Array Insertion at Middle Index in DSA Java

Mental Model

To insert a new value in the middle of an array, we shift all elements from that position to the right by one place to make space.

Analogy: Imagine a row of chairs where you want to add a new person in the middle. Everyone from that spot onward stands up and moves one chair to the right to make room.

Index: 0   1   2   3   4
Array: [1]->[2]->[3]->[4]->[5]
Insert new value at index 2 here ↑

Dry Run Walkthrough

Input: array: [1, 2, 3, 4, 5], insert value 99 at index 2

Goal: Insert 99 at index 2 and shift elements right without losing data

Step 1: Start shifting elements from the last index to index 2 one position to the right

[1]->[2]->[3]->[4]->[5]->null (original array, space assumed at end)

Why: We need to move elements to make space at index 2

Step 2: Move element at index 4 (5) to index 5

[1]->[2]->[3]->[4]->[5]->[5]

Why: Shift last element right to avoid overwriting

Step 3: Move element at index 3 (4) to index 4

[1]->[2]->[3]->[4]->[4]->[5]

Why: Continue shifting elements right

Step 4: Move element at index 2 (3) to index 3

[1]->[2]->[3]->[3]->[4]->[5]

Why: Shift element at insertion index right

Step 5: Insert new value 99 at index 2

[1]->[2]->[99]->[3]->[4]->[5]

Why: Place new value in the freed space

Result:

Final array: [1]->[2]->[99]->[3]->[4]->[5]

Annotated Code

DSA Java

public class ArrayInsertion {
    public static int[] insertAtMiddle(int[] arr, int value, int index) {
        int n = arr.length;
        int[] newArr = new int[n + 1];
        for (int i = 0; i < index; i++) {
            newArr[i] = arr[i];
        }
        for (int i = n - 1; i >= index; i--) {
            newArr[i + 1] = arr[i];
        }
        newArr[index] = value;
        return newArr;
    }

    public static void printArray(int[] arr) {
        for (int i = 0; i < arr.length; i++) {
            System.out.print(arr[i]);
            if (i != arr.length - 1) {
                System.out.print(" -> ");
            }
        }
        System.out.println(" -> null");
    }

    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 4, 5};
        int value = 99;
        int index = 2;
        int[] result = insertAtMiddle(arr, value, index);
        printArray(result);
    }
}

for (int i = n - 1; i >= index; i--) { newArr[i + 1] = arr[i]; }

Shift elements from the end to index right by one to make space

newArr[index] = value;

Insert new value at the middle index

Output

1 -> 2 -> 99 -> 3 -> 4 -> 5 -> null

Complexity Analysis

Time: O(n) because we must shift elements from the insertion index to the end once

Space: O(n) because we create a new array with one extra space

vs Alternative: In-place insertion in arrays is costly due to shifting; linked lists can insert in O(1) without shifting but have other trade-offs

Edge Cases

Insert at index 0 (start of array)

All elements shift right, new value placed at start

DSA Java

for (int i = n - 1; i >= index; i--) {
    newArr[i + 1] = arr[i];
}

Insert at index equal to array length (end)

No shifting needed, new value appended at end

DSA Java

// No shifting loop needed when index == n
newArr[index] = value;

Empty array insertion

New array with single element is created

DSA Java

int[] newArr = new int[n + 1];

When to Use This Pattern

When you need to add an element inside an array and keep order, think of shifting elements right to make space before insertion.

Common Mistakes

Mistake: Shifting elements from left to right causing overwriting data
Fix: Shift elements from right to left starting at the end to avoid overwriting

Mistake: Not creating a new array or resizing, causing index out of bounds
Fix: Create a new array with size +1 to hold the extra element

Summary

It inserts a new value at a given middle index by shifting elements right to make space.

Use it when you want to add an element inside an array without losing existing data.

Remember to shift elements from the end backward to avoid overwriting during insertion.

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