📡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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