Bird
Raised Fist0
DSA Cprogramming~10 mins

Array Insertion at Middle Index in DSA C - Execution Trace

Choose your learning style10 modes available
LearnWhyDeepVisualTryChallengeProjectRecallTime
Concept Flow - Array Insertion at Middle Index
Start with array and size
↓
Calculate middle index = size/2
↓
Shift elements right from end to middle
↓
Insert new element at middle index
↓
Increase size by 1
↓
Array updated with new element in middle
The flow shows how to find the middle, shift elements right to make space, insert the new element, and update the size.
Execution Sample
DSA C
int arr[6] = {1, 2, 4, 5, 6};
int size = 5;
int val = 3;
int mid = size / 2;
// Shift and insert
for (int i = size; i > mid; i--) arr[i] = arr[i-1];
arr[mid] = val;
size++;
This code inserts value 3 into the middle of the array by shifting elements right and placing 3 at index 2.
Execution Table
StepOperationArray StatePointer ChangesVisual State
1Start with array and size[1, 2, 4, 5, 6, _]size=5, mid=21 -> 2 -> 4 -> 5 -> 6 -> _
2Calculate middle indexSamemid=21 -> 2 -> 4 -> 5 -> 6 -> _
3Shift element at index 4 to 5[1, 2, 4, 5, 6, 6]i=51 -> 2 -> 4 -> 5 -> 6 -> 6
4Shift element at index 3 to 4[1, 2, 4, 5, 5, 6]i=41 -> 2 -> 4 -> 5 -> 5 -> 6
5Shift element at index 2 to 3[1, 2, 4, 4, 5, 6]i=31 -> 2 -> 4 -> 4 -> 5 -> 6
6Insert 3 at index 2[1, 2, 3, 4, 5, 6]arr[2]=31 -> 2 -> 3 -> 4 -> 5 -> 6
7Increase size by 1size=6size=61 -> 2 -> 3 -> 4 -> 5 -> 6
8EndFinal arraysize=61 -> 2 -> 3 -> 4 -> 5 -> 6
💡 All elements shifted right from index 2, new element inserted, size updated to 6
Variable Tracker
VariableStartAfter Step 3After Step 4After Step 5After Step 6Final
arr[1, 2, 4, 5, 6, _][1, 2, 4, 5, 6, 6][1, 2, 4, 5, 5, 6][1, 2, 4, 4, 5, 6][1, 2, 3, 4, 5, 6][1, 2, 3, 4, 5, 6]
size555556
mid222222
iN/A543N/AN/A
Key Moments - 3 Insights
Why do we shift elements starting from the end towards the middle?
Shifting from the end prevents overwriting elements before they are moved, as shown in steps 3 to 5 in the execution_table.
Why is the middle index calculated as size/2?
Because integer division truncates, it gives the correct middle position to insert the new element, as seen in step 2.
What happens if we insert without shifting elements?
Without shifting, the element at the middle index would be overwritten and lost, breaking the array order, which is avoided by steps 3-5.
Visual Quiz - 3 Questions
Test your understanding
Look at the execution_table at step 5, what is the array state?
A[1, 2, 4, 5, 5, 6]
B[1, 2, 4, 5, 6, 6]
C[1, 2, 4, 4, 5, 6]
D[1, 2, 3, 4, 5, 6]
💡 Hint
Check the 'Array State' column at step 5 in the execution_table.
At which step is the new value 3 inserted into the array?
AStep 6
BStep 5
CStep 4
DStep 7
💡 Hint
Look for the operation mentioning 'Insert 3' in the execution_table.
If the initial size was 6 instead of 5, how would the middle index change?
AIt would be 2
BIt would be 3
CIt would be 4
DIt would be 5
💡 Hint
Middle index is calculated as size/2; check variable_tracker for mid value.
Concept Snapshot
Array Insertion at Middle Index:
- Calculate middle index as size/2
- Shift elements right from end to middle
- Insert new element at middle index
- Increase size by 1
- Maintains order and space in array
Full Transcript
This visual trace shows how to insert a new element into the middle of an array. First, we calculate the middle index by dividing the size by two. Then, starting from the end of the array, we shift each element one position to the right until we reach the middle index. This creates space to insert the new element without overwriting existing data. After inserting the new value at the middle index, we increase the size of the array by one. The execution table tracks each step, showing the array state and pointer changes. Key moments clarify why shifting starts from the end and why the middle index is calculated as size divided by two. The quiz tests understanding of array states and index calculations. This method keeps the array ordered and correctly sized after insertion.

Practice

(1/5)
1. What is the correct middle index to insert an element in an array of size 6?
easy
A. 5
B. 2
C. 4
D. 3

Solution

  1. Step 1: Calculate middle index for even-sized array

    For an array of size 6, middle index is size/2 = 6/2 = 3.

