorted sizes will decrease.

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So similarly, three will come.

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So the index will become two.

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You can see child index will become two, then you can swap that image.

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Three will come here, five will come here.

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And similarly then you will take one.

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So basically one will be in this position.

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So the child index is three.

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You can see the child index is three.

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Again, you will do the comparison to one and it will get stabbed.

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So this is when then one in three will get stabbed.

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So this is one and this is three.

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And similarly, you will take the element forward into consideration.

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So finally, I he will increase and the state will that elements.

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There are no certain elements left.

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So you will push six at this position, so finally, what is my hip?

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So this is one, then it is three, then it is five, then it is containing eight and this is six.

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Simple.

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So basically, if you will convert it into an area again to the level of traversal.

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So this is one.

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This is three.

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This is five.

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This is it.

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And this is six.

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So very, very simple.

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But we have to do we will just reverse from index one.

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We will start reversing from the next one and we can copy paste our old code, our old insert function

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called now one thing we cannot directly use our one triadic because our own priorities create a vector.

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So we cannot just blindly copy our code.

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So we just have to do minor, minor changes.

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Now let's start writing the code.

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So this is my file and the place he ordered CPV, so let us create a function so that the function will

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be void because it will just sort Meyera, it will not do anything else.

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It will not return anything.

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So the name of the function is inbuilt.

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He sort.

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So what this function will take, so this function will take two things as input to first, one is actually

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the Eddie.

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And let's say the name of that is in and the second one is actually the size of the ADL and.

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So what is the first step?

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So first step is very simple, but we are going to do it so we will build keep inside the area.

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So first step is very simple.

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We will build deep.

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In prudery.

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We will not use any extra space for building the ship, I told you, we have to start our trading from

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next one because we can assume that index zero is already in a heap.

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So you can see.

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So the value at index zero is already inside the heap, we have to start premising from index one and

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then we will take element in to help one by one.

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Simple.

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So I will start from one and then I listened and I placeless, simple.

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So what they have to do, we have to copy paste of something called.

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So this is my Plotka class, this is insert function now what we are doing, so this is our elemental

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I was pushing elements.

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Now, this thing is not required.

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We have to copy paste only this much amount of gold.

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So after copy pasting the code.

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So this is the situation.

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Now we have to do some changes, for example, because nothing peculiar to say so.

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This is my child index, so what is child index, you can see so the index is nothing.

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It is just the index of the area, so it will start from one, then it'll become two, then it will

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become three and so on.

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So first, I have to update my child index.

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Which index is actually the index of the area.

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So this is simply a.

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So first thing.

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So this is very simple, very minor change, so child indexes, a and one more thing.

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So here, the name of the is input, but we are actually using Piku everywhere, so let's change the

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name to the name of the input that is Piku.

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So first I am finding out the child index, which is correct, then this is the standard code.

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So while Calix is good then it'll find the index, then do the comparison.

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If the child index is Milovan, so do the part then update the child index as to the break.

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So this break is actually.

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So this statement actually makes the inner most loop.

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So the innermost loop is by loop.

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So this big statement can only break by loop.

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It cannot break for loop.

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So this thing you should remember now after doing this.

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So actually, our help is right now, our help is ready then what we have to do.

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So basically now we have to call the remove minimum function and times.

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So what we have to do, we have to call remove minimum function and times now what we are doing here.

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So after this cold, so let's say my input at is actually, let's say five eight, one, two and zero.

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So let's say this is my input area five it one to zero.

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So after passing through this code, what will happen to.

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This is the value of I, so I will vary from index one to the last index.

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So index is nothing.

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It is just.

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The value of the added, just the index of the area, so first five elements.

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So this element, five inside our heap, so take element.

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So nothing will happen.

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So this condition will become false and we will break.

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So I have element one Satake element one, then compared element one with element five.

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So we will do the stripping then you will take element to so element to it.

