Array Insertion at Middle Index in DSA Java - Execution Trace

Concept Flow - Array Insertion at Middle Index

Start with array

↓

Find middle index

↓

Shift elements right from middle

↓

Insert new element at middle

↓

Array updated with new element

This flow shows how to insert a new element in the middle of an array by first finding the middle index, shifting elements to the right, and then placing the new element.

Execution Sample

DSA Java

int[] arr = {1, 2, 4, 5, 0};
int newElement = 3;
int mid = arr.length / 2;
// Shift elements right
for (int i = arr.length - 2; i >= mid; i--) {
    arr[i + 1] = arr[i];
}
arr[mid] = newElement;

This code inserts the element 3 into the middle of the array {1, 2, 4, 5} by shifting elements and placing 3 at the middle index.

Execution Table

Step	Operation	Index/Pointer	Array State	Pointer Changes	Visual State
1	Start with array	-	[1, 2, 4, 5, _]	head at 0	1 -> 2 -> 4 -> 5 -> _
2	Calculate middle index	mid = 2	-	-	Index 2 selected for insertion
3	Shift element at index 3 to 4	i=3	[1, 2, 4, 5, 5]	arr[4] = arr[3]	1 -> 2 -> 4 -> 5 -> 5
4	Shift element at index 2 to 3	i=2	[1, 2, 4, 4, 5]	arr[3] = arr[2]	1 -> 2 -> 4 -> 4 -> 5
5	Insert new element at mid	mid=2	[1, 2, 3, 4, 5]	arr[2] = 3	1 -> 2 -> 3 -> 4 -> 5
6	Insertion complete	-	[1, 2, 3, 4, 5]	-	Final array with new element inserted

💡 All elements shifted right from middle index, new element inserted at index 2, array updated.

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 5	Final
arr	[1, 2, 4, 5, _]	[1, 2, 4, 5, _]	[1, 2, 4, 4, 5]	[1, 2, 3, 4, 5]	[1, 2, 3, 4, 5]
mid	undefined	2	2	2	2
i	undefined	undefined	2	undefined	undefined

Key Moments - 3 Insights

Why do we shift elements starting from the end towards the middle?

We shift elements from the end to the middle (see steps 3 and 4) to avoid overwriting elements before they are moved. Shifting forward would overwrite data.

Why is the middle index calculated as length / 2?

The middle index (step 2) divides the array roughly in half. For even length arrays, it points to the first element of the second half where insertion happens.

What happens if we don't have space at the end of the array?

In this example, the array has an extra space (_) at the end. Without extra space, shifting elements would overwrite data or cause errors.

Visual Quiz - 3 Questions

Look at the execution table, what is the array state after step 4?

A [1, 2, 4, 4, 5]

B [1, 2, 4, 5, 5]

C [1, 2, 3, 4, 5]

D [1, 2, 4, 5, _]

💡 Hint

Check the 'Array State' column for step 4 in the execution_table.

At which step is the new element inserted into the array?

A Step 3

B Step 5

C Step 4

D Step 6

💡 Hint

Look for the operation 'Insert new element at mid' in the execution_table.

If the array had no extra space at the end, what would happen during shifting?

A Elements would shift without issues

B Middle index would change

C Last element would be overwritten or lost

D New element would be inserted at the end

💡 Hint

Refer to the key_moments about array space and shifting.

Concept Snapshot

Array Insertion at Middle Index:
- Find middle index: mid = length / 2
- Shift elements right from end to mid
- Insert new element at mid
- Requires extra space at array end
- Prevents overwriting by shifting backward

Full Transcript

This visualization shows how to insert an element into the middle of an array. First, we find the middle index by dividing the array length by two. Then, starting from the last element, we shift each element one position to the right until we reach the middle index. This creates space at the middle index where we insert the new element. The array must have extra space at the end to avoid overwriting data. The execution table tracks each step, showing how the array changes and how pointers move. Key moments clarify why shifting starts from the end and why the middle index is chosen this way. The visual quiz tests understanding of array states and insertion steps.

Practice

1. What happens when you insert an element at the middle index of an array in Java?

easy

A. The array size automatically increases without shifting elements.

B. The new element replaces the element at the middle index without shifting.

C. Elements after the middle index shift right to make space for the new element.

D. The element is added at the end of the array regardless of the index.

Solution

1. Step 1: Understand array insertion behavior

   In Java, arrays have fixed size, so to insert an element in the middle, elements from that index onward must shift right to make space.

2. Step 2: Effect of insertion at middle index

   The element at the middle index and all elements after it move one position to the right, making room for the new element.

