How many hashes (#) will the following snippet send to the console? lst = [[x for x in range(3)] for y in range(3)] for r in range(3): for c in range(3): if lst[r][c] % 2 != 0: print(“#”)

lst = [[x for x in range(3)] for y in range(3)]
 
for r in range(3):
    for c in range(3):
        if lst[r][c] % 2 != 0:
            print("#")

• three
• nine
• six
• zero

Explanation: Let’s analyze the code:

• Line 1: we’re using a list comprehension to create the following two-dimensional array:
  [[0, 1, 2], [0, 1, 2], [0, 1, 2]]
• the program enters the first for loop with the r iteration variable taking the value of the first item inside the sequence generated by the range(3) function, which is 0. Then the program enters the second (nested) for loop with the c iteration variable taking the value of the first item inside the sequence generated by the range(3) function, which is 0. The following condition is checked:
  if lst[0][0] % 2 != 0, then print a hash (#) to the screen. The value returned from the lst list stored under the [0][0] indexes is 0, so the condition evaluates to false (0 % 2 != 0 → False), which means a hash is not printed to the screen at this time, and the program execution jumps back to the outer loop. The iterations are repeated for the following index pairs: [0][1], [0][2], [1][2], [1][0], [1][1], [1][2], [2][2], [2][0], [2][1], and [2][2].
• the print() function will print a hash when the following index pairs become part of the conditional check:
• [0][1] → 1 because if 1 % 2 != 0 evaluates to True
• [1][1] → 1 because if 1 % 2 != 0 evaluates to True
• [2][1] → 1 because if 1 % 2 != 0 evaluates to True
• Therefore, the print() function will output three hashes to the screen.

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