PPA-3

Question

Accept an integer as input and print the time of the day. Use the following table for reference.

Input	Output
\(T < 0\)	INVALID
\(0 \leq T \leq 5\)	NIGHT
\(6 \leq T \leq 11\)	MORNING
\(12 \leq T \leq 17\)	AFTERNOON
\(18 \leq T \leq 23\)	EVENING
\(T \geq 24\)	INVALID

The input will be a single line containing an integer. The output should be one of these strings: NIGHT, MORNING, AFTERNOON, EVENING, INVALID.

Hint

How is this problem related to the previous problem? Do they have the same structure?

Think about the following code snippets:

Snippet-1

if T < 0:
    print('INVALID')
elif 0 <= T <= 5:
    print('NIGHT')

Snippet-2

if T < 0:
    print('INVALID')
if 0 <= T <= 5:
    print('NIGHT')

Come up with two different solutions to the problem, one that develops snippet-1 and the other which develops snippet-2. What is the difference between these two solutions that you arrive at? Which one is better?

Solutions

Solution-1

t = int(input())
if 0 <= t <= 5:
    print('NIGHT')
elif 6 <= t <= 11:
    print('MORNING')
elif 12 <= t <= 17:
    print('AFTERNOON')
elif 18 <= t <= 23:
    print('EVENING')
else:
    print('INVALID')

Solution-2

It is better to stick to if-elif-else ladders wherever possible and avoid a sequence of if conditions. The if-elif-else ladder is more efficient. As soon as one of the conditions is satisfied, the control will exit from the ladder. An if-if-if ladder on the other hand will end up checking every one of the if conditions for every possible input.

t = int(input())
if t < 0:
    print('INVALID')
if 0 <= t <= 5:
    print('NIGHT')
if 6 <= t <= 11:
    print('MORNING')
if 12 <= t <= 17:
    print('AFTERNOON')
if 18 <= t <= 23:
    print('EVENING')
if t >= 24:
    print('INVALID')

Video Solution