Array Insertion at Middle Index in DSA Java - Time & Space Complexity

Time Complexity: Array Insertion at Middle Index

O(n)

Understanding Time Complexity

When we insert an element in the middle of an array, the time it takes depends on how many elements need to move.

We want to know how the work grows as the array gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public void insertMiddle(int[] arr, int n, int value) {
    int mid = n / 2;
    for (int i = n - 1; i >= mid; i--) {
        arr[i + 1] = arr[i];
    }
    arr[mid] = value;
}
    

This code inserts a new value in the middle of an array by shifting elements to the right.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that shifts elements one by one.
• How many times: About half of the array elements are moved.

How Execution Grows With Input

As the array size grows, the number of elements to shift grows roughly half as fast.

Input Size (n)	Approx. Operations
10	5 shifts
100	50 shifts
1000	500 shifts

Pattern observation: The work grows linearly with the size of the array.

Final Time Complexity

Time Complexity: O(n)

This means the time to insert grows in direct proportion to the array size.

Common Mistake

[X] Wrong: "Inserting in the middle is always fast because we just put the value there."

[OK] Correct: We must move many elements to make space, which takes time proportional to the number of elements shifted.

Interview Connect

Understanding how array insertions scale helps you explain trade-offs between arrays and other data structures like linked lists.

Self-Check

"What if we inserted at the end of the array instead of the middle? How would the time complexity change?"

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