🚨 Last call the deadline is today

💻A2SV prep: Sliding window problem

🟡 2024. Maximize the Confusion of an Exam

Straight forward solution

class Solution:
    def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:
        n = len(answerKey)
        max_length = 0

        flips = k
        l, r = 0, 0
        # F -> T
        while r < n:
            if answerKey[r] == 'F':
                flips -= 1
            if flips < 0:
                if answerKey[l] == 'F':
                    flips += 1
                l += 1

            max_length = max(max_length, r - l + 1)
            r += 1

        flips = k
        l, r = 0, 0
        # T -> F
        while r < n:
            if answerKey[r] == 'T':
                flips -= 1
            if flips < 0:
                if answerKey[l] == 'T':
                    flips += 1
                l += 1

            max_length = max(max_length, r - l + 1)
            r += 1

        return max_length

As you have noticed there are some repetions in the code so I have modularized it into one single function that can calculate the possible maximum length.

class Solution:
    def maxConsecutiveAnswers(self, answerKey: str, k: int) -> int:

        def calculateLength(ans: str):
            flips = k 
            max_length = 0

            n = len(answerKey)
            l, r = 0, 0

            while r < n:
                if answerKey[r] == ans:
                    flips -= 1
                if flips < 0:
                    if answerKey[l] == ans:
                        flips += 1
                    l += 1

                max_length = max(max_length, r - l + 1)
                r += 1

            return max_length

        return max(calculateLength('T'), calculateLength('F'))

https://leetcode.com/problems/maximize-the-confusion-of-an-exam/description/

📣 A2SV Interview Format 📣

Get ready for an exciting and comprehensive interview process! Here’s what to expect:

🕒 Total Duration: 1 Hour 30 Minutes

💎 Behavioral Interview (First 45 Minutes)
This part of the interview will focus on your past experiences, skills, and motivations.
Be prepared to discuss situations where you demonstrated leadership, teamwork, and problem-solving abilities.
They might ask questions like:
⭐️ Tell me about yourself.
⭐️ What do you know about A2SV and why do you want to join us?
⭐️ Describe a time when you had to step out of your comfort zone to achieve something.

💎 Technical Interview (Remaining 45 Minutes)
This segment will test your coding and problem-solving skills.
You’ll be given programming challenges to solve in real-time.
Much of this interview will utilize sharepad.io as the collaborative coding editor.
Example questions might include:
Write a function to solve a particular algorithm problem.
Explain your approach to solving a complex data structure challenge.

🔧 Preparation Tips:

💎 Practice Coding: Brush up on your data structures and algorithms. Websites like LeetCode and codeforce are great for practice.

💎 Stay Calm and Confident: Remember, they’re not just looking for the right answers but also how you approach problems and handle stress.


Prepare well and best of luck to everyone!🧨🚀

✅ Almost Identical question with the above 👆very easy if you did the above

💻A2SV prep: Sliding window problem
🟡 1004. Max Consecutive Ones III

⏳2 min on the clock PR for me 😅

class Solution:
    def longestOnes(self, nums: List[int], k: int) -> int:
        max_length = 0
        l = 0
        for r in range(len(nums)):
            if nums[r] == 0:
                k -= 1
            while k < 0:
                if nums[l] == 0:
                    k += 1
                l += 1

            max_length = max(max_length, r - l + 1)

        return max_length

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https://leetcode.com/problems/max-consecutive-ones-iii/description/

👨‍💻 PROGRESS ☕️

📡COA - Computer Organization and Architecture

Hardwired vs. Microprogrammed (Softwired) Logic

✅ The Teacher was stressing this topic (💰likely to come up on the exam)

Hardwired Logic:
Functionality is embedded directly into the hardware using fixed circuits. These systems are fast and efficient but lack flexibility, as changes require hardware redesign. This is common in specific-purpose computers, like appliances or embedded systems optimized for specific tasks (e.g., calculators).

Microprogrammed (Softwired) Logic:
Here, functionality is defined by software, making these systems flexible and adaptable but slightly slower due to software execution overhead. Softwired logic is a hallmark of general-purpose computers, like laptops or servers, which can handle multiple tasks by running different programs.

Takeaway:
Hardwired logic is best for specialized, high-speed tasks, while softwired logic is ideal for versatile, multi-functional systems.

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COA questions:

part 1 👉 click here
part 2 👉 click here

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https://youtu.be/IrUHoXLx0G4?si=oiQZ5-lnf5iLNw9V

💻 A2SV prep: Sliding window problem

✅ The following questions should be done sequentially. Level 1 and Level 2

🟢 3206. Alternating Groups I Easy
🔗Link: Click here!

python 
class Solution:
    def numberOfAlternatingGroups(self, colors: List[int]) -> int:
        count = 0
        # check the end colors 
        if len(colors) > 2:
            if colors[0] != colors[1] and colors[1] == colors[-1]:
                count += 1
            if colors[-2] != colors[-1] and colors[-2] == colors[0]:
                count += 1
        
        l, r = 0, 0
        k = 3
        for r in range(len(colors)):
            if r - l + 1 == 3:
                l += 1
            if r < len(colors) - 1 and colors[l] == colors[r+1] and colors[l] != colors[r]:
                count += 1

        return count

📗 A more optimized version

class Solution:
    def numberOfAlternatingGroups(self, colors: List[int]) -> int:
        n = len(colors) # keep the orginal length for latter
        count = 0

        # extend the colors array to simulate a circular pattern 
        colors.extend(colors[:2])

        for r in range(n): 
            if colors[r] == colors[r+2] and colors[r] != colors[r+1]:
                count += 1
                
        return count

🟡 3208. Alternating Groups II ~ 1800 ( Hard medium )
🔗 Link: Click here!

python 
class Solution:
    def numberOfAlternatingGroups(self, colors: List[int], k: int) -> int:
        count = 0

        colors.extend(colors[:k-1]) # if k = 3 we add 2 numbers like the above question
        print(colors)

        L = 0
        for R in range(1, len(colors)):
            
            if colors[R] == colors[R-1]:
                L = R
            if R - L + 1 == k:
                count += 1
                L += 1

        return count

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✅ Explanation for 3208. Alternating Groups II

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💻 A2SV prep: Sliding window problem

🟡 3191. Minimum Operations to Make Binary Array Elements Equal to One I
🔗 Link: Click here

class Solution:
    def minOperations(self, nums: List[int]) -> int:
        n = len(nums)
        count = L =0
        for R in range(n - 2): 
            if nums[R] == 0:  
                nums[R] = 1 - nums[R]
                nums[R + 1] = 1 - nums[R + 1]
                nums[R + 2] = 1 - nums[R + 2]
                count += 1

            if nums[L] == 1: L+= 1
        
        for i in range(len(nums)):
            if nums[i] == 0:
                return -1
        return count

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🔥 Gear Up for Java Exam! QUIZ Time 🔥

Get ready for a detailed and comprehensive Java prep with tricky questions.

💡 Don’t miss out—share with friends and tackle it together! 🚀

#javaexam #oopexam #oopquiz #javaquiz #oopmidexam #javamidexam

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