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Hello everyone.



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Welcome to this video.



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So in this session what we do we will solve one more problem with the help of occasion.



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OK so the name of the problem is count digits.



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Okay so what is the problem.



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The problem is very simple.



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Suppose you will be given a number let's say the number is 4 0 2 5 and you have to tell the number of



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digits in the numbers so number of digits is food.



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Okay.



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Somalia.



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The number is 125.



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Number of digits is three.



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If the number is 20 the number of digits is 2.



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Okay.



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So you have to count the number of digits present in the number.



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Okay.



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Now what I have to do.



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So I have to write the function count what it will take it will take a number as input and you have



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to tell how many digits are present in this number.



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And now there is a constraint on the value of.



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So the end will be good then articles 2 1.



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Okay.



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So the value of in the end what is then and will be a natural number.



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Okay.



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So this is our constraint and is a natural number.



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Okay.



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That means and cannot be zero and and basically cannot be left not of course to zero and will always



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be a positive number.



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Okay.



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So okay.



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So we have to calculate the number of digits present in item but now how going to calculate the number



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of digits.



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Okay.



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So what we will do we will take the help of recursion.



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Okay.



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If you remember we have already solved this problem with the help of four loop by little but here since



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we're using regression things we are learning recursion.



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So we will try to solve this problem with the help of recursion.



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Now how we can solve this problem with the help of the occasion.



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So what will happen.



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Okay so this is our number 4 0 2 5.



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So what I'm planning to do here is what I will do I will split this number okay.



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I will tell the recursion to solve this problem for me.



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So recursion what would load it occasion will turn me down.



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I said three and then 12 what I will do I will add plus 1.



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So output will be for okay.



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So what I will do I will break my number.



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And then I will tell regression.



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Then I will tell revision to solve this problem.



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So the question will give me tree and I will add plus 1.



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Okay let's take one more example so suppose let's take a bigger example so let's say 1 2 3 4 and 5 and



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6.



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So how many digits are there in this number.



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So there are six digits.



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Okay.



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What I will do I will break this number and then I will tell the equation to solve this smaller problem



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for me.



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So the occasion will return me 5 and then I will add plus 1.



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So it output will become 6.



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Okay let's try it in on one example.



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So suppose let's say our example is 4 0 2 5.



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Okay.



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So the value of end is good then goes to 1.



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So this number qualifies for our problem.



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Okay so what I'm planning to do here is I will break this number at this part okay.



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And I will tell recursion to solve this problem for me.



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Okay.



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So I will reach for zero to number and then again recursion big problem.



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I will call on this part so I will call on 40 again.



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We will break and we will call on this part so I will call on for again I will break at this part and



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I will call here.



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So this is zero basically.



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So I am calling 1 0.



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So this will be our base case okay.



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This will be our base case.



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So what it will done.



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It will return 0 now.



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Our form levels this part because it will give me the answer of this part and I will add plus 1.



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Okay so 0 plus 1.



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So I will return one here okay.



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So the onset of this part is 1 and I will add plus 1.



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So I will have done 2 Okay so does sort of this part of the equation told me downs of this part is too



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so and I will add plus one so I will done three.



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So that onset of this part discussion has told me the downside of this part is three and three plus



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one.



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So our output will be full which is our correct output which is the correct output.



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Okay.



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So if this is n this whole number is n so 4 0 2 5.



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If this is n now what is 4 0 2 so 4 0 2.



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This is and by then.



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Okay.



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So in Deja upon Deja will we end in Deja.



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So 4 0 2 5 4 so 4 0 2 5 4 divided by 10 will not be 4 0 2.5.



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It will be for it'll do we.



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Because we remember it digit upon integer will we in danger.



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Okay.



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Digital divide won't be there will be in okay.



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So I will write a function count.



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So this current function will take and as input what we have to do we have to break our problem.



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So what this current function do is it will take an integer n and it will tell me the number of digits



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present in n.



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So what I will do I will give.



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I will call the function on end by 10.



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So count will count.



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Function will give me how many digits are present in that number and by then basically leave the value



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of end is 4 0 2 5.



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Then what I'm calling I'm calling 1 4 0 2.



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So the equation will tell me that the number of digits column function will tell me the number of digits



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present in this is 3.



