Merge Two Sorted Arrays Without Extra Space in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of merging two sorted arrays without extra space

What is the output of the following Java code that merges two sorted arrays without using extra space?

DSA Java

public class MergeSortedArrays {
    public static void merge(int[] arr1, int[] arr2) {
        int n = arr1.length;
        int m = arr2.length;
        int i = 0, j = 0, k = n - 1;
        while (i <= k && j < m) {
            if (arr1[i] < arr2[j]) {
                i++;
            } else {
                int temp = arr2[j];
                arr2[j] = arr1[k];
                arr1[k] = temp;
                j++;
                k--;
            }
        }
        java.util.Arrays.sort(arr1);
        java.util.Arrays.sort(arr2);
    }

    public static void main(String[] args) {
        int[] arr1 = {1, 5, 9, 10, 15, 20};
        int[] arr2 = {2, 3, 8, 13};
        merge(arr1, arr2);
        System.out.print(java.util.Arrays.toString(arr1) + " ");
        System.out.print(java.util.Arrays.toString(arr2));
    }
}

A [1, 2, 3, 5, 8, 9] [10, 13, 15, 20]

B [1, 3, 5, 8, 9, 10] [2, 13, 15, 20]

C [1, 2, 3, 5, 8, 9] [10, 15, 20, 13]

D [1, 2, 3, 5, 9, 10] [8, 13, 15, 20]

💡 Hint

Focus on how elements are swapped between the end of the first array and the start of the second array, then both arrays are sorted.

The code swaps elements from the end of the first array with smaller elements from the second array, then sorts both arrays. This results in the first array having the smallest elements and the second array having the larger elements, both sorted.

🧠 Conceptual

intermediate

Understanding the key idea behind merging without extra space

What is the main idea behind merging two sorted arrays without using extra space?

A Use a linked list to merge elements in sorted order.

B Copy all elements into a new array and sort it.

C Swap elements between arrays to ensure smaller elements are in the first array, then sort both arrays.

D Sort both arrays independently without swapping elements.

💡 Hint

Think about how to avoid extra memory by rearranging elements inside the given arrays.

The approach swaps elements between the two arrays so that the first array contains smaller elements and the second array contains larger elements, then sorts both arrays to complete the merge without extra space.

🔧 Debug

advanced

Identify the error in this merge implementation

What error will occur if we replace the line 'while (i <= k && j < m)' with 'while (i < k && j <= m)' in the merge method?

DSA Java

public static void merge(int[] arr1, int[] arr2) {
    int n = arr1.length;
    int m = arr2.length;
    int i = 0, j = 0, k = n - 1;
    while (i < k && j <= m) {
        if (arr1[i] < arr2[j]) {
            i++;
        } else {
            int temp = arr2[j];
            arr2[j] = arr1[k];
            arr1[k] = temp;
            j++;
            k--;
        }
    }
    java.util.Arrays.sort(arr1);
    java.util.Arrays.sort(arr2);
}

A NullPointerException

B No error, code runs correctly

C Infinite loop

D ArrayIndexOutOfBoundsException

💡 Hint

Check the loop condition and array index usage carefully.

The condition 'j <= m' allows j to become equal to m, which is out of bounds for arr2 (indices go from 0 to m-1), causing ArrayIndexOutOfBoundsException.

❓ Predict Output

advanced

Result after merging arrays with duplicates

What is the output of the following code after merging two sorted arrays with duplicate elements without extra space?

