Merge Two Sorted Arrays Without Extra Space in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to swap two elements in the arrays.

DSA Java

int temp = arr1[[1]];
arr1[[1]] = arr2[j];
arr2[j] = temp;

Drag options to blanks, or click blank then click option'

Aj

Bi

C0

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong index variable for arr1.

Swapping with an incorrect element from arr2.

We swap elements at index i in arr1 with element at index j in arr2 to maintain order.

2fill in blank

medium

Complete the code to compare elements from arr1 and arr2.

DSA Java

if (arr1[i] > arr2[[1]]) {
    // swap logic
}

Drag options to blanks, or click blank then click option'

Alength

Bi

Cj

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing arr1[i] with arr1[j] instead of arr2[j].

Using an incorrect index for arr2.

We compare arr1[i] with arr2[j] to decide if swapping is needed.

3fill in blank

hard

Fix the error in the loop condition to iterate over arr1 correctly.

DSA Java

for (int i = 0; i < [1]; i++) {
    // loop body
}

Drag options to blanks, or click blank then click option'

Aarr1.length

Barr2.length

Carr1.length - 1

Darr2.length - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using arr2.length which may be different.

Using length - 1 which misses the last element.

We loop through all elements of arr1 using arr1.length as the limit.

4fill in blank

hard

Fill both blanks to complete the insertion sort step for arr2.

DSA Java

int key = arr2[[1]];
int j = [2] - 1;
while (j >= 0 && arr2[j] > key) {
    arr2[j + 1] = arr2[j];
    j--;
}
arr2[j + 1] = key;

Drag options to blanks, or click blank then click option'

Ai

Bj

Ci + 1

Darr2.length

Attempts:

3 left

💡 Hint

Common Mistakes

Using j instead of i for key index.

Starting j from wrong index.

We pick the element at index i in arr2 as key and start comparing from i-1.

5fill in blank

hard

Fill all three blanks to complete the merge function without extra space.

DSA Java

public static void merge(int[] arr1, int[] arr2) {
    int n = arr1.length;
    int m = arr2.length;
    for (int i = 0; i < n; i++) {
        if (arr1[i] > arr2[[1]]) {
            int temp = arr1[i];
            arr1[i] = arr2[[2]];
            arr2[[3]] = temp;
            // insertion sort step for arr2
            int key = arr2[[2]];
            int j = [2] - 1;
            while (j >= 0 && arr2[j] > key) {
                arr2[j + 1] = arr2[j];
                j--;
            }
            arr2[j + 1] = key;
        }
    }
}

Drag options to blanks, or click blank then click option'

A0

Bj

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing indices i and j incorrectly.

Not updating arr2 properly after swap.

We compare and swap elements at index i in arr1 with index j in arr2, then insert arr2[j] correctly.

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