Computer organization and architecture questions

 UNIT–1 (6 MARK QUESTIONS)

1. Explain the generations of computers with features.

2. Draw and explain the functional block diagram of a computer.

3. Differentiate between hardware and software.

4. Explain machine level, assembly level and high-level languages.

5. What is a digital computer? Explain its characteristics.

6. Explain number systems used in computers.

7. Perform binary addition and binary subtraction with example.

8. Explain binary to decimal and decimal to binary conversion.

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✅ UNIT–2 (6 MARK QUESTIONS)

1. Explain basic logic gates with truth tables.

2. Explain NAND and NOR gates and why they are called universal gates.

3. Explain Boolean algebra laws and theorems.

4. What are minterms and maxterms? Explain with example.

5. Explain SOP and POS forms.

6. Explain Karnaugh Map (K-Map) and its advantages.

7. Explain Gray code and its applications.

8. Explain BCD and Excess-3 code.

9. Explain code conversion (Binary ↔ Gray or BCD ↔ Excess-3).

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✅ UNIT–3 (6 MARK QUESTIONS)

1. Explain Half Adder and Full Adder with logic diagram.

2. Explain 4-bit parallel adder.

3. Explain Half Subtractor and Full Subtractor.

4. Explain Multiplexer with diagram.

5. Explain Demultiplexer with diagram.

6. Explain Encoder and Decoder.

7. Explain SR flip-flop.

8. Explain JK flip-flop with truth table.

9. Explain D flip-flop and T flip-flop.

10. Explain Ripple counter.

11. Explain Ring counter.

12. What is state reduction and state assignment?

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✅ UNIT–4 (6 MARK QUESTIONS)

1. What is a microprocessor? Explain its basic elements.

2. Explain essential and non-essential elements of a microprocessor.

3. Explain addressing modes.

4. What are interrupts? Explain their types.

5. Explain the instruction cycle.

6. Differentiate between instruction cycle, machine cycle and T-state.

7. Explain pipelining.

8. What is superscalar pipeline?

9. Explain pipeline hazards.

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✅ UNIT–5 (6 MARK QUESTIONS)

1. Explain memory hierarchy with diagram.

2. Explain cache memory.

3. Explain direct mapping technique in cache memory.

4. Explain associative mapping in cache memory.

5. Explain memory mapped I/O.

6. Explain DMA (Direct Memory Access).

7. Explain memory interleaving.

8. Explain 2-D and 3-D memory organization.

9. Explain memory management.

10.

✅ Topics for Day 6 DSA with Java

Focus: Two Pointer Technique

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🔁 1. Learn & Implement Two Pointer Technique

• Understand how the two-pointer technique works on sorted arrays.

• Implement the following problems:

// Java Example: Two Sum (Sorted Array)
public int[] twoSum(int[] numbers, int target) {
    int left = 0, right = numbers.length - 1;
    while (left < right) {
        int sum = numbers[left] + numbers[right];
        if (sum == target) {
            return new int[] { left + 1, right + 1 }; // 1-indexed
        } else if (sum < target) {
            left++;
        } else {
            right--;
        }
    }
    return new int[] {}; // no solution
}

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🧠 2. Solve 2 LeetCode Problems Using Two Pointers

1. LeetCode 167. Two Sum II – Input Array Is Sorted

2. LeetCode 283. Move Zeroes

   • Optimize with minimal swaps using the two-pointer pattern.

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✨ Bonus Challenges (Optional but Recommended)

• LeetCode 26. Remove Duplicates from Sorted Array

• LeetCode 844. Backspace String Compare

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📊 Explore Visualization

Try two-pointer visualizations at:
👉 https://visualgo.net/en/list

---

🧩 Concept Summary

• Two pointers are most useful for:

  • Sorted arrays

  • Reverse traversal

  • Sliding window (upcoming)

  • Partitioning logic

✅ Day 5 Solution - DSA with Java

 

