Divisible array problem

#1


test case 3 is not passing giving incorrect anser

#2

array2[i]=(array[i-1] + array2[i-1])%n;
We need to take modulo at this step also.
This will help in preventing overflow, also in case where sum comes to be a big negative number this will help in bringing it down,
Eg array2[i]=-150, n=51
now array3[i]=(-150+51)%51 =-48(but we need positive number so)
so
array2[i]=array2[i]%n
this will make
array2[i]=-48
array3[i]=(array2[i]+n)%n=3

Please mark your doubt as resolved if this solves your doubt.

#3

can you send the test case 3 inputs

#4

also when i did not took modulo there the index may come negative of pre array so why does other test cases passes

#5

Every testcase does not contain every cases and corner cases, that is why there are multiple test case.

#6

So what is the main aim to take modulo there only one reason overfolw ??

#7

overflow and negative number as explained above, when you write code you need to write it keeping in mind every possible case(even if the occurrence of case is rare)

#8

IMG_20200314_172903
IMG_20200314_1729033000Ă—632 589 KB

According to video in the line 23 is we can take modulo of sum till that index like this… How this is working…

#9

What you have written is essentially the same, you just used an entire array for calculating sum, and have shifted the entire sum position by 1 which does not matter.

#10

I am talking about what is written in the video that line I want to. understand from line 23 to 25 how this is working…

#11

We take the input, then we directly add it to sum, now sum at this point will contain the same value as that of array2[i+1], now we take the modulo of sum, similar to taking mod of array2[i+1] then again we take mod by adding n to adjust negative number same as (array2[i+1]+n)%n, now it does pre[sum]++ that is equivalent to array3[array2[i+1]]++.
As I said both are doing exactly the same thing, this code is just a bit optimised it is saving space and extra for loops.
If this solves your doubt mark it as resolved.

1 Like
#12

thanks sir for your crystal clear explanantion…

#13

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