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AP CSA Unit-Level MCQ

2D Array Traversal

Practice mode with 15 Java-focused questions, immediate answer checks, and explanations.

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What 2d array traversal covers

Row-major traversal puts the row loop outside and the column loop inside, visiting an entire row before moving down. Swapping the loops gives column-major order, which visits one column top to bottom before moving right. The enhanced form nests a loop over each row array inside a loop over the grid. Per-row work such as a row sum requires resetting the accumulator at the top of each outer iteration.

Where students lose points

Using the outer loop variable to index columns, or bounding the inner loop with grid.length instead of grid[r].length, breaks any non-square grid. Declaring a per-row accumulator before the outer loop makes it a grand total instead of a row total, which is a common silent wrong answer.

How it shows up on the AP exam

Unit 4 material combining 2D arrays with the nested loops from Unit 2. Traversal order questions and per-row computation are both standard FRQ tasks.

For practice use only.

2D Array Traversal MCQ Practice

AP CSA 2D array traversal practice: nested row-major loops, column-major order, enhanced for loops over rows, and per-row accumulators.

Question 1 of 15

Answered 0 of 15

Choose one answer.

Code Tracing 01 - 2D Array Traversal: track the final printed value.

int[][] grid = {
  {2, 5, 1},
  {4, 3, 6},
  {7, 0, 8}
};
String out = "";
for (int c = 0; c < grid[0].length; c++) {
  for (int r = 0; r < grid.length; r++) {
    out += grid[r][c] + " ";
  }
}
System.out.print(out.trim());

All 15 2d array traversal questions

Work through the interactive quiz above first. This is the full 2d array traversal question set with worked solutions, so students can review any question after attempting it.

  1. 1.Code Tracing 01 - 2D Array Traversal: track the final printed value.

    int[][] grid = {
  {2, 5, 1},
  {4, 3, 6},
  {7, 0, 8}
};
String out = "";
for (int c = 0; c < grid[0].length; c++) {
  for (int r = 0; r < grid.length; r++) {
    out += grid[r][c] + " ";
  }
}
System.out.print(out.trim());
    • 8 6 1 0 3 5 7 4 2
    • 2
    • 2 4 7 5 3 0 1 6 8
    • 2 5 1 4 3 6 7 0 8
    Show worked solution

    Correct answer: 2 4 7 5 3 0 1 6 8

    Putting the column loop on the outside walks the grid column by column instead of row by row.

  2. 2.Code Tracing 12 - 2D Array Traversal: follow the variable updates.

    int[][] grid = {
  {4, 7, 3},
  {6, 5, 8},
  {9, 2, 10}
};
int total = 0;
for (int r = 0; r < grid.length; r++) {
  total += grid[r][r];
}
System.out.print(total);
    • 54
    • 14
    • 23
    • 19
    Show worked solution

    Correct answer: 19

    Using the same variable for both subscripts walks the main diagonal of the grid.

  3. 3.Code Tracing 28 - 2D Array Traversal: trace the branch and loop path.

    int[][] grid = {
  {7, 6, 1},
  {6, 8, 11},
  {7, 1, 13}
};
int best = 0;
int bestSum = 0;
for (int r = 0; r < grid.length; r++) {
  int sum = 0;
  for (int c = 0; c < grid[r].length; c++) sum += grid[r][c];
  if (r == 0 || sum > bestSum) { bestSum = sum; best = r; }
}
System.out.print(best + ":" + bestSum);
    • 25:1
    • 0:14
    • 2:21
    • 1:25
    Show worked solution

    Correct answer: 1:25

    The inner loop resets its accumulator on every row, so each row total is independent.

  4. 4.Code Tracing 39 - 2D Array Traversal: watch the index changes.

    int[][] grid = {
  {9, 8, 3},
  {5, 10, 7},
  {9, 3, 15}
};
int count = 0;
for (int[] row : grid) {
  for (int v : row) {
    if (v % 2 == 0) count++;
  }
}
System.out.print(count);
    • 2
    • 7
    • 9
    • 3
    Show worked solution

    Correct answer: 2

    The outer enhanced for loop hands back each row as an int array, which the inner loop then walks.

  5. 5.Code Tracing 50 - 2D Array Traversal: evaluate the state change step by step.

    int[][] grid = {
  {11, 6, 5},
  {4, 12, 9},
  {11, 1, 17}
};
int total = 0;
for (int r = 0; r < grid.length; r++) {
  for (int c = 0; c < grid[r].length; c++) {
    if ((r + c) % 2 == 0) total += grid[r][c];
  }
}
System.out.print(total);
    • 57
    • 56
    • 20
    • 76
    Show worked solution

    Correct answer: 56

    Testing the parity of r + c selects a checkerboard pattern of cells rather than whole rows or columns.