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So then you will find out your paint index.

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So this is element to the index is basically.

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So this is index zero one, two and three.

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Again you have to do the slapping part of this is it.

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And this is to again take the elements.

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And also this is element zero.

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You will do the slapping.

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Part of this is two.

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This is.

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So this is zero, then you will slap one in zero, so when will come in and zero will come here?

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So finally, this is your area.

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So this is 01, then I have five here, then I have it here and then I have two here.

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So this is your.

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Minimum heap, so this school is actually for minimum.

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We are trying to implement our minimum priority queue and if you convert converted into an area.

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So this is zero, then I have one, then I have five.

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Then I have it.

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And then I have two.

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So 015 eight two.

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Now, this is our input area.

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So this area has now been converted into this area and now we have to do so.

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We just have to call the remove minimum function and times.

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Or you can also call and sometimes both will be correct, because if you recall, and minus one times,

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so what will happen, only this element will be left.

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So it is your choice whether to.

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So this is not irrelevant.

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So basically, the size will be one, size will be one.

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And it is your choice because the element is going to establish itself.

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So it doesn't make any sense to call.

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And times you can also call and minus one times, let's say we will call and times.

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So if you remember how they removed minimum function, Wilberg, so this was my area zero, then one.

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So this was my minimum 015 eight and do so I have to call the minimum function.

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So it will going it will Sub-Zero and also zero will come here and two will come here.

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Then actually in the original code we were trying to delete element zero, but here we will not delete

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zero.

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We will just assume that element zero has been removed.

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We just have to adjust.

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So what we will do, I will take a variable size and instead of doing B good artpop back.

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So work bob back function do so it also decrease the size it removed the elemental back function, do

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two things so it will remove the element and it would also decrease the size by one.

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So we will not use proper function material.

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So we will just reduce size minus minus.

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So size minus minus indicates that the element has been removed.

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Because we do not have to actually remove the element, we just have to assume so for assumption, we

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will just take a variable size and we will lose size minus minus and size minus minus indicates that

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the element is removed.

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Simple.

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So let's do this.

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And also, after stepping, we have to put two added side operations here compared to with one in five.

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And again, I will say absolutely will come here.

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And when will commercial now to compare to with it, but do not compare two with zero because this is

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removed, this element has been removed.

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So that's why I will lose size minus.

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Minus.

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So what we have to do, so let's take a variable size, so initially my initial my area is containing

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N elements.

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So what we have to do so first thing is basically we have to swab the first element with the last element.

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Because I want to remove medium function and time so that the first element, which is zero with the

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last element to what is the last element, so I will use size minus one to size minus one will give

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me the index of the last element and then if you will see.

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So here's what I was doing, so I was calling the function back, but here I will just write says minus

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minus Simbel.

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So this indicates that the element has been removed, but this is only assumption to element is removed.

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Now you can just copy paste the code.

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Because we have to don epiphytes.

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It's down here, Biffy.

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OK, so this is the code for down here, Biffy.

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So let me check the loops first.

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OK, so this is correct.

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So, again, we have to do down here.

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If I saw the paint index zero, correct.

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While Drew, I am finding out the left index, right index, then the index is actually the paint index,

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then left index is less than I should be good outside.

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So here I will not use peculiarities.

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I will simply use the size variable.

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So the size variable.

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Now this is correct.

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So again, instead of peculiar because we do a simple vector here, so simple simply I will use the

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same variable.

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So this variable.

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Then this is correct.

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So I'm scrapping and then Bendix is getting ablated with the index, so correct.

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So this is one time I have to perform it and times I have to call remove Menom function and times.

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So let's do it.

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So now to call the function and times what I will do.

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So while size is guerdon or goes to one.

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We have to call the movement function and.

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So, yes, everything seems fine.

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So as discussed, how many times we have to remove minimum function, so we have to remove minimum function

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and times, so the size, the value of sizes and while size is bigger than order, close to one end

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size, minus minus.

