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
LIVE interactive session
Limited seats only (registration will close soon)
Date: 16th April
TIME: 8 PM IST / 7:30 AM PST
Duration: 1 hour approx
Fee: 99 INR / 10 USD
CONTACT US (Whatsapp): +91 8918633037
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
LIVE interactive session
Limited seats only (registration will close soon)
Date: 16th April
TIME: 8 PM IST / 7:30 AM PST
Duration: 1 hour approx
Fee: 99 INR / 10 USD
CONTACT US (Whatsapp): +91 8918633037
Count Good Triplets in an Array | Leetcode 2179 :
https://youtu.be/QBhfiHswL1Y
https://youtu.be/QBhfiHswL1Y
YouTube
Count Good Triplets in an Array | Leetcode 2179 | Segment Tree
This video explains Count Good Triplets in an Array using the most optimal segment tree hashmap approach.
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Count and Say:
https://youtu.be/htSwmFGdFUI
https://youtu.be/htSwmFGdFUI
YouTube
Count and Say | Leetcode 38
This video explains Count and Say using the simulation approach.
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🟣 JOIN our 𝐋𝐈𝐕𝐄 𝐢𝐧𝐭𝐞𝐫𝐯𝐢𝐞𝐰 𝐭𝐫𝐚𝐢𝐧𝐢𝐧𝐠…
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🟣 JOIN our 𝐋𝐈𝐕𝐄 𝐢𝐧𝐭𝐞𝐫𝐯𝐢𝐞𝐰 𝐭𝐫𝐚𝐢𝐧𝐢𝐧𝐠…
Count the Number of Fair Pairs:
https://youtu.be/xe6y4aEfvDw
https://youtu.be/xe6y4aEfvDw
YouTube
Count the Number of Fair Pairs | Leetcode 2563
This video explains count the number of fair pairs using the most optimal binary search technique using lower bound and upper bound.
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Rabbits in Forest:
https://youtu.be/Do2coBlSIFo
https://youtu.be/Do2coBlSIFo
YouTube
Rabbits in Forest | Leetcode 781
This video explains Rabbits in Forest using the most optimal group counting approach.
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🟣 JOIN our…
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🟣 JOIN our…
Count the Hidden Sequences:
https://youtu.be/z0cdra6jHs4
https://youtu.be/z0cdra6jHs4
YouTube
Count the Hidden Sequences | Leetcode 2145
This video explains Count the Hidden Sequences using the most optimal single parse counting solution.
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Count the Number of Ideal Arrays:
https://youtu.be/FGf2EPyxcWg
https://youtu.be/FGf2EPyxcWg
YouTube
Count the Number of Ideal Arrays | Leetcode 2338
This video explains Count the Number of Ideal Arrays
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🟣 JOIN our 𝐋𝐈𝐕𝐄 𝐢𝐧𝐭𝐞𝐫𝐯𝐢𝐞𝐰 𝐭𝐫𝐚𝐢𝐧𝐢𝐧𝐠 𝐩𝐫𝐨𝐠𝐫𝐚𝐦…
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🟣 JOIN our 𝐋𝐈𝐕𝐄 𝐢𝐧𝐭𝐞𝐫𝐯𝐢𝐞𝐰 𝐭𝐫𝐚𝐢𝐧𝐢𝐧𝐠 𝐩𝐫𝐨𝐠𝐫𝐚𝐦…
Which architecture would you choose for read-heavy system ?
Anonymous Quiz
29%
Multi-Master Slave Architecture
51%
Single-Master Multi-Slave Architecture
18%
P2P Architecture
2%
Pipe-Filter Architecture
Count Complete Subarrays in an array:
https://youtu.be/Taj4VhCycRU
https://youtu.be/Taj4VhCycRU
YouTube
Count Complete Subarrays in an Array | Leetcode 2799
This video explains Count Complete Subarrays in an Array using the most optimal sliding window two pointer approach.
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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;
}
};
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
https://youtu.be/kGTqCOdXrhE
YouTube
Find Minimum Time to Reach Last Room 1 & 2 | Leetcode 3341 and 3342
This video explains the minimum time to reach last room problem which is from leetcode 3341 and Leetcode 3342 using the most optimal dijkstra algorithm.
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Find Minimum Time to Reach Last Room 1 & 2:
https://youtu.be/kGTqCOdXrhE
https://youtu.be/kGTqCOdXrhE
YouTube
Find Minimum Time to Reach Last Room 1 & 2 | Leetcode 3341 and 3342
This video explains the minimum time to reach last room problem which is from leetcode 3341 and Leetcode 3342 using the most optimal dijkstra algorithm.
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Count Number of Balanced Permutations:
https://youtu.be/bNnXXa2a-Jo
https://youtu.be/bNnXXa2a-Jo
YouTube
Count Number of Balanced Permutations | Leetcode 3343
This video explains Count Number of Balanced Permutations using the most optimal dynamic programming recursion with memoization technique.
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