Form Number Divisible by 3 | Sliding Window Challenge | C++ Placement Course | Lecture 30.3

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Notes of this Lecture:

[@anshulsachdeva9981]

In this que, we were supposed to calculate sum of digits and not individual sums. The code ran correctly because in the sum of 12,56,84 is divisible by 3. Another function may be used for calculation of digits to correct it.

[@ritwikdutta6914]

Actually if sum is divisible by 3 then the the number written sequentially is also divisible by 3 . It's fact.
I was also confused at start

[@kashishjain4634]

The condition used here is true. This can be proved by:
Lets say 3 numbers sum is divisible by three ie (a+b+c)/3=q
Now we want to check divisibility for concatenated number formed by a,b,c ie 100a+10b+c
So, (100a+10b+c)/3 = (99a+9b + a+b+c)/3= 33a+3b+q
Hence concatenated number 100a+10b+c is also divisible by 3, if a+b+c is divisible by 3

[@piyushverma8207]

But we did not use the fact!!. We did not add the digits of individual elements, rather we just added the numbers and nothing else. Can someone tell me what I am missing?

[@tommylommy1333]

There is also another fact of divisible to 3:
If you having n numbers let's suppose n1, n2, n3......
And if you want to make a number with all n numbers divisible by three so for that condition sum all numbers should be divisible by three

[@sanketrajgiri3239]

@@tommylommy1333 What will be the complexity if we use the traditional adding all digits approach ?

[@tommylommy1333]

@@sanketrajgiri3239 but in this video method it was O(1) time complexity to add only array elements

[@tommylommy1333]

@@sanketrajgiri3239 i hope i cleared your doubt

[@harshrathi8955]

Fact was not used right…instead of individual digits we just added numbers in array?

[@ashishrao5323]

Chanell Monitized. 👍👍.

[@vissamsettyroja7914]

Can u make a video/podcast on importance of doing MS degree in foreign countries

[@aloksharma9706]

th-cam.com/video/6siaSZNg7L8/w-d-xo.html

[@GAURAVPASWAN-g1d]

kindly provide notes of each lecture

[@0xsh1v4m9]

Please upload videos fast. I have already completed all 85 videos of this playlist.

[@aloksharma9706]

th-cam.com/video/6siaSZNg7L8/w-d-xo.html

[@rutvikrana512]

Then go to GFG or HackerEarth and solve problems, there are many posts on GFG also which are helpful. There are many things yet remained in this series, Hash, Heap, Graph, DP etc. don’t just rely on videos 😊

[@0xsh1v4m9]

@@rutvikrana512 already done that bro and also completed all 129 videos. Btw thanks for help

[@rutvikrana512]

@@0xsh1v4m9 ...Wow I now noticed '3 week ago' 😂

[@mdekramuddin8534]

Hello..I want to know that does this playlist contains complete c++ and DSA ?

[@k.murari]

Use keypad for better understanding. And code writing approach is not so good.

[@sankalpsharma4890]

Great work 👍

[@mohakgidwani1505]

Amazing Lecture as always

[@hustlewithVaibhav]

Thank you 🙏🙏

[@praveenanand5689]

sum= sum+arr[j]-arr[j-k];
yeh arr[j-k] ->yeh toh hamesa arr[0] element ko point karega
phir next element kaise subtract honge

[@akashchoudhary6876]

j will increament by each loop and so will j-k

[@dhruvsakariya3129]

Boolean solution
bool isPossible(vector &nums, int k)
{
int len = nums.size();
int windowSum = 0;
for (int i = 0; i < k; i++)
{
windowSum += nums[i];
}
for (int j = k; j < len; j++)
{
if (windowSum % 3 == 0)
{
return true;
};
windowSum = windowSum + nums[j] - nums[j - k];
}
return windowSum % 3 == 0 ? true : false;
};

[@iamnaveenshankar]

Is this 130 video playlist the end of placement series??

[@aryandeswal1013]

No

[@KeshavSharma-ng4yr]

Java lecture upload krna start kr dijiye !!

[@aloksharma9706]

th-cam.com/video/6siaSZNg7L8/w-d-xo.html

[@techflying8660]

Bhaiya aaj mera brithday hai 22 march

[@sohamDevRel]

Happy birthday bro!

[@techflying8660]

@@sohamDevRel thank you

[@smackedup7657]

hbd

[@aloksharma9706]

th-cam.com/video/6siaSZNg7L8/w-d-xo.html

[@prateekrajput2646]

@@aloksharma9706 lol

[@Npanywayme]

FOR BETTER EXPAINATION VISIT ::: th-cam.com/play/PL_z_8CaSLPWeM8BDJmIYDaoQ5zuwyxnfj.html
//-->Check if a subarray of size K exists whose elements form a number divisible by 3
// C++ implementation of the above approach
#include
using namespace std;
// Function to find the
// K size subarray
void findSubArray(vector arr, int n, int size)
{
pair ans;
int i = 0, j = 0, sum = 0;
bool found = 0;
while (j < n)
{
sum = sum + arr[j];
if (j - i + 1 < size)
{
j++;
}
else if (j - i + 1 == size)
{
if (sum % 3 == 0)
{
ans = make_pair(i, j);
found = true;
break;
}
sum = sum - arr[i];
i++;
j++;
}
}
// If no such subarray is found
if (found == 0)
ans = make_pair(-1, 0);
if (ans.first == -1)
{
cout