BESTMOVIE - Editorial

editorial
, ,
1

PROBLEM LINK:

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

Author: raysh07
Tester: sushil2006
Editorialist: iceknight1093

DIFFICULTY:

Cakewalk

PREREQUISITES:

None

PROBLEM:

You have N movies to choose from to rent. The i-th movie has a rating of A_i and a cost of B_i.
You will only pick a movie with a rating of at least 7. What’s the minimum cost of renting a movie?

EXPLANATION:

It suffices to directly implement the condition outlined in the statement: only consider movies with a rating (A_i) that’s \geq 7, and take the minimum cost (B_i) among them.

That gives the following algorithm:

  • Initialize \text{ans} = \infty
    In practice, you can just set it to some number that’s larger than any B_i, like 1001.
  • Next, loop over i from 1 to N.
    • If A_i \lt 7, ignore this movie.
    • If A_i \geq 7, set \text{ans} = \min(\text{ans}, B_i)

In the end, print \text{ans}.
If there’s no movie with rating \geq 7, the value of \text{ans} will be what you initialized it to, i.e. a value greater than 1000. In this case, output -1 instead.

TIME COMPLEXITY:

\mathcal{O}(N) per testcase.

CODE:

Editorialist's code (PyPy3) 
for _ in range(int(input())):
    n = int(input())
    ans = 2000
    for i in range(n):
        a, b = map(int, input().split())
        if a >= 7: ans = min(ans, b)
    if ans == 2000: ans = -1
    print(ans)