3. Final Answer:

   Elements after the middle index shift right to make space for the new element. -> Option C

4. Quick Check:

   Insertion shifts elements right [OK]

Hint: Insertion shifts elements right from middle index [OK]

Common Mistakes:

• Thinking the new element replaces existing one without shifting
• Assuming array size grows automatically
• Believing insertion adds element at the end always

2. Which of the following Java code snippets correctly inserts an element `x` at the middle index of an integer array `arr` of size `n`?

easy

A. int mid = n / 2; for (int i = n - 1; i >= mid; i--) arr[i + 1] = arr[i]; arr[mid] = x;

B. int mid = n / 2; for (int i = 0; i < mid; i++) arr[i] = arr[i + 1]; arr[mid] = x;

C. int mid = n / 2; arr[mid] = x; for (int i = mid + 1; i < n; i++) arr[i] = arr[i];

D. int mid = n / 2; arr[mid + 1] = x; for (int i = mid; i < n; i++) arr[i] = arr[i + 1];

Solution

1. Step 1: Identify correct shifting direction

   To insert at middle, elements from middle to end must shift right, so loop runs backward from end to middle.

2. Step 2: Verify code correctness

   int mid = n / 2; for (int i = n - 1; i >= mid; i--) arr[i + 1] = arr[i]; arr[mid] = x; correctly shifts elements right starting from the last index to middle, then inserts x at mid.

3. Final Answer:

   int mid = n / 2; for (int i = n - 1; i >= mid; i--) arr[i + 1] = arr[i]; arr[mid] = x; -> Option A

4. Quick Check:

   Backward loop shifts right, then insert [OK]

Hint: Shift elements right backward from end to middle [OK]

Common Mistakes:

• Shifting elements left instead of right
• Looping forward causing overwrite
• Assigning new element before shifting

3. What is the output of the following Java code?

int[] arr = {1, 2, 3, 4, 5, 0};
int n = 5;
int x = 99;
int mid = n / 2;
for (int i = n - 1; i >= mid; i--) {
    arr[i + 1] = arr[i];
}
arr[mid] = x;
for (int i = 0; i <= n; i++) {
    System.out.print(arr[i] + " ");
}

medium

A. 1 2 99 3 4 5

B. 1 2 3 99 4 5

C. 5 4 3 99 2 1

D. 1 2 99 3 4 5 0

Solution

1. Step 1: Calculate middle index and shift elements

   n=5, mid=2. Loop shifts elements from index 4 to 2 right by one: arr[5]=arr[4]=5, arr[4]=arr[3]=4, arr[3]=arr[2]=3.

2. Step 2: Insert x at middle and print array

   arr[2]=99. Final array: [1, 2, 99, 3, 4, 5]. Printing elements from 0 to 5 outputs: 1 2 99 3 4 5 0

3. Final Answer:

   1 2 99 3 4 5 0 -> Option D

4. Quick Check:

   Shift right + insert at mid = 1 2 99 3 4 5 0 [OK]

Hint: Shift right from end to mid, then insert new element [OK]

Common Mistakes:

• Forgetting to shift elements before insertion
• Printing extra zero at end
• Incorrect middle index calculation

4. The following code attempts to insert element `x` at the middle index of array `arr` but produces incorrect output. What is the error?

int mid = arr.length / 2;
for (int i = mid; i < arr.length; i++) {
    arr[i + 1] = arr[i];
}
arr[mid] = x;

medium

A. Loop shifts elements left instead of right causing overwrite.

B. Loop runs forward causing overwriting of elements during shift.

C. Middle index calculation is incorrect for even length arrays.

D. New element is inserted before shifting elements.

Solution

1. Step 1: Analyze loop direction and effect

   The loop runs forward from mid to end, assigning arr[i+1] = arr[i], which overwrites elements before they are copied.

2. Step 2: Correct approach for shifting elements

   To avoid overwrite, elements must be shifted right starting from the end backward to mid.

3. Final Answer:

   Loop runs forward causing overwriting of elements during shift. -> Option B

4. Quick Check:

   Shift must go backward to avoid overwrite [OK]

Hint: Shift elements backward from end to mid to avoid overwrite [OK]

Common Mistakes:

• Shifting elements forward causing data loss
• Misunderstanding loop boundaries
• Inserting element before shifting

5. You have an array `arr` of size `n` with an extra space at the end (size `n+1`). Write the steps to insert an element `x` exactly in the middle of the array, shifting elements correctly, and print the final array. Which of the following sequences of steps is correct?

hard

A. Calculate middle index; shift elements right from end to middle; insert x at middle; print array from 0 to n.

B. Calculate middle index; insert x at middle; shift elements right from middle to end; print array from 0 to n.

C. Calculate middle index; shift elements left from middle to end; insert x at middle; print array from 0 to n-1.

D. Calculate middle index; insert x at end; shift elements right from middle to end; print array from 0 to n.

Solution

1. Step 1: Calculate middle index and prepare for insertion

   Find middle index as n/2. Since array has extra space, shifting elements right from end to middle creates space.

2. Step 2: Shift elements right and insert new element

   Shift elements starting from last valid index backward to middle index to avoid overwriting. Insert x at middle index.

3. Step 3: Print the updated array

   Print elements from index 0 to n (inclusive) to show the array with inserted element.

4. Final Answer:

   Calculate middle index; shift elements right from end to middle; insert x at middle; print array from 0 to n. -> Option A

5. Quick Check:

   Shift right backward, insert at mid, print full array [OK]

Hint: Shift right backward, insert at mid, print full array [OK]

Common Mistakes:

• Inserting before shifting elements
• Shifting elements forward causing overwrite
• Printing wrong array range