Array Deletion at Middle Index in DSA Go - Time & Space Complexity

Time Complexity: Array Deletion at Middle Index

O(n)

Understanding Time Complexity

We want to understand how long it takes to remove an element from the middle of an array.

How does the work grow as the array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

package main

import "fmt"

func deleteMiddle(arr []int, index int) []int {
    for i := index; i < len(arr)-1; i++ {
        arr[i] = arr[i+1]
    }
    return arr[:len(arr)-1]
}

func main() {
    arr := []int{1, 2, 3, 4, 5}
    arr = deleteMiddle(arr, 2) // delete element at index 2
    fmt.Println(arr)
}

This code removes the element at the given middle index by shifting all elements after it one step left.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that shifts elements left starting from the deletion index.
• How many times: It runs from the deletion index to the end of the array minus one, roughly n - index - 1 times.

How Execution Grows With Input

As the array size grows, the number of shifts grows roughly with the size of the array.

Input Size (n)	Approx. Operations (shifts)
10	About 5 (if deleting near middle)
100	About 50
1000	About 500

Pattern observation: The number of operations grows roughly linearly with the array size.

Final Time Complexity

Time Complexity: O(n)

This means the time to delete grows in a straight line as the array gets bigger.

Common Mistake

[X] Wrong: "Deleting an element from the middle of an array is very fast, like O(1)."

[OK] Correct: Because after deletion, all elements after the deleted spot must move left, which takes time proportional to the number of elements shifted.

Interview Connect

Understanding this helps you explain why arrays are fast for access but slower for middle deletions, a key insight in many coding problems.

Self-Check

"What if we used a linked list instead of an array? How would the time complexity of deleting a middle element change?"

Practice

1. What happens when you delete the middle element from an array with an odd number of elements?

easy

A. The array size stays the same but the middle element becomes zero.

B. The middle element is removed and the array size decreases by one.

C. The array is reversed after deletion.

D. The first element is removed instead of the middle.

Solution

1. Step 1: Understand array deletion

   Deleting an element removes it and shifts later elements left.

2. Step 2: Effect on array size

   Removing one element reduces the array length by one.

3. Final Answer:

   The middle element is removed and the array size decreases by one. -> Option B

4. Quick Check:

   Delete middle element = size - 1 [OK]

Hint: Deleting removes element and shrinks array size [OK]

Common Mistakes:

• Thinking array size stays the same after deletion
• Assuming deletion replaces element with zero
• Confusing middle deletion with first element deletion

2. Which Go code snippet correctly deletes the middle element from an array named arr with odd length?

easy

A. arr = arr[len(arr)/2+1:]

B. arr = append(arr[1:], arr[:len(arr)-1]...)

C. arr = append(arr[:len(arr)/2], arr[len(arr)/2+1:]...)

D. arr = arr[:len(arr)/2] + arr[len(arr)/2+1:]

Solution

1. Step 1: Identify correct slicing syntax

   In Go, to delete an element at index i, use append(arr[:i], arr[i+1:]...).

2. Step 2: Apply to middle index

   Middle index is len(arr)/2 for odd length, so arr = append(arr[:len(arr)/2], arr[len(arr)/2+1:]...).

3. Final Answer:

   arr = append(arr[:len(arr)/2], arr[len(arr)/2+1:]...) -> Option C

4. Quick Check:

   Use append with slices excluding middle [OK]

Hint: Use append(arr[:i], arr[i+1:]...) to delete index i [OK]

Common Mistakes:

• Using + operator to join slices (invalid in Go)
• Not using ... to expand second slice in append
• Deleting wrong index or slice ranges

3. What is the output of this Go code?

arr := []int{10, 20, 30, 40, 50}
mid := len(arr) / 2
arr = append(arr[:mid], arr[mid+1:]...)
fmt.Println(arr)

medium

A. [10 20 40 50]

B. [10 20 30 40]

C. [20 30 40 50]

D. [10 30 40 50]

Solution

1. Step 1: Calculate middle index

   len(arr) = 5, so mid = 5 / 2 = 2 (integer division).

2. Step 2: Delete element at index 2

   Element at index 2 is 30. After deletion, arr = [10, 20] + [40, 50] = [10 20 40 50].

3. Final Answer:

   [10 20 40 50] -> Option A

4. Quick Check:

   Removed 30 at index 2 [OK]

Hint: Middle index = length/2; remove that element [OK]

Common Mistakes:

• Using float division instead of integer
• Deleting wrong index element
• Not expanding slice with ... in append

4. Find the error in this Go code that tries to delete the middle element from arr:

arr := []int{1, 2, 3, 4, 5}
mid := len(arr) / 2
arr = append(arr[:mid], arr[mid:]...)
fmt.Println(arr)

medium

A. The second slice should start from mid+1, not mid.

B. The middle index calculation is wrong.

C. append cannot be used with slices.

D. The array length is even, so no middle element.

Solution

1. Step 1: Check slice ranges in append

   The code uses arr[mid:] which includes the middle element again, not deleting it.

2. Step 2: Correct slice to exclude middle element

   To delete middle element at mid, use arr[mid+1:] to skip it.

3. Final Answer:

   The second slice should start from mid+1, not mid. -> Option A

4. Quick Check:

   Exclude middle element by slicing from mid+1 [OK]

Hint: Second slice must start at mid+1 to delete middle [OK]

Common Mistakes:

• Using arr[mid:] includes middle element again
• Miscalculating middle index for odd length
• Thinking append can't join slices

5. Given an array arr := []int{5, 10, 15, 20, 25, 30, 35}, write the Go code to delete the middle element and then insert 100 at the same middle index. What will be the final array?

hard

A. [5 10 15 20 100 25 30 35]

B. [5 10 15 20 25 30 35 100]

C. [5 10 15 20 25 30 35]

D. [5 10 15 100 25 30 35]

Solution

1. Step 1: Delete middle element

   Length is 7, middle index = 7/2 = 3. Delete element at index 3 (value 20): arr = append(arr[:3], arr[4:]...) = [5 10 15 25 30 35].

2. Step 2: Insert 100 at middle index

   Insert 100 at index 3: arr = append(arr[:3], append([]int{100}, arr[3:]...)...) = [5 10 15 100 25 30 35].

3. Step 3: Verify final array

   Final array is [5 10 15 100 25 30 35].

4. Final Answer:

   [5 10 15 100 25 30 35] -> Option D

5. Quick Check:

   Delete middle, insert 100 at same index [OK]

Hint: Delete middle then insert at same index using append [OK]

Common Mistakes:

• Not updating array after deletion before insertion
• Inserting 100 at wrong index
• Appending 100 at end instead of middle