Strategy: Draw the 2D array as a grid on paper. Label rows (0, 1, 2...) on the left and columns (0, 1, 2...) on top. Trace through the nested loops writing down which cell is visited at each step.
Problem 1
Row-Major Traversal Output
int[][] grid = {{1, 2, 3},
{4, 5, 6}};
for (int r = 0; r < grid.length; r++)
{
for (int c = 0; c < grid[0].length; c++)
{
System.out.print(grid[r][c] + " ");
}
}
Answer: 1 2 3 4 5 6
Row-major: visits all columns in row 0 first (1 2 3), then all columns in row 1 (4 5 6).
Row-major: visits all columns in row 0 first (1 2 3), then all columns in row 1 (4 5 6).
Problem 2
Column-Major Traversal Output
int[][] grid = {{1, 2, 3},
{4, 5, 6}};
for (int c = 0; c < grid[0].length; c++)
{
for (int r = 0; r < grid.length; r++)
{
System.out.print(grid[r][c] + " ");
}
}
Answer: 1 4 2 5 3 6
Column-major: visits all rows in column 0 (1 4), then all rows in column 1 (2 5), then column 2 (3 6).
Column-major: visits all rows in column 0 (1 4), then all rows in column 1 (2 5), then column 2 (3 6).
Problem 3
Row Sum Accumulation
int[][] data = {{2, 4, 6},
{1, 3, 5},
{7, 8, 9}};
for (int r = 0; r < data.length; r++)
{
int rowSum = 0;
for (int c = 0; c < data[0].length; c++)
{
rowSum += data[r][c];
}
System.out.println("Row " + r + ": " + rowSum);
}
Answer:
Row 0: 12
Row 1: 9
Row 2: 24
The accumulator
Row 0: 12
Row 1: 9
Row 2: 24
The accumulator
rowSum resets to 0 for each row. Row 0: 2+4+6=12. Row 1: 1+3+5=9. Row 2: 7+8+9=24.Problem 4
Diagonal Access
int[][] matrix = {{10, 20, 30},
{40, 50, 60},
{70, 80, 90}};
int sum = 0;
for (int i = 0; i < matrix.length; i++)
{
sum += matrix[i][i];
}
System.out.println(sum);
Answer: 150
Main diagonal: matrix[0][0]=10, matrix[1][1]=50, matrix[2][2]=90. Sum = 10+50+90 = 150. This only works on square grids.
Main diagonal: matrix[0][0]=10, matrix[1][1]=50, matrix[2][2]=90. Sum = 10+50+90 = 150. This only works on square grids.
Problem 5
Modify Grid In Place
int[][] grid = {{1, 2},
{3, 4}};
for (int r = 0; r < grid.length; r++)
{
for (int c = 0; c < grid[0].length; c++)
{
grid[r][c] = grid[r][c] * (r + c);
}
}
// What is grid now?
Answer: {{0, 2}, {3, 8}}
grid[0][0] = 1*(0+0) = 0. grid[0][1] = 2*(0+1) = 2. grid[1][0] = 3*(1+0) = 3. grid[1][1] = 4*(1+1) = 8.
grid[0][0] = 1*(0+0) = 0. grid[0][1] = 2*(0+1) = 2. grid[1][0] = 3*(1+0) = 3. grid[1][1] = 4*(1+1) = 8.
Crush Every 2D Array Problem
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