Minimum Operations to Make a Uni-Value Grid:
https://youtu.be/JXlM58dalbw

Minimum Index of a valid split:
https://youtu.be/VIygRO5dHqQ

NOTE:
Resuming video making today.
Today's video will come by 2 PM IST

Solving Questions With Brainpower:
https://youtu.be/k2JC9CZ0UGU

Maximum Value of an Ordered Triplet I & II:
https://youtu.be/qzJJdIZGjfA

Interview Expectations in 2025
LIVE session

Learn :-
1. The current marking standards of top companies
2. What to expect in changing focus from online to offline (on-site) interviews
3. Interview level from OA to on-site
4. Strategies which works in on-site rounds
5. Difficulty level analysis round-wise
6. How DSA and System Design expectations are different
+MUCH MORE…

We will have a QnA session too, where you can ask any of your questions too.

JOIN us today: https://chat.whatsapp.com/H9Yae2Kjw3J...

Date: 9th April
Day: wednesday
Time: 7:30 PM IST / 7 AM PST
Fee: 99 INR / 2 USD

Maximum Value of an Ordered Triplet I and II:
https://youtu.be/qzJJdIZGjfA

Lowest Common Ancestor of Deepest Leaves:
https://youtu.be/vXTlRSr-zfY

Even though I had mentioned not to make EASY problems starting this month, still I see that today's problem is interesting, hence it will be uploaded by afternoon.

Sum of All Subset XOR Totals:
https://youtu.be/TnpowktUe1E

Today's video will be released by 3 PM.
Meanwhile you can watch the prerequisite:
https://youtu.be/mouCn3CFpgg?si=k3jMqD_EzPE41lG3

Longest Divisible Subset:
https://youtu.be/q8UcbJDHp3M

Today's video:
Partition equal subset sum: https://youtu.be/obhWqDfzwQQ

Main Problem: https://www.youtube.com/watch?v=34l1kTIQCIA

Today's question is EASY and should be solvable :)
Hence, no upload today.

Count the Number of Powerful Integers:
https://youtu.be/pPm3Pf3oh5Q

No Video today (EASY problem)
2843. Count Symmetric Integers

CPP reference code:-
class Solution {
bool checkSymmetric(string no){
int size = no.size();
if(size&1)
return false;

int left_sum = 0;
int right_sum = 0;
for(int i=0;i<(size+1)/2;++i){
left_sum += no[i];
right_sum += no[size-i-1];
}
return left_sum==right_sum;
}
public:
int countSymmetricIntegers(int low, int high) {
int count = 0;
for(int no=low;no<=high;++no){
if(checkSymmetric(to_string(no)))
count++;
}
return count;
}
};

Find the Count of Good Integers:
https://youtu.be/dsHmPTHuIkA

Count Good Numbers:
https://youtu.be/ZhLyJpBVU1s