💻 A2SV prep: Two pointers

🟢 Two sum
🟡 Two sum ll input array is sorted

👆The above two are easy; warm up

🟡 3sum Click here

Solution:

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        res = []
        nums.sort()
        n = len(nums)
        for i in range(n-2):
            if nums[i] > 0:
                break
            if i > 0 and nums[i] == nums[i-1]:
                continue
            
            L, R = i + 1, n - 1
            while L < R:
                sum = nums[i] + nums[L] + nums[R]
                if sum == 0:
                    res.append([nums[i], nums[L], nums[R]])
                    L +=1 
                    R -=1
                    while L < R and nums[L] == nums[L-1]:
                        L += 1
                    while L < R and nums[R] == nums[R+1]:
                        R -= 1
                    
                elif sum > 0:
                    R -= 1
                else:
                    L += 1
    
        return res

🟡 3sum closest :- click here

Solution:

python 
class Solution:
    def threeSumClosest(self, nums: List[int], target: int) -> int:
        nums.sort()
        res = sum(nums[:3])
        for i in range(len(nums)):
            L, R = i+1, len(nums)-1

            while L < R:
                closest = nums[i] + nums[L] + nums[R]
                if abs(target - closest) < abs(target - res):
                    res = closest
                
                if closest > target:
                    R -= 1
                elif closest < target:
                    L += 1
                else:
                    return closest
        
        return res

🟡 4Sum :- click here

Solution:

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        nums.sort()
        n = len(nums)
        res = []
        
        for i in range(n-3):
            if i > 0 and nums[i] == nums[i-1]:
                continue

            for j in range(i+1, n-2):
                if j > i+1 and nums[j] == nums[j-1]:
                    continue

                L, R = j + 1, n - 1
                while L < R:
                    four_sum = nums[i] + nums[j] + nums[L] + nums[R]
                    if four_sum == target:
                        res.append([nums[i], nums[j], nums[L], nums[R]])
                    
                        L += 1
                        R -= 1
                        while L < R and nums[L] == nums[L-1]:
                            L += 1
                        while L < R and nums[R] == nums[R+1]:
                            R -= 1

                    elif four_sum > target:
                        R -= 1
                    else:
                        L += 1

        return res

🌟🚀 @AceCoding Presents! 🚀🌟

✅ 100% Yesified 😂

Theory vs Practice

@AceCoding

🚨 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

🌟🚀 @AceCoding Presents! 🚀🌟

💻 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)

🌟🚀 @AceCoding Presents! 🚀🌟

🔥 This 3D website is crazy. check it out.

⚒️Built with React.js and Three.js

👉 https://bfcm.shopify.com

@AceCoding

💻 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

🌟🚀 @AceCoding Presents! 🚀🌟

📢 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