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Content
0 -> Hello Whats up guys, Anuj here and welcome to the DSA 1 course
2.352 -> And today in this video we are going to talk about buy and sell stock part-2
5.812 -> In the last video we had seen part 1, in which we could buy stock only one time and sell it
10.994 -> In today's problem we can buy and sell stock how much ever time we want
15.316 -> So we can do buy operation and sell operation as much we wish to
18.165 -> And we have to maximize our profit
19.976 -> There is a catch in this, which is, after every buy you have to sell
24.299 -> This means you can't buy continuously 2 times, that means at a time you should have only 1 stock, you cannot purchase 2 stocks
30.986 -> So, basically this is the problem, that again you are given an array
34.787 -> And in this array, it is specified that on each day what is the price of stock
37.828 -> So, there is one stock only, whose multiple are already given, you already know the future price
43.055 -> And you can buy and sell any number of stocks, but you should always have 1 stock
48.317 -> It is not possible that you have 2 stocks at a time
51.043 -> This means you have to sell immediately after buy
53.229 -> So how the profit will maximize?
54.659 -> Suppose , we purchased on the day with price 2, and sold on the day with price 7, so this can be 1 transaction
60.94 -> Next transaction can be we purchased on the day with price 3, and sold on the day with price 6
63.976 -> So this can be our transaction, and what will be the next transaction?
67.041 -> We purchased on the day with price 1, and sold on the dat with price 4
69.747 -> So from here what profit are we getting?
71.49 -> We get 7-2=5., 3-6=3, 4-1=3
76.787 -> If we add them 5+3+3=11
78.747 -> So from here the maximum profit we get from here is 11
81.581 -> We will not get more profit then this from here
83.172 -> So this the problem and now I think you might have already understood from here that how we have to solve it
89.933 -> But still think how can we solve it
91.495 -> And after that we will move ahead
92.997 -> So I have represented the same thing in a graphical form, so that you can clearly understand the proposed solution which I will tell you all
99.404 -> So basically, what we will do is, that whenever the bottom will come we will sell it, and whenever it will be at peak we will purchase it
106.466 -> So, I think you might have understood from here only that how it will be done
110.309 -> So whenever we are getting a local minimum, when ever we are getting local minimum of stock price, we will purchase the stock
116.228 -> And as we will come to know that the peak has came, we will sell it
119.294 -> And this the ideal situation of any stock market, that when the market will dip, when the market is down, at that time we purchase the stocks
128.874 -> And when the market is at the top, when it is going to crash, we sell it
132.815 -> But we don't know about future that when it will be at bottom and when it will be at top
136.633 -> So that is why over here, if you know the past trend in stock market, then it is more helpful
143.273 -> And there are many other factors also
145.786 -> But over here we will come out of it, we will come back to our problem
148.713 -> So, when our market will be down we will purchase it, and when it will be on top we will sell it
154.305 -> So we purchased at 2, sold at 7
155.714 -> So we got the profit of 5
157.194 -> So from this one we have got the profit 5
160.838 -> Then next local minimum is this, when will be the local minimum, when the left and right prices are bigger than the current price
166.598 -> This is the way to identify the local minimum
168.261 -> What is the way to identify the local minimum that, the left one and right one both are greater
171.159 -> And what is the identification of local maximum?
173.101 -> That the left one is also smaller and the right one is also smaller
174.748 -> So we purchased at 3 and sold at 6, because 6 is local maximum, how?
177.835 -> Because left price from 6 is smaller and the price on right side of 6 is also smaller, 3 and 1
181.711 -> So from here we get 6-3=3
186.778 -> Then we again local minimum because, left one is also greater and right one is also greater, so we purchased it on 1
192.336 -> But we will not sell it on 2, because 2 is not local maximum, we will well it at 4
197.805 -> Because left one from is smaller and there is nothing on right, so 4 is our local maxima
201.564 -> And 4-1=3, fine?