DSA Java

public class MergeSortedArrays {
    public static void merge(int[] arr1, int[] arr2) {
        int n = arr1.length;
        int m = arr2.length;
        int i = 0, j = 0, k = n - 1;
        while (i <= k && j < m) {
            if (arr1[i] < arr2[j]) {
                i++;
            } else {
                int temp = arr2[j];
                arr2[j] = arr1[k];
                arr1[k] = temp;
                j++;
                k--;
            }
        }
        java.util.Arrays.sort(arr1);
        java.util.Arrays.sort(arr2);
    }

    public static void main(String[] args) {
        int[] arr1 = {1, 3, 5, 7, 9};
        int[] arr2 = {3, 5, 7, 11, 13};
        merge(arr1, arr2);
        System.out.print(java.util.Arrays.toString(arr1) + " ");
        System.out.print(java.util.Arrays.toString(arr2));
    }
}

A [1, 3, 5, 7, 7] [3, 5, 9, 11, 13]

B [1, 3, 3, 5, 5] [7, 7, 9, 11, 13]

C [1, 3, 3, 5, 7] [5, 7, 9, 11, 13]

D [1, 3, 5, 7, 9] [3, 5, 7, 11, 13]

💡 Hint

Duplicates are swapped and sorted just like other elements.

After swapping and sorting, duplicates are distributed between the two arrays maintaining sorted order without extra space.

🧠 Conceptual

expert

Why is sorting both arrays after swapping necessary?

In the merge without extra space approach, why do we sort both arrays after swapping elements?

A Because swapping only ensures elements are roughly partitioned, sorting finalizes the order within each array.

B Because sorting is required to remove duplicates after swapping.

C Because sorting is needed to merge arrays into a single array.

D Because swapping sorts the arrays automatically, so sorting is redundant.

💡 Hint

Think about what swapping achieves and what sorting completes.

Swapping places smaller elements in the first array and larger in the second but does not guarantee order inside each array. Sorting each array after swapping ensures each is sorted individually.

Practice

1. What is the main goal when merging two sorted arrays without extra space?

easy

A. Combine both arrays into one sorted sequence without using extra memory

B. Copy all elements into a new array and sort it

C. Sort each array separately without merging

D. Use a temporary array to hold merged elements

Solution

1. Step 1: Understand the problem requirement

   The problem asks to merge two sorted arrays into a sorted sequence without using extra space, meaning no additional arrays or large memory allocation.

2. Step 2: Identify the correct approach

   Using in-place techniques like swapping elements between arrays and sorting them ensures no extra space is used.

3. Final Answer:

   Combine both arrays into one sorted sequence without using extra memory -> Option A

4. Quick Check:

   Merge without extra space = Combine in-place [OK]

Hint: Remember: no extra array means in-place merging [OK]

Common Mistakes:

• Thinking extra array is allowed
• Sorting arrays separately without merging
• Using extra memory for temporary storage

2. Which of the following Java code snippets correctly swaps elements between two arrays arr1 and arr2 at indices i and j without extra space?

easy

A. arr1[i] = arr1[j]; arr2[j] = arr2[i];

B. arr1[i] = arr2[j]; arr2[j] = arr1[i];

C. int temp = arr2[j]; arr1[i] = arr2[j]; arr2[j] = temp;

D. int temp = arr1[i]; arr1[i] = arr2[j]; arr2[j] = temp;

Solution

1. Step 1: Check the swap logic

   Swapping requires a temporary variable to hold one value while assigning the other to avoid overwriting.

2. Step 2: Verify correct use of temp variable

   int temp = arr1[i]; arr1[i] = arr2[j]; arr2[j] = temp; uses a temp variable to hold arr1[i], then assigns arr2[j] to arr1[i], and finally assigns temp to arr2[j], correctly swapping values.

3. Final Answer:

   int temp = arr1[i]; arr1[i] = arr2[j]; arr2[j] = temp; -> Option D

4. Quick Check:

   Swap needs temp variable = int temp = arr1[i]; arr1[i] = arr2[j]; arr2[j] = temp; [OK]

Hint: Always use a temp variable to swap two values safely [OK]

Common Mistakes:

• Assigning without temp causes overwriting
• Swapping elements within the same array incorrectly
• Confusing indices during swap

3. Given two sorted arrays arr1 = [1, 5, 9, 10] and arr2 = [2, 3, 8, 13], after applying the merge without extra space algorithm, what will be the final state of arr1 and arr2?

medium

A. arr1 = [1, 2, 3, 8], arr2 = [5, 9, 10, 13]

B. arr1 = [1, 2, 3, 5], arr2 = [8, 9, 10, 13]

C. arr1 = [1, 3, 5, 8], arr2 = [2, 9, 10, 13]

D. arr1 = [1, 2, 5, 8], arr2 = [3, 9, 10, 13]

Solution

1. Step 1: Compare and swap elements between arrays

   Start comparing last element of arr1 with first of arr2. Swap if arr1 element is greater. After swaps, sort both arrays.