✅ 1. Implement Merge Sort and Quick Sort

🔹 Merge Sort

public class MergeSort {
    public static void mergeSort(int[] arr, int left, int right) {
        if (left < right) {
            int mid = (left + right) / 2;

            mergeSort(arr, left, mid);
            mergeSort(arr, mid + 1, right);

            merge(arr, left, mid, right);
        }
    }

    public static void merge(int[] arr, int left, int mid, int right) {
        int n1 = mid - left + 1, n2 = right - mid;
        int[] L = new int[n1];
        int[] R = new int[n2];

        for (int i = 0; i < n1; i++) L[i] = arr[left + i];
        for (int j = 0; j < n2; j++) R[j] = arr[mid + 1 + j];

        int i = 0, j = 0, k = left;
        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) arr[k++] = L[i++];
            else arr[k++] = R[j++];
        }

        while (i < n1) arr[k++] = L[i++];
        while (j < n2) arr[k++] = R[j++];
    }
}

---

🔹 Quick Sort

public class QuickSort {
    public static void quickSort(int[] arr, int low, int high) {
        if (low < high) {
            int pi = partition(arr, low, high);

            quickSort(arr, low, pi - 1);
            quickSort(arr, pi + 1, high);
        }
    }

    public static int partition(int[] arr, int low, int high) {
        int pivot = arr[high];
        int i = low - 1;

        for (int j = low; j < high; j++) {
            if (arr[j] < pivot) {
                i++;
                // Swap
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }

        // Final pivot swap
        int temp = arr[i + 1];
        arr[i + 1] = arr[high];
        arr[high] = temp;

        return i + 1;
    }
}

---

✅ 2. LeetCode Problems

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🔸 Merge Sorted Array

Merge nums2 into nums1 sorted in-place.

public class MergeSortedArray {
    public static void merge(int[] nums1, int m, int[] nums2, int n) {
        int i = m - 1, j = n - 1, k = m + n - 1;

        while (i >= 0 && j >= 0) {
            nums1[k--] = (nums1[i] > nums2[j]) ? nums1[i--] : nums2[j--];
        }

        while (j >= 0) {
            nums1[k--] = nums2[j--];
        }
    }
}

---

🔸 Sort an Array

Use any sorting method. Here's Merge Sort used:

public class SortAnArray {
    public static int[] sortArray(int[] nums) {
        mergeSort(nums, 0, nums.length - 1);
        return nums;
    }

    public static void mergeSort(int[] arr, int l, int r) {
        if (l < r) {
            int m = (l + r) / 2;
            mergeSort(arr, l, m);
            mergeSort(arr, m + 1, r);
            merge(arr, l, m, r);
        }
    }

    public static void merge(int[] arr, int l, int m, int r) {
        int n1 = m - l + 1, n2 = r - m;
        int[] L = new int[n1];
        int[] R = new int[n2];
        for (int i = 0; i < n1; ++i) L[i] = arr[l + i];
        for (int j = 0; j < n2; ++j) R[j] = arr[m + 1 + j];

        int i = 0, j = 0, k = l;
        while (i < n1 && j < n2)
            arr[k++] = (L[i] <= R[j]) ? L[i++] : R[j++];
        while (i < n1) arr[k++] = L[i++];
        while (j < n2) arr[k++] = R[j++];
    }
}

---

✅ 3. Non-Recursive Quick Sort (Iterative)

import java.util.Stack;

public class QuickSortIterative {
    public static void quickSort(int[] arr) {
        Stack<int[]> stack = new Stack<>();
        stack.push(new int[]{0, arr.length - 1});

        while (!stack.isEmpty()) {
            int[] range = stack.pop();
            int low = range[0], high = range[1];

            if (low < high) {
                int pi = partition(arr, low, high);
                stack.push(new int[]{low, pi - 1});
                stack.push(new int[]{pi + 1, high});
            }
        }
    }

    private static int partition(int[] arr, int low, int high) {
        int pivot = arr[high], i = low - 1;
        for (int j = low; j < high; j++) {
            if (arr[j] < pivot) {
                i++;
                // Swap
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
        // Final pivot placement
        int temp = arr[i + 1];
        arr[i + 1] = arr[high];
        arr[high] = temp;
        return i + 1;
    }
}

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✅ 4. Visualize Merge vs Quick Sort

You can interactively visualize how Merge and Quick Sort work at:

🔗 Visualgo.net – Sorting Visualizations

Click:

• Merge Sort → observe divide → merge steps

• Quick Sort → watch pivot partitioning → recursive reduction

💡 Use small array sizes (e.g. 10 elements) to step-through for learning.