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Now what I have to do I have to add plus 1.



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So if I will add plus 1 I will get the problem.



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I will get the solution of the bigger problem let's count often.



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Okay so this will be our recursive case okay.



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And that will be a success if the value of any 0.



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I have to return 0.



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Okay now one thing that you may can get confused is if the value find the 0 0 5 you are not returning



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when we are in order.



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Don't need one here because initially I took a look at the value a friend will be good then order goes



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to 1.



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So this is basically invalid input.



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Okay this will be invalid.



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That's why I'm returning 0 here okay.



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So let's write the code and then we will draw this diagram again.



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So what do we that are done by Britain.



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I will be in danger okay because I have got it done.



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How many did your child present.



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So let's write the function count it will take a number n and we have to find how many digits out peasant



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in the number and so base case will be so this is our base case and base case is very simple.



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If the value of any 0.



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So how many did you decide peasant 0 digits at present.



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Okay because this is invalid input then again recursive case



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so recursive case small output small answer so a small answer is if you want to calculate number of



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digits in n then calculate the number of digits and then by 10.



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Okay.



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And then the calculation part



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so conclusion part is very simple.



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After calculating this more long said I have to add plus one and they're done.



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And let's test our function.



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So I'm calling this function count for let's say 7 8 2 0.



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Okay I'm calling this function 4 7 2 0.



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Now let's run our file so our output should be for data four digits.



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Okay



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so I would put this forward civil code is working fine now after writing the code with the help of PMI



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with what we will do we will try to do that and our goal.



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Basically we will try to draw the diagram so that we can have a better understanding.



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Okay so 7 8 2 0 the value of land is 7 8 2 0 so 7 8 2 0 okay.



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So this is not the case 2 0 7 0 2 0 is not close to zero.



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So line number 10.



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I will call an end by 10.



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So I will call on 7 8 2 then again I will come out this line.



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So basically 7 8 2 0 is waiting at line number 10.



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Senator 2 is waiting at line number 10.



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Then I will call in 7.



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Again I will call on this function.



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So it will wait at line number 10.



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It will call on 7 7 is not to question 0.



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This also will headline number 10.



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So it will call long 0 okay and by 10 so 7 by 10 will be 0.



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So if the value of any 0 I returning 0.



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So I will return 0.



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So 7 was waiting at line number 10.



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Salvador the real number Densmore lines up plus 1.



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Okay so the value of small arms that was 0 so what 7 Miller done small Lancer plus 1.



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So it will return 1.



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Now here the value of small lines that is 1 again I have done smaller small said plus 1 so I will return



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to the value of small lines.



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That is to say I have the right and small answer plus 1 so I will return 3.



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Then I have to write down small lines that plus 1.



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So I hate the value of small lines that is 3 so I have to attend 3 plus 1 which is 4.



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Okay so 4 is our answer 4 7 8 2 0.



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Okay so I have 70 to 0.



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Then I beg this number I.



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I call that equation on this part of the equation.



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So this is 7 8 2.



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I big this number.



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Recursion I call the equation on this part.



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So this is 78.



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I big this number.



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The equation will call on this part I break this number recursion will call on this part so it will



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be 0 so 0 will return 0 from it.



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And 0 plus 1.



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So I will return when one plus one.



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So I will return to 2 plus 1.



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I will return 3 3 plus 1.



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I will return food.



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Okay so in short you can understand it like this way.



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I want to calculate the number of digits present in this part present in the numbers 7 8 2 0.



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I will break the number because the questions is vague the number will big the bigger problem and the



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smaller problem and I am calling recursion on this part.



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So the equation will tell me you do not we have to assume okay.



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I told you we have to assume so we have to assume that the equation will tell me the answer of this



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party's tree.



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Okay now you do not have to think well how does that tree assume that the recursion will give me the



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answer tree.



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So recursion gives me the answer tree and then I will add plus 1 so 3 plus 1 and the answer is four.



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Okay so I am assuming here that and by then common function will give me the value.



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Give me the answer for this part.



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So the answer for this one will come out to be tree and then I am returning 3 plus 1 which is 4.



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Okay so I hope you understood the code.



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Okay.



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The problem was very very simple.



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Okay thank you.