  6. 6.What is printed by the following code segment?

    int[][] g = {{1, 2, 3}, {4, 5, 6}};
String out = "";
for (int c = 0; c < g[0].length; c++)
{
    for (int r = 0; r < g.length; r++)
    {
        out += g[r][c];
    }
}
System.out.print(out);
    • 123456
    • 412536
    • 654321
    • 142536
    Show worked solution

    Correct answer: 142536

    Putting the column loop outside walks the grid column by column.

  7. 7.What is printed by the following code segment?

    int[][] g = {{1, 2}, {3, 4}, {5, 6}};
int t = 0;
for (int r = 0; r < g.length; r++)
{
    int rs = 0;
    for (int c = 0; c < g[r].length; c++) { rs += g[r][c]; }
    if (rs % 2 == 1) { t += rs; }
}
System.out.print(t);
    • 21
    • 0
    • 11
    • 3
    Show worked solution

    Correct answer: 21

    Every row total is odd here, so all three are added.

  8. 8.What is printed by the following code segment?

    int[][] g = {{1, 0}, {0, 1}};
int c = 0;
for (int r = 0; r < g.length; r++)
{
    for (int k = 0; k < g[r].length; k++)
    {
        if (g[r][k] == 1 && r == k) { c++; }
    }
}
System.out.print(c);
    • 1
    • 2
    • 4
    • 0
    Show worked solution

    Correct answer: 2

    Only the diagonal cells hold 1, and both satisfy the combined condition.

  9. 9.What is printed by the following code segment?

    int[][] g = {{1, 2, 3}, {4, 5, 6}};
int t = 0;
for (int r = 0; r < g.length; r++)
{
    for (int c = 0; c < g[r].length; c++)
    {
        if ((r + c) % 2 == 0) { t += g[r][c]; }
    }
}
System.out.print(t);
    • 12
    • 21
    • 9
    • 10
    Show worked solution

    Correct answer: 9

    The cells whose row and column indexes sum to an even number hold 1, 3, and 5.

  10. 10.What is printed by the following code segment?

    int[][] g = new int[3][3];
for (int i = 0; i < 3; i++)
{
    g[i][2 - i] = 1;
}
int t = 0;
for (int[] row : g)
{
    for (int v : row) { t += v; }
}
System.out.print(t);
    • 1
    • 9
    • 0
    • 3
    Show worked solution

    Correct answer: 3

    One cell per row is marked along the opposite diagonal.

  11. 11.What is printed by the following code segment?

    int[][] g = {{2, 4}, {6, 8}};
for (int r = 0; r < g.length; r++)
{
    for (int c = 0; c < g[r].length; c++)
    {
        g[r][c] /= 2;
    }
}
System.out.print(g[0][0] + "" + g[1][1]);
    • 14
    • 28
    • 13
    • 24
    Show worked solution

    Correct answer: 14

    Every cell is halved in place, and the two reported cells become 1 and 4.

  12. 12.What is printed by the following code segment?

    int[][] g = {{1, 2}, {3, 4}};
int[][] h = new int[2][2];
for (int r = 0; r < 2; r++)
{
    for (int c = 0; c < 2; c++)
    {
        h[c][r] = g[r][c];
    }
}
System.out.print(h[0][1] + " " + h[1][0]);
    • 2 3
    • 3 2
    • 1 4
    • 4 1
    Show worked solution

    Correct answer: 3 2

    Swapping the indexes on the write transposes the grid.

  13. 13.What is printed by the following code segment?

    int[][] g = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
int t = 0;
for (int r = 0; r < g.length; r++)
{
    for (int c = 0; c < g[r].length; c++)
    {
        if (r == 0 || r == g.length - 1 || c == 0
                || c == g[r].length - 1) { t += g[r][c]; }
    }
}
System.out.print(t);
    • 45
    • 30
    • 40
    • 20
    Show worked solution

    Correct answer: 40

    Every cell except the centre lies on the border, so the total is 45 minus 5.

  14. 14.What is printed by the following code segment?

    int[][] g = {{1, 2}, {3, 4}};
String out = "";
for (int r = g.length - 1; r >= 0; r--)
{
    for (int c = g[r].length - 1; c >= 0; c--)
    {
        out += g[r][c];
    }
}
System.out.print(out);
    • 1234
    • 3412
    • 2143
    • 4321
    Show worked solution

    Correct answer: 4321

    Both loops count down, so the grid is visited in fully reversed order.

  15. 15.What is printed by the following code segment?

    int[][] g = {{1, 1}, {1, 1}};
int c = 0;
boolean stop = false;
for (int r = 0; r < g.length && !stop; r++)
{
    for (int k = 0; k < g[r].length && !stop; k++)
    {
        if (g[r][k] == 1) { c++; }
        if (c == 3) { stop = true; }
    }
}
System.out.print(c);
    • 3
    • 4
    • 2
    • 1
    Show worked solution

    Correct answer: 3

    A flag tested in both loop conditions stops the traversal cleanly as soon as the count reaches three.