🚀 Day 5: Merge Sort & Quick Sort in Java 🎯 Goals for Today

• Understand the Divide and Conquer strategy

• Learn and implement:

  • Merge Sort

  • Quick Sort

• Analyze Time & Space Complexities

• Solve intermediate-level sorting problems

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📘 1. Merge Sort

A stable, divide-and-conquer sorting algorithm.
Time Complexity: O(n log n) in all cases
Space Complexity: O(n) (extra array)

✅ Logic:

• Divide the array into two halves

• Sort each half recursively

• Merge the two sorted halves

🔹 Java Implementation

public class MergeSort {
    public static void mergeSort(int[] arr, int left, int right) {
        if (left < right) {
            int mid = (left + right) / 2;

            // Sort the left and right halves
            mergeSort(arr, left, mid);
            mergeSort(arr, mid + 1, right);

            // Merge them
            merge(arr, left, mid, right);
        }
    }

    public static void merge(int[] arr, int left, int mid, int right) {
        int n1 = mid - left + 1;
        int n2 = right - mid;

        int[] L = new int[n1];
        int[] R = new int[n2];

        for (int i = 0; i < n1; i++) L[i] = arr[left + i];
        for (int j = 0; j < n2; j++) R[j] = arr[mid + 1 + j];

        int i = 0, j = 0, k = left;

        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) arr[k++] = L[i++];
            else arr[k++] = R[j++];
        }

        while (i < n1) arr[k++] = L[i++];
        while (j < n2) arr[k++] = R[j++];
    }
}

---

⚡ 2. Quick Sort

Quick and efficient, especially on average cases.
Time Complexity:

• Best/Average: O(n log n)

• Worst: O(n²) (when pivot is worst chosen)
  Space: O(log n) (due to recursive calls)

✅ Logic:

• Choose a pivot

• Partition the array: elements < pivot go left, > pivot go right

• Recursively sort left and right

🔹 Java Implementation

public class QuickSort {
    public static void quickSort(int[] arr, int low, int high) {
        if (low < high) {
            int pi = partition(arr, low, high);

            quickSort(arr, low, pi - 1);
            quickSort(arr, pi + 1, high);
        }
    }

    public static int partition(int[] arr, int low, int high) {
        int pivot = arr[high];
        int i = low - 1; // Index of smaller element

        for (int j = low; j < high; j++) {
            if (arr[j] < pivot) {
                i++;
                // Swap
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }

        // Swap pivot to the right place
        int temp = arr[i + 1];
        arr[i + 1] = arr[high];
        arr[high] = temp;

        return i + 1;
    }
}

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📊 3. Merge Sort vs Quick Sort

Feature	Merge Sort	Quick Sort
Time (Best)	O(n log n)	O(n log n)
Time (Worst)	O(n log n)	O(n²)
Space	O(n)	O(log n)
Stable	✅ Yes	❌ No
In-place	❌ No	✅ Yes
Use Case	Linked lists	Arrays

🧠 4. Practice Problems

Try these problems:

1. ✅ Merge Sorted Array

2. ✅ Sort an Array

3. 🔁 Kth Largest Element in Array – Use Quick Select

4. 🧠 Count Inversions in Array – Use Merge Sort logic

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📚 Homework for Day 5

• Implement Merge Sort and Quick Sort

• Solve 2 LeetCode problems involving sorting or partitioning

• Try to write non-recursive quick sort (advanced)

• Visualize Merge vs Quick: Try https://visualgo.net/en/sorting

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🔜 Coming Up on Day 6:

• Recursion Mastery

• Backtracking intro (Subset, Permutations)

• Understanding call stack and dry runs

✅ Day 4 Solution - DSA with Java

 

