Question

Consider the following code segment: int[][] mystery = new int[3][3]; int counter = 0; while(counter < mystery.length) { for(int col = 0; col < mystery[counter].length; col++) { if(counter == 0) { mystery[col][counter] = 1; } else if(counter == 1) { mystery[counter][col] = 2; } else { mystery[counter][counter] = 3; } } counter++; }

What will the value of each element in mystery be after the execution of the code segment?

a) {{1, 0, 0} {2, 2, 2} {1, 0, 3}}

b) {{1, 1, 1} {2, 2, 2} {3, 3, 3}}

c) {{1, 2, 0} {1, 2, 0} {1, 2, 3}}

d) {{3, 0, 0} {2, 2, 2} {1, 0, 3}}

e) {{1, 2, 3} {1, 2, 3} {1, 2, 3}}

Answer 1

a) {{1, 0, 0} {2, 2, 2} {1, 0, 3}}

Fo the first iteration of while loop, counter will be 0, so in for loop if condition will be executed only because counter will be equal to 0.

counter == 0 returns true,

So, mystery[col][counter] = 1; => mystery[0][0] = 1;

mystery[col][counter] = 1; => mystery[1][0] = 1;

mystery[col][counter] = 1; => mystery[2][0] = 1;

Fo the second iteration of while loop, counter will be 1, so in for loop else if condition will be executed only because counter will be equal to 1.

counter == 1 returns true,

So, mystery[counter][col] = 2; => mystery[1][0] = 2;

mystery[counter][col] = 2; => mystery[1][1] = 2;

mystery[counter][col] = 2; => mystery[1][1] = 2;

Fo the third iteration of while loop, counter will be 2, so in for loop else condition will be executed only because counter will be equal to 2.

counter == 2 returns true,

So, mystery[counter][counter] = 3; => mystery[2][2] = 3;

Therefore, At the end mystery array will be {{1, 0, 0} {2, 2, 2} {1, 0, 3}} .