CHKEV - Editorial

editorial
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PROBLEM LINK:

Practice
Contest: Division 1
Contest: Division 2
Contest: Division 3
Contest: Division 4

Author: iceknight1093
Tester: sushil2006
Editorialist: iceknight1093

DIFFICULTY:

Cakewalk

PREREQUISITES:

None

PROBLEM:

You’re given L and R.
Does there exist an even integer among the values L, L+1, \ldots, R?

EXPLANATION:

If L \lt R the answer is always Yes, because we have both L and L+1 with us, and one of them is definitely even.

If L = R we only have a single integer with us, so the answer is Yes if and only if it’s even.

Thus,

  • If L = R and L is odd, the answer is No
  • In every other case the answer is Yes.

TIME COMPLEXITY:

\mathcal{O}(1) per testcase.

CODE:

Editorialist's code (PyPy3) 
l, r = map(int, input().split())
if l == r and l%2 == 1: print('No')
else: print('Yes')
2

I have seen people write if-else conditions within print, reduces the need to write print twice for smaller codes…

Alternate Approach with same logic

l, r = map(int, input().split())
print(“No” if l == r and l & 1 else “Yes”)