✅ 1. Implement All Three Sorting Algorithms

🔹 Bubble Sort

public class BubbleSort {
    public static void bubbleSort(int[] arr) {
        int n = arr.length;
        for (int i = 0; i < n - 1; i++) {
            boolean swapped = false;
            for (int j = 0; j < n - i - 1; j++) {
                if (arr[j] > arr[j + 1]) {
                    // Swap
                    int temp = arr[j];
                    arr[j] = arr[j + 1];
                    arr[j + 1] = temp;
                    swapped = true;
                }
            }
            if (!swapped) break; // Optimization
        }
    }
}

---

🔹 Selection Sort

public class SelectionSort {
    public static void selectionSort(int[] arr) {
        int n = arr.length;
        for (int i = 0; i < n - 1; i++) {
            int minIdx = i;
            for (int j = i + 1; j < n; j++) {
                if (arr[j] < arr[minIdx]) {
                    minIdx = j;
                }
            }
            // Swap
            int temp = arr[i];
            arr[i] = arr[minIdx];
            arr[minIdx] = temp;
        }
    }
}

---

🔹 Insertion Sort

public class InsertionSort {
    public static void insertionSort(int[] arr) {
        for (int i = 1; i < arr.length; i++) {
            int key = arr[i];
            int j = i - 1;

            while (j >= 0 && arr[j] > key) {
                arr[j + 1] = arr[j];
                j--;
            }
            arr[j + 1] = key;
        }
    }
}

---

✅ 2. LeetCode Problems

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🔸 Problem 1: Sort Colors

Sort an array containing only 0s, 1s, and 2s.
Use Dutch National Flag Algorithm (O(n) time, O(1) space)

public class SortColors {
    public static void sortColors(int[] nums) {
        int low = 0, mid = 0, high = nums.length - 1;

        while (mid <= high) {
            if (nums[mid] == 0) {
                // Swap with low
                int temp = nums[low];
                nums[low] = nums[mid];
                nums[mid] = temp;
                low++;
                mid++;
            } else if (nums[mid] == 1) {
                mid++;
            } else {
                // nums[mid] == 2
                int temp = nums[mid];
                nums[mid] = nums[high];
                nums[high] = temp;
                high--;
            }
        }
    }
}

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🔸 Problem 2: Check if Array Is Sorted and Rotated

Return true if array is sorted and rotated at most once.

🔍 Logic:

Count how many times arr[i] > arr[i+1]. If it happens more than once, it's not valid.

public class CheckSortedRotated {
    public static boolean check(int[] nums) {
        int count = 0;
        int n = nums.length;

        for (int i = 0; i < n; i++) {
            if (nums[i] > nums[(i + 1) % n]) {
                count++;
            }
        }

        return count <= 1;
    }

    public static void main(String[] args) {
        int[] arr = {3, 4, 5, 1, 2};
        System.out.println("Is Sorted and Rotated? " + check(arr)); // true
    }
}

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🧠 Summary of Day 4

• ✅ Bubble, Selection, Insertion Sort implemented

• ✅ LeetCode problems solved

  • Dutch National Flag (0s, 1s, 2s)

  • Check Sorted & Rotated

✅ Day 4: Sorting Algorithms (Basic to Intermediate)

🎯 Goals for Today

• Understand how sorting works and why it’s needed

• Learn and implement:

  • Bubble Sort

  • Selection Sort

  • Insertion Sort

• Analyze time & space complexities

• Practice beginner sorting problems

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📘 1. Why Sorting Matters

Sorting helps:

• Optimize searching (e.g., binary search)

• Solve problems like duplicates, frequency, etc.

• Prepare data for algorithms like two pointers, greedy, divide & conquer, etc.

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🔁 2. Sorting Algorithms

🔹 Bubble Sort

Repeatedly swap adjacent elements if they are in the wrong order.

public class BubbleSort {
    public static void bubbleSort(int[] arr) {
        int n = arr.length;
        for (int i = 0; i < n - 1; i++) {
            boolean swapped = false;
            for (int j = 0; j < n - i - 1; j++) {
                if (arr[j] > arr[j + 1]) {
                    // swap
                    int temp = arr[j];
                    arr[j] = arr[j + 1];
                    arr[j + 1] = temp;
                    swapped = true;
                }
            }
            if (!swapped) break;
        }
    }
}

