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

Interview Preparation strategy for META
1. Talk directly with Software Engineer at META
2. Get the study plan
3. Learn the challenges she faced and the effort she put on a daily basis
4. Know the preparation timeline and what worked for her
5. Get to know the difficulty level in the entire interview process for both DSA & System Design
6. Validate your approach for any future interview offer
7. Get to know the common interview preparation pitfalls for META
8. Know how the interview at META is different from other MAANG companies
9. Ask me anything is included with the session

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Count Good Triplets in an Array | Leetcode 2179 :
https://youtu.be/QBhfiHswL1Y

Count and Say:
https://youtu.be/htSwmFGdFUI

Count the Number of Fair Pairs:
https://youtu.be/xe6y4aEfvDw

Rabbits in Forest:
https://youtu.be/Do2coBlSIFo

Count the Hidden Sequences:
https://youtu.be/z0cdra6jHs4

Count the Number of Ideal Arrays:
https://youtu.be/FGf2EPyxcWg

No video today for EASY question :)

Count Complete Subarrays in an array:
https://youtu.be/Taj4VhCycRU

Today's problem is same as yesterday's and many last few problems.
Hence, not making video today.

REFERENCE CODE:-
class Solution {
#define ll long long
public:
long long countSubarrays(vector<int>& nums, int k) {
ll valid_subarrays = 0;
int win_count = 0;
int left = 0,right = 0;

int maximum = *max_element(nums.begin(), nums.end());
int n = nums.size();
while(left<n){
while(right<n and win_count<k){
if(nums[right]==maximum)
win_count++;
right++;
}

if(win_count==k)
valid_subarrays += n-right+1;

if(nums[left]==maximum)
win_count--;
left++;
}
return valid_subarrays;
}
};

Find Minimum Time to Reach Last Room 1 & 2 :
https://youtu.be/kGTqCOdXrhE

Find Minimum Time to Reach Last Room 1 & 2:
https://youtu.be/kGTqCOdXrhE

Count Number of Balanced Permutations:
https://youtu.be/bNnXXa2a-Jo