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So let me save this loop.

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So this law will run and time's simple, this loop is going to run and times are good, then are close

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to one end size, minus minus.

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And see what I was trying to tell you, that instead of using this condition, says the Noriko's to

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one, you can also write the condition, says guerdon one.

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Why?

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Because see if the value of sales is one, then what is happening here.

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So the value is one.

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So one minus one that is zero and this is also zero.

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So what you are doing, you're swapping AOF zero with zero.

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Where is the eddy?

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So you are swiping the same two elements.

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So that's why I told you that you can also write is lower than when both are correct if you will is

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good then one.

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So the least value of sales will be to and in that case you will be swapping the element of zero with

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each to both are correct.

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It is your choice.

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So this is our standard called.

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We do not have to do anything here, we just have to right size here instead of peculiarities, you

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will use the size, variable and similarly size.

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So everything will remain the same.

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So afraid of the cold, we have to test whether our goal is working, correct or not.

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So let's make our input very.

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So let's say the elements are five one, two zero and eight, so this is zero and eight.

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So I have five elements then I have to call the function in place.

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So let's copy the name I have to pass to I have to pass booting the area.

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And this is.

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So the name of the ad is and put.

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And there are five elements.

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And now let's try to pin down our elements.

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So this is my input today, I am calling the function in place, he observed, and we know it is advised

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by a defense.

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And in C++, that is our best bet, if it's by default and then I'm bending the elements.

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So what will be my output?

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So basically, we have written the code for implementing minimum prior queue, so virtually all parts

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of my output should be it, then it should be five, then it should be two, then it should be one,

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and then it should be zero.

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So basically our output will be in decreasing order.

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Why it is in decreasing order, because if you will see the last video, so in the last video, I already

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explained to you that the output is coming out to be in decreasing order.

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So you can see if we are going to implement the minimum protocol, then basically our output is going

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to be sorted in descending order.

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There are is going to be in descending order if you want Iraq to be in ascending order.

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So you have two options first implemented prior to go and then apply the reverse function, you can

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just reverse the error.

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And the second way is just write the code for the maximum priority queue.

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So this should be my output, eight five two one zero.

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So let's see.

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So you can see the output is coming out in descending order.

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So eight five two one zero.

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So our goal is working fine.

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Now, again, if you want, we can analyze the time and the space complexity.

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So, first of all, the space complexity and this is very simple.

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We are just creating very few variables.

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So the space complexity is constrained.

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If you want to analyze the time complexity.

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So let me do that.

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So we have to function first inside function and then we have removed minimum function, so we are using

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to loop.

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So this is the first loop.

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So this loop will run and times and this is the insert function called insert at the right place.

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And we know insertion takes longer in time.

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So this is a login so and login for building the heap and login for building the heap in the input area.

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And then we have to call remove minimum function and times.

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So I am calling the function and times and each time removal.

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So we are doing down here biffy and here we are doing up by so end times and each time login time so

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and login.

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And if you will add these two to the time complexities and login and the space complexity is very simple.

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This is big of one.

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So this is the code for the minimum priority queue.

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And the output is going to be in descending order.

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So if you want to implement maximum protocol, so what changes you have to do?

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So first of all, you have to change this code, building the heap so you will just change a few variables.

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So, for example, instead of less than you will.

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Right, good.

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Then then similarly, you have to do minor minor changes in remove function, for example, where you

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are writing the condition less.

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You will write the conditions there then.

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So these are very few minor changes that you have to do for implementing maximum protocol.

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And I have confidence in you guys that you can definitely implement, mix and protect yourself.

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So if you will implement makes improved, you are getting the time and the space complexity.

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So they will be same time is going to be and Logan and the space is going to be big often.

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And basically your input array will be started in ascending order.

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So I will share this quote with you.

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You can analyze this called.

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So I will see you in the next one.

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Bye bye.