Key skill: After every
remove(), all elements to the right shift left by 1. After every add(index, element), all elements at that index and beyond shift right by 1. Write the list state after each line.Problem 1
Add and Remove Sequence
ArrayListlist = new ArrayList (); list.add("A"); list.add("B"); list.add("C"); list.add("D"); list.remove(1); list.add(1, "X"); System.out.println(list);
Answer: [A, X, C, D]
After adds: [A, B, C, D]. Remove index 1 (B): [A, C, D]. Add X at index 1: [A, X, C, D].
After adds: [A, B, C, D]. Remove index 1 (B): [A, C, D]. Add X at index 1: [A, X, C, D].
Problem 2
Remove in Forward Loop (Bug)
ArrayListnums = new ArrayList (); nums.add(1); nums.add(2); nums.add(3); nums.add(4); nums.add(5); for (int i = 0; i < nums.size(); i++) { if (nums.get(i) % 2 == 0) { nums.remove(i); } } System.out.println(nums);
Answer: [1, 3, 5]
Remove index 1 (value 2): list becomes [1, 3, 4, 5], i becomes 2. At i=2, value is 4 (even), remove: [1, 3, 5], i becomes 3, size is 3, loop ends. Both even numbers happen to be removed here, but this pattern can skip elements in other cases.
Remove index 1 (value 2): list becomes [1, 3, 4, 5], i becomes 2. At i=2, value is 4 (even), remove: [1, 3, 5], i becomes 3, size is 3, loop ends. Both even numbers happen to be removed here, but this pattern can skip elements in other cases.
Problem 3
set() Does Not Shift
ArrayListcolors = new ArrayList (); colors.add("red"); colors.add("green"); colors.add("blue"); colors.set(1, "yellow"); System.out.println(colors.size() + " " + colors.get(1));
Answer: 3 yellow
set(1, "yellow") replaces index 1 in place. Size stays 3. No shifting occurs — this is different from add(1, "yellow") which would shift and grow the list to size 4.Problem 4
Integer remove() Ambiguity
ArrayListdata = new ArrayList (); data.add(10); data.add(20); data.add(30); data.remove(1); System.out.println(data);
Answer: [10, 30]
remove(1) removes the element at index 1 (value 20), not the value 1. For ArrayList, the int parameter matches the index overload.Problem 5
Backward Removal Trace
ArrayListitems = new ArrayList (); items.add("cat"); items.add("dog"); items.add("cat"); items.add("bird"); items.add("cat"); for (int i = items.size() - 1; i >= 0; i--) { if (items.get(i).equals("cat")) { items.remove(i); } } System.out.println(items);
Answer: [dog, bird]
Backward traversal safely removes all occurrences of “cat” at indices 4, 2, and 0 without skipping any elements.
Backward traversal safely removes all occurrences of “cat” at indices 4, 2, and 0 without skipping any elements.
Need More ArrayList Practice?
The Cram Kit includes targeted ArrayList drills with step-by-step trace walkthroughs.
Get the Cram Kit — $29.99 1-on-1 Tutoring