• Time Complexity: Worst → O(n²), Best → O(n) (already sorted)

• Space: O(1)

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🔹 Selection Sort

Find the minimum element and put it at the beginning.

public class SelectionSort {
    public static void selectionSort(int[] arr) {
        int n = arr.length;
        for (int i = 0; i < n - 1; i++) {
            int minIndex = i;
            for (int j = i + 1; j < n; j++) {
                if (arr[j] < arr[minIndex]) {
                    minIndex = j;
                }
            }
            // swap
            int temp = arr[i];
            arr[i] = arr[minIndex];
            arr[minIndex] = temp;
        }
    }
}

• Time Complexity: Always O(n²)

• Space: O(1)

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🔹 Insertion Sort

Build the sorted array one element at a time by inserting elements at the correct position.

public class InsertionSort {
    public static void insertionSort(int[] arr) {
        for (int i = 1; i < arr.length; i++) {
            int current = arr[i];
            int j = i - 1;

            while (j >= 0 && arr[j] > current) {
                arr[j + 1] = arr[j];
                j--;
            }
            arr[j + 1] = current;
        }
    }
}

• Time Complexity: Best → O(n), Worst → O(n²)

• Space: O(1)

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🧠 3. Sorting Algorithm Comparison Table

Algorithm	Time Complexity (Best / Avg / Worst)	Space	Stable?
Bubble Sort	O(n) / O(n²) / O(n²)	O(1)	✅
Selection Sort	O(n²) / O(n²) / O(n²)	O(1)	❌
Insertion Sort	O(n) / O(n²) / O(n²)	O(1)	✅

🔍 4. Practice Problems

Try these:

• Sort an array of 0s, 1s, and 2s (Dutch National Flag)

• Check if an array is sorted and rotated

• Count the number of swaps required to sort the array

• Sort a nearly sorted (k-sorted) array

You can also implement a custom comparator or reverse order sort (forward-thinking Java practice).

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📚 Homework for Day 4

✅ Implement all three sorting algorithms
✅ Solve 2 sorting-based problems from LeetCode:

• Sort Colors

• Check if Array Is Sorted and Rotated

Would you like solutions for these 2 LeetCode problems?

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⏭️ What’s Coming on Day 5:

• Advanced Sorting: Merge Sort, Quick Sort

• Time Complexity Analysis

• Recursion involved in sorting

✅ Solution Implement Both Linear and Binary Search Day 3 DSA with Java

🔹 Linear Search in Java

public class LinearSearch {
    public static int linearSearch(int[] arr, int key) {
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] == key) return i;
        }
        return -1; // Not found
    }

    public static void main(String[] args) {
        int[] arr = {5, 3, 8, 2, 9};
        int key = 8;
        int index = linearSearch(arr, key);
        System.out.println("Element found at index: " + index);
    }
}

---

🔹 Binary Search in Java (Iterative)

public class BinarySearch {
    public static int binarySearch(int[] arr, int key) {
        int start = 0, end = arr.length - 1;
        while (start <= end) {
            int mid = start + (end - start) / 2;
            if (arr[mid] == key) return mid;
            else if (arr[mid] < key) start = mid + 1;
            else end = mid - 1;
        }
        return -1;
    }

    public static void main(String[] args) {
        int[] arr = {1, 3, 5, 7, 9};
        int key = 5;
        int index = binarySearch(arr, key);
        System.out.println("Element found at index: " + index);
    }
}

---

✅ 2. LeetCode Problems

🔸 Search Insert Position

Problem: Return the index if the target is found. If not, return the index where it would be inserted.

public class SearchInsertPosition {
    public static int searchInsert(int[] nums, int target) {
        int low = 0, high = nums.length - 1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (nums[mid] == target) return mid;
            else if (nums[mid] < target) low = mid + 1;
            else high = mid - 1;
        }
        return low;
    }

    public static void main(String[] args) {
        int[] nums = {1, 3, 5, 6};
        int target = 2;
        System.out.println("Insert position: " + searchInsert(nums, target));
    }
}

---

🔸 First Bad Version

Assume isBadVersion(version) API exists.

public class FirstBadVersion {
    static int bad = 4; // Let's assume version 4 is the first bad version

    public static boolean isBadVersion(int version) {
        return version >= bad;
    }

    public static int firstBadVersion(int n) {
        int low = 1, high = n;
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (isBadVersion(mid)) high = mid;
            else low = mid + 1;
        }
        return low;
    }

    public static void main(String[] args) {
        System.out.println("First bad version: " + firstBadVersion(10));
    }
}

---

🌱 3. Explore: Lower Bound & Upper Bound

These are binary search variants often used in competitive programming.