206.757 -> So this is basically our solution
208.498 -> That whenever we get a local minimum we will purchase it, and whenever we get local maximum we will sell it
214.259 -> And if we will keep on doing like this, then we will get our answer
217.229 -> Now, how we will find local minimum and maximum?
220.894 -> Basically, we have to iterate the whole array one time
223.689 -> And whenever you get an element that can be a local minimum then you purchase it
227.444 -> And when you have already purchased it then find local maximum
231.966 -> So in this way also you can move ahead
233.387 -> Or there is one more very simple code, when I will tell it in code form
237.631 -> So over her I have coded this, and if you will see over here
240.415 -> Then, I have not tried to find local minimum or local maximum
244.23 -> Basically, I have applied a small trick in this
247.39 -> Due to that trick I did not have to do the cumbersome code , a very simple and sweet code is ready over here
252.009 -> And the trick basically is, that suppose 5,2
255.594 -> Nothing will happen on 5, when we came at 2, then nothing will happen at 2 also
258.835 -> When we came at 6, I can see that 6 is greater then the value of yesterday's price
264.382 -> So 6>2, so I will purchase it over here
266.657 -> And I will do 6-2, so we will get 4
269.22 -> Now, I will move ahead, I will see what the current value is greater than yesterday's value? No
274.498 -> I am just checking with yesterday's value, I am not checking anything else
277.395 -> I am only checking with yesterday
278.724 -> Now I will come over here, Is this value greater than yesterday's value? Yes it is
281.62 -> So I will do 4-1=3
283.055 -> Now, I will move ahead and check i 7>4? Yes it is
288.207 -> So I will again do 7-4=3
290.847 -> So can you see over here, If I would be doing local minimum and local maximum then what would have happened
295.35 -> I would say local minimum is 1, so I will purchase it at 1,and 7 is local maxima so I will sell at 7
300.589 -> There is no need of it, I am getting 7-1=6, and by doing this 2 transaction also I am getting 6
305.687 -> I can do as much as transaction I want
306.669 -> So I am taking benefit of that over here
308.257 -> So 4-1=3, 7-4=3, and 3+3=6
313.498 -> So I am doing that only, I am checking it with the previous price, due to which my code is simple now, logic is still the same
318.187 -> Time complexity is still the same
319.873 -> We will move ahead, next is 3, is 3>7? No
322.924 -> Is 6>3? Yes
324.651 -> So we will sell at 6 which was purchased at 3, so 6-3=3
329.972 -> So we are getting 13 as our profit from here
335.497 -> I have done the same code over here, everytime I am tracking it with yesterday's price
340.679 -> So I have made a variable profit, and we will start from i=1
346.163 -> Now a[i]>a[i-1], is 2>5?No
350.716 -> Then It will not go in this condition
352.833 -> loop will come at 6
353.516 -> is 6>2? is a[i]>a[i-1]?yes 6 is greater than 2, so we will add in to profit 6-2=4, so profit will be 4
363.437 -> So now profit=4
364.853 -> Now we will move ahead, now we are on the day when price is 1
369.523 -> Is 1>6? No
371.733 -> So we will move on next day, 4>1? yes
374.861 -> So we will in profit 4-1=3, so now profit will be 4+3=7
381.795 -> Now next day, price is 7, 7>4? yes
385.202 -> Then 7-4=3, add 3 in profit
388.544 -> So 3+7=10, so profit will be 10
392.724 -> Now next day is with price 3, so 3>7? no
395.279 -> So, it will not go in this, we will come at 6, 6>3? yes
399.707 -> So we will add 6-3=3 in profit, so profit will be 10+3=13
405.49 -> So in this way without calculating local minimum and local maximum, and by applying a small trick I have simplified this code
413.529 -> So with this this problem finished over here
416.652 -> After that we will move ahead in array and solve some more problems
419.675 -> With this I'm signing off, If you liked this video then press the like button