2. Step 2: Final sorted arrays after merging

   After swapping and sorting, arr1 becomes [1, 2, 3, 5] and arr2 becomes [8, 9, 10, 13].

3. Final Answer:

   arr1 = [1, 2, 3, 5], arr2 = [8, 9, 10, 13] -> Option B

4. Quick Check:

   In-place merge sorts arrays = arr1 = [1, 2, 3, 5], arr2 = [8, 9, 10, 13] [OK]

Hint: Swap bigger arr1 elements with smaller arr2, then sort both [OK]

Common Mistakes:

• Not sorting arrays after swapping
• Swapping incorrectly leading to unsorted arrays
• Mixing indices during swaps

4. Identify the error in the following code snippet for merging two sorted arrays without extra space:

for (int i = 0; i < arr1.length; i++) {
  if (arr1[i] > arr2[0]) {
    int temp = arr1[i];
    arr1[i] = arr2[0];
    arr2[0] = temp;
    Arrays.sort(arr2);
  }
}

medium

A. Using Arrays.sort on arr1 instead of arr2 is required

B. Swapping elements without checking arr2 length causes error

C. Sorting arr2 inside the loop causes unnecessary repeated sorting

D. Loop should run until arr2.length instead of arr1.length

Solution

1. Step 1: Analyze sorting inside the loop

   The code sorts arr2 inside the loop every time a swap happens, which is inefficient and slows down the algorithm.

2. Step 2: Suggest optimization

   Sorting arr2 once after all swaps or using a better approach like gap method improves performance.

3. Final Answer:

   Sorting arr2 inside the loop causes unnecessary repeated sorting -> Option C

4. Quick Check:

   Repeated sorting inside loop = inefficiency [OK]

Hint: Avoid sorting inside loops; sort once after swaps [OK]

Common Mistakes:

• Sorting arr2 repeatedly inside loop
• Confusing loop limits between arrays
• Ignoring performance impact of repeated sorting

5. You have two sorted arrays arr1 = [1, 4, 7, 8, 10] and arr2 = [2, 3, 9]. Using the gap method to merge without extra space, what is the final state of both arrays?

hard

A. arr1 = [1, 2, 3, 4, 7], arr2 = [8, 9, 10]

B. arr1 = [1, 2, 3, 4, 9], arr2 = [7, 8, 10]

C. arr1 = [1, 3, 4, 7, 8], arr2 = [2, 9, 10]

D. arr1 = [1, 2, 4, 7, 8], arr2 = [3, 9, 10]

Solution

1. Step 1: Calculate initial gap and compare elements

   Total length is 8, initial gap = ceil(8/2) = 4. Compare elements at gap distance and swap if needed, then reduce gap.

2. Step 2: Perform gap reductions and swaps

   After successive gap reductions (4, 2, 1), arrays become sorted as arr1 = [1, 2, 3, 4, 7] and arr2 = [8, 9, 10].

3. Final Answer:

   arr1 = [1, 2, 3, 4, 7], arr2 = [8, 9, 10] -> Option A

4. Quick Check:

   Gap method sorts arrays in-place = arr1 = [1, 2, 3, 4, 7], arr2 = [8, 9, 10] [OK]

Hint: Use gap method: reduce gap, swap if needed, repeat until gap=0 [OK]

Common Mistakes:

• Not reducing gap correctly
• Swapping wrong elements across arrays
• Confusing gap calculation with array lengths