🚨 A2SV G6 Remote Education Recruitment Update 🚨

It’s been an incredible journey since we began interviewing candidates for G6 Remote Education. After posting the application form, we were overwhelmed by the sheer number of applications! 🎉

We’re committed to reviewing every application fairly. Due to limited interview slots, only applicants with the highest grades will proceed to the interview stage, and invitations will be sent out soon.

For those who have already completed their interviews, your interviews are being graded. The final results for all applicants will be released in three weeks, once the entire process is concluded.

Thank you for your patience and the effort you’ve put into this journey. 🙌

#A2SV #RemoteEducation #RecruitmentUpdate

💻 A2SV prep: Two pointers

🟢 1995. Count Special Quadruplets

Solution: This is a good example of questions with a small amount of input, or constraints that can be solved using a simple brute force.

class Solution:
    def countQuadruplets(self, nums: List[int]) -> List[List[int]]:
        res = 0 
        n = len(nums)
        for i in range(n-3):
            for j in range(i+1, n-2):
                for k in range(j +1, n-1):
                    for l in range(k +1,n):
                        if nums[i] + nums[k] + nums[j] - nums[l] == 0:
                            res += 1
        return res

🟡 454. 4Sum II

Intuition : from the first two arrays find the count of sum of paris in nums1 and nums2 next get the compliment from the last two arrays nums3 and nums4

class Solution:
    def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
        sum_count = Counter(a + b for a in nums1 for b in nums2)
        count = 0
        for c in nums3:
            for d in nums4:
                complement = -(c + d)
                count += sum_count.get(complement, 0)
        
        return count

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💻 A2SV prep: Sliding window problem

🟡 1248. Count Number of Nice Subarrays

Intuition: we want to find the count of the subarrays with exactly k odd numbers so our window should have exactly k odd numbers. To get the exact count we can just count the number of sub arrays with at most k odd numbers and subtract the count of sub arrays with at most k - 1 odd numbers which will give us the count of substrings with exact k odd numbers in it.

class Solution:
    def numberOfSubarrays(self, nums: List[int], k: int) -> int:
        def atMost(k):
            count = 0
            L, R = 0, 0
            for R in range(len(nums)):
                if nums[R] % 2 != 0:
                    k -= 1
                while k < 0:
                    if nums[L] % 2 != 0:
                        k += 1
                    L += 1
                count += R - L + 1
            return count
            
        return atMost(k) - atMost(k-1)

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💻 A2SV prep: Sliding window problem

🟡 3. Longest Substring Without Repeating Characters

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        maxlen = 0
        L = R = 0
        hashset = set()
        while R < len(s):
            if s[R] in hashset:
                maxlen = max(maxlen, R - L) # maxlen = max(maxlen, len(hashset))
                while s[L] != s[R]:
                    hashset.remove(s[L])
                    L += 1
                L += 1

            hashset.add(s[R])
            R += 1
            maxlen = max(maxlen, R - L)
        
        return maxlen

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📢 Announcement: A2SV G6 In-Person Application Update

Dear Applicants,

Thank you for submitting your applications for the A2SV G6 recruitment process. After carefully reviewing all submissions, we are excited to announce that the shortlist of candidates selected for interviews will be shared very soon! 🎉 Please check your emails regularly for updates. 📧

Starting Monday, shortlisted candidates will be able to schedule their technical interviews. 📅

We highly recommend that you begin preparing for the technical interview to ensure your success. 🚀 The key topics to focus on include:
- Sorting
- Two Pointers
- Prefix Sum
- Sliding Window
- Stack & Queues

Best of luck to all applicants, and we look forward to meeting you during the interview process! 💪

Sincerely,
The A2SV Team

Here are the resources for the topics covered so far. We've also linked them in the Resources tab of the Community Sheet

• Python Learning Path
• Time and Space Complexity Learning Path
• Sorting Learning Path
• Stack and Queue Learning Path
• Recursion Learning Path
• LinkedList Learning Path
• Sliding Window and Two Pointers Learning Path
• Prefix Sum Learning Path
• Heaps Learning Path

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📢 Important Update for A2SV G6 In person Education Cohort!

If you've received an email from A2SV, congratulations! 🎉 You’ve been shortlisted for an interview for the A2SV G6 Cohort!

🗓 Interviews Begin: on Tuesday
📌 Focus Areas: Make sure to prepare thoroughly on the topics mentioned in the email.

If you haven’t checked your email yet, go check it now and start your preparation! 📨

🚀 @AceCoding Presents! 🚀

🧠 Python Quiz: Predict the Output

arr = [[0]] * 5  
print(arr)  
arr[0][0] = 1  
print(arr)  

What will the output be? You can choose your answer below.

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