  2. Step 2: Confirm insertion position

    Inserting at index 3 places the element exactly in the middle.

  3. Final Answer:

    3 -> Option D

  4. Quick Check:

    Middle index = size/2 = 3 [OK]
Hint: Middle index = array size divided by 2 [OK]
Common Mistakes:
  • Using size-1 as middle index
  • Using size/2 - 1 for even size
  • Confusing zero-based indexing
2. Which of the following C code snippets correctly inserts an element at the middle index of an array arr of size n?
easy
A. int mid = n / 2; for(int i = 0; i < mid; i++) arr[i] = arr[i+1]; arr[mid] = x;
B. int mid = n / 2; for(int i = n; i > mid; i--) arr[i] = arr[i-1]; arr[mid] = x;
C. int mid = n / 2; arr[mid] = x;
D. int mid = n / 2; for(int i = mid; i < n; i++) arr[i] = arr[i+1]; arr[mid] = x;

Solution

  1. Step 1: Identify correct shifting direction

    To insert at middle, elements from middle to end must shift right to make space.

  2. Step 2: Check code correctness

    int mid = n / 2; for(int i = n; i > mid; i--) arr[i] = arr[i-1]; arr[mid] = x; shifts elements right from end to middle, then inserts x at mid. Others shift incorrectly or overwrite.

  3. Final Answer:

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

  4. Quick Check:

    Shift right from end to mid, then insert [OK]
Hint: Shift elements right from end to middle before insertion [OK]
Common Mistakes:
  • Shifting elements left instead of right
  • Not shifting elements before insertion
  • Overwriting elements without shifting
3. What will be the content of the array after inserting 99 at the middle index of arr = {10, 20, 30, 40, 50} using the correct insertion method?
medium
A. [10, 20, 30, 40, 99, 50]
B. [10, 20, 99, 30, 40, 50]
C. [10, 20, 30, 99, 40, 50]
D. [10, 20, 30, 40, 50, 99]

Solution

  1. Step 1: Calculate middle index

    Array size is 5, middle index = 5/2 = 2 (integer division).

  2. Step 2: Shift elements right and insert 99

    Shift elements from index 4 to 2 right by one position. Insert 99 at index 2.

  3. Final Answer:

    [10, 20, 99, 30, 40, 50] -> Option C

  4. Quick Check:

    Insert at index 2 shifts elements right [OK]
Hint: Middle index = size/2; shift right, then insert [OK]
Common Mistakes:
  • Inserting at wrong index
  • Not shifting elements before insertion
  • Overwriting elements
4. Identify the error in the following code snippet that inserts an element x at the middle of array arr of size n: 
int mid = n / 2;
for(int i = mid; i < n; i++) {
    arr[i+1] = arr[i];
}
arr[mid] = x;
medium
A. Loop should run backwards from n to mid
B. Index mid is incorrect for insertion
C. No need to shift elements before insertion
D. Array size n should be incremented before insertion

Solution

  1. Step 1: Analyze loop direction

    The loop shifts elements from mid to n forward, but does so from low to high index, causing overwrite.

  2. Step 2: Correct loop direction

    Loop must run backwards from n to mid to shift elements right safely without overwriting.

  3. Final Answer:

    Loop should run backwards from n to mid -> Option A

  4. Quick Check:

    Shift right must be done backwards [OK]
Hint: Shift elements right by looping backwards from end [OK]
Common Mistakes:
  • Looping forward causing data overwrite
  • Not updating array size after insertion
  • Incorrect middle index calculation
5. Given an array arr = {5, 10, 15, 20, 25, 30} of size 6, write the final array after inserting 99 at the middle index and then inserting 77 at the new middle index of the updated array.
hard
A. [5, 10, 15, 77, 99, 20, 25, 30]
B. [5, 10, 77, 99, 15, 20, 25, 30]
C. [5, 10, 99, 15, 77, 20, 25, 30]
D. [5, 10, 15, 99, 77, 20, 25, 30]

Solution

  1. Step 1: Insert 99 at middle of original array

    Original size = 6, middle index = 3. Insert 99 at index 3 shifting elements right: [5, 10, 15, 99, 20, 25, 30]

  2. Step 2: Insert 77 at middle of updated array

    New size = 7, middle index = 7/2 = 3. Insert 77 at index 3 shifting elements right: [5, 10, 15, 77, 99, 20, 25, 30]

  3. Final Answer:

    [5, 10, 15, 77, 99, 20, 25, 30] -> Option A

  4. Quick Check:

    Insert twice at middle recalculating index each time [OK]
Hint: Recalculate middle after each insertion before next insert [OK]
Common Mistakes:
  • Not recalculating middle index after first insertion
  • Overwriting elements without shifting
  • Inserting at wrong index second time
Recommended

Ready to prove it?

Pass the test and this pattern moves unknown strong while earning +1 Grip toward your certificate.

10 questions~5 mineasy hard