Java.Collections.Suggest an efficient algorithm for removing several adjacent elements from the middle of a list implemented by ArrayList.

If you want to efficiently remove several adjacent elements from the middle of an ArrayList, there’s a much better way than calling remove(index) in a loop.

Let’s go step by step 👇

❌ Inefficient way (common mistake):

for (int i = 0; i < count; i++) {
    list.remove(startIndex); // each remove() causes a shift!
}

This is O(n × k) time, where:

• n is the size of the list
• k is the number of elements being removed

Because each call to remove() shifts elements left one-by-one.

✅ Efficient way: Use subList().clear()

list.subList(startIndex, endIndex).clear();

Removes all elements in one bulk operation

Time complexity: O(n – endIndex) – shifts remaining tail only once

List<String> list = new ArrayList<>(
    Arrays.asList("A", "B", "C", "D", "E", "F", "G")
);

// Remove 3 elements from index 2 (i.e., "C", "D", "E")
int start = 2;
int count = 3;
list.subList(start, start + count).clear();

System.out.println(list); // [A, B, F, G]

🧠 Why is this efficient?

• subList() gives you a view of part of the list.
• clear() on the view:
  • Removes all elements in one go.
  • Shifts tail elements only once.
  • No multiple reallocations or shifting.

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🚀 Summary:

Method	Time Complexity	Efficiency
Loop with remove(index)	❌ O(n × k)	Slow
subList(start, end).clear()	✅ O(n – end)	Fast