Array Insertion at Middle Index in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does 'Array Insertion at Middle Index' mean?

It means adding a new element into an array at the position in the middle, shifting the existing elements to the right to make space.

beginner

Why do we need to shift elements when inserting in the middle of an array?

Because arrays have fixed size and continuous memory, to insert a new element in the middle, we move elements after that position one step right to avoid overwriting data.

intermediate

In Java, what happens if you try to insert an element in a full array without resizing?

You cannot insert because the array size is fixed; you must create a new larger array or use a dynamic structure like ArrayList.

intermediate

What is the time complexity of inserting an element at the middle of an array?

The time complexity is O(n) because you may need to shift about half of the elements to the right.

beginner

Show the steps to insert element 99 at middle index of array [10, 20, 30, 40, 50].

1. Find middle index (length 5 → middle 2).
2. Shift elements from index 2 to right: 30→index3, 40→index4, 50→index5.
3. Insert 99 at index 2.
Result: [10, 20, 99, 30, 40, 50]

What is the first step when inserting an element at the middle of an array?

A Find the middle index

B Shift elements to the left

C Delete the last element

D Resize the array

You first find the middle index where the new element will be inserted.

Why do we shift elements to the right during insertion in the middle of an array?

A To sort the array

B To remove duplicates

C To make space for the new element

D To reduce array size

Shifting elements right creates space at the insertion index.

What is the time complexity of inserting an element in the middle of an array?

A O(1)

B O(n^2)

C O(log n)

D O(n)

Insertion requires shifting elements, which takes linear time.

If an array has 6 elements, what is the middle index for insertion?

A 2

B 3

C 4

D 5

Middle index is length/2, so 6/2 = 3.

What happens if you try to insert in a full fixed-size array without resizing?

A Insertion fails or overwrites data

B Array automatically resizes

C Array elements are deleted

D Insertion succeeds without issues

Fixed-size arrays cannot grow; insertion fails or overwrites existing data.

Explain how to insert an element at the middle index of an array step-by-step.

Think about making space and where to put the new element.

Describe why inserting in the middle of an array is slower than inserting at the end.

Consider what happens to existing elements.

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