🔸 Lower Bound (First element ≥ target)

public static int lowerBound(int[] arr, int target) {
    int low = 0, high = arr.length;
    while (low < high) {
        int mid = low + (high - low) / 2;
        if (arr[mid] < target) low = mid + 1;
        else high = mid;
    }
    return low; // returns the index
}

🔸 Upper Bound (First element > target)

public static int upperBound(int[] arr, int target) {
    int low = 0, high = arr.length;
    while (low < high) {
        int mid = low + (high - low) / 2;
        if (arr[mid] <= target) low = mid + 1;
        else high = mid;
    }
    return low; // returns the index
}

You can test both with:

int[] arr = {1, 3, 3, 5, 7};
System.out.println("Lower Bound of 3: " + lowerBound(arr, 3)); // Output: 1
System.out.println("Upper Bound of 3: " + upperBound(arr, 3)); // Output: 3

🚀 Day 3: Searching Algorithms in Java

🎯 Goals:

• Understand Linear Search and Binary Search

• Learn when and how to use each

• Solve real interview problems using searching techniques

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🔍 1. Linear Search

📘 Definition:

Linearly checks each element one by one. Used when the array is unsorted.

✅ Code Example:

public static int linearSearch(int[] arr, int key) {
    for (int i = 0; i < arr.length; i++) {
        if (arr[i] == key) return i;
    }
    return -1; // Not found
}

---

🧠 2. Binary Search

📘 Definition:

Used on sorted arrays. It divides the search space in half each time — O(log N) time complexity.

✅ Code (Iterative Approach):

public static int binarySearch(int[] arr, int key) {
    int start = 0, end = arr.length - 1;
    while (start <= end) {
        int mid = start + (end - start) / 2;

        if (arr[mid] == key) return mid;
        else if (arr[mid] < key) start = mid + 1;
        else end = mid - 1;
    }
    return -1;
}

✅ Code (Recursive Approach):

public static int binarySearchRec(int[] arr, int key, int start, int end) {
    if (start > end) return -1;

    int mid = start + (end - start) / 2;
    if (arr[mid] == key) return mid;
    else if (arr[mid] < key) return binarySearchRec(arr, key, mid + 1, end);
    else return binarySearchRec(arr, key, start, mid - 1);
}

---

💡 3. Binary Search Use Cases

• Search in sorted array

• Find first/last occurrence of an element

• Search in rotated sorted arrays

• Peak elements in mountain arrays

---

✍️ 4. Practice Problems

🔹 Write functions to:

• Find first and last occurrence of an element (e.g. in sorted duplicates)

• Count how many times a number appears in sorted array

• Search in a rotated sorted array

• Find square root using binary search (approximate to floor)

---

🔥 5. BONUS – Search in 2D Matrix

🔹 Use binary search logic on 2D:

public static boolean searchMatrix(int[][] matrix, int target) {
    int rows = matrix.length;
    int cols = matrix[0].length;
    int low = 0, high = rows * cols - 1;

    while (low <= high) {
        int mid = low + (high - low) / 2;
        int value = matrix[mid / cols][mid % cols];

        if (value == target) return true;
        else if (value < target) low = mid + 1;
        else high = mid - 1;
    }
    return false;
}

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📚 Homework for Day 3:

• Implement both linear and binary search

• Solve 2 problems from LeetCode:

  • Search Insert Position

  • First Bad Version

• Explore: Lower Bound, Upper Bound (optional but forward-thinking)

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⏭️ What’s Coming on Day 4?

• Sorting Algorithms: Bubble, Selection, Insertion (with time/space analysis)