Java ArrayList Remove Elements - Understanding the Output

Let's analyze the output of this Java code snippet:

List<Integer> nums = new ArrayList<>();
nums.add(0);
nums.add(1);
nums.add(2);
nums.add(3);
nums.add(4);
nums.add(5);
nums.add(6);
for (int i = 0; i < nums.size(); i++) {
    System.out.println(nums.remove(i));
}

The output is:

0
2
4
6

Explanation:

The code iterates through the ArrayList nums using a for loop. Inside the loop, nums.remove(i) removes the element at index i and returns the removed element, which is then printed. However, the removal process changes the size and indices of the ArrayList, leading to unexpected behavior.

Here's a step-by-step breakdown:

1. Iteration 1:
   • i = 0: nums.remove(0) removes 0 and prints it. The ArrayList now contains [1, 2, 3, 4, 5, 6].2. Iteration 2:
   • i = 1: nums.remove(1) removes 2 and prints it. The ArrayList becomes [1, 3, 4, 5, 6]. Note that the index of 3 is now 1 due to the removal.3. Iteration 3:
   • i = 2: nums.remove(2) removes 4 and prints it. The ArrayList becomes [1, 3, 5, 6].4. Iteration 4:
   • i = 3: nums.remove(3) removes 6 and prints it. The ArrayList becomes [1, 3, 5]. The loop terminates because i is now equal to nums.size().

Key Takeaway: When removing elements from an ArrayList while iterating, be mindful that the indices shift after each removal. This can result in unexpected behavior if you're not careful.