programmingevan Does this challenge seem oddly easy to anyone else? Like a warmup exercise?
Unchained "Warmup" is an over estimation.
Omnikron13 You are not alone. As far as programming is concerned this is just I/O and a single addition, so... 'challenge' isn't the word I'd use to describe this. =P
chestaisokay This was very very easy . I hope there are some good challenges ahead .
c650Alpha I agree. I saw this problem, thought I was overthinking it, and then checked the Discussion to see if I was right.
_faker_ Exactly the same happpened with me!
aloksingh3007 same here !
RubberSideDown The solution to this challenge is an excellent, basic example of an application of the pigeonhole principle, which definitely Falls under the fundamentals category.
jpjaspreetmahal I am here just to read comments XD
AllisonP This should really be called Maximum Draws (because it's worst-case scenario). Minimum draw would always be 2.
lvsz_ I like the idea of a best worst case scenario.
aigookenzaki ahhmm ahh can you help me to this account cuase i dont understand the terms can you help me plzzz
whyzee Thank you. Thats what I thought too.
Debu_Basu Yeah exactly! It occured to me once, but then it was clear when I saw the sample I/O :)
aigookenzaki ahh i am new in this account i need help cuase i dont understand can you help me pls
Neon Simple 3 pairs : R,B,G Draw 1 : R1 Draw 2 : B1 Draw 3 : G1 Draw 4 {R2,B2,G2} any of the pick makes a pair with previously picked color :)
revernus Thanks, I was able to understand the problem only after reading your explanation.
gokuvsvegita lol
panindra_ganni27 lol
sakib_muhit Perfect... :D
chvnikhil good
codermehraj thanks
sankireddyvinee1 nice..
panditsaurav6 Just a request is there any way we can write it as permutation and combination problem.
foxtrot I'm surprised nobody mentioned the Pigeonhole Principle :-P
thakursaurabh98 That was the main concept behind the question but still simpler ways! :D
sachingadagi Am I the only one who did (pairs+ 1) ? :P
shunn789 No.
trendsetter37 No.
M4GNV5 No. ++ ?
badillosoft Yes, you're right :D
num of pairs plus one, because in the worst case you'll get off each one of the different pairs, thus, you should get one more.
TheVoidMaster i did that too.
wjm91 This one is really easy. If you have n-pairs of socks, that means that worst case you pick one sock from each pair. From there any sock you choose has to be a match.
So by the pidgeon hole principle we get the answer to just be n+1
Surprisingly basic to be honest
markbinder1234 I have absolutley NO CLUE how to solve this! I'am just shocked at just how many people say it's easy and do it. This problem is just unbielievebly confusing and impossible to understand. I'm new. Probably that's why but still HackerRank should make questions just a tad bit more understandable for people who just joined HackerRank. Please whoever is reading this just please explain the "thing" about HackerRank that everyone knows and uses it to solve these kind of stuff. Fyi I know programming very well. - Mark
GRAVI7APS import java.io.; import java.util.;
public class Solution {
public static void main(String[] args) { Scanner scan = new Scanner(System.in); // just to remove the first variable int numberOfLines = scan.nextInt(); // the number of draws is the number of pairs + 1 while(scan.hasNext()) { System.out.println(scan.nextInt()+1); } }}
GRAVI7APS the maximum number of draws, just try to count the maximum number of draws when you have 1 pair of socks, the 2 and so on and you will quickly get the pattern. Don't worry, buddy. Sometimes you get stuck with a problem, who anybody seems to solve on the fly and sometimes you solve the hardest problem, that anybody is able to solve.
techniker Well Initially I though that the person has to remove all the socks from the drawer in order to get the last matching sock. Then test cases failed. Thought harder & got it right.
aoa1usa Draw a picture and keep track of your pairs and draws you need. You should see a pattern after about three rows of data.
codespinach One line of Python 3:
[print(int(input()) + 1) for i in range(int(input()))]canhascodez Ruby
puts (1..gets.to_i).map {|_| gets.to_i.succ }
Alternately:
puts Array.new(gets.to_i) {|_| gets.to_i.succ }
Or:
gets.to_i.times { puts gets.to_i.succ }
The last is probably most idiomatic.
ajinkyapathak import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;
public class Solution {
public static void main(String[] args) { /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */ Scanner sc=new Scanner(System.in); int t=sc.nextInt(); while(t-->0) { int p=sc.nextInt(); int a=p+1; System.out.println(""+a); } }}
This Code Runs All Test Cases...
arno_lee It takes more to reason about possible pitfals, because you're like "it cannot be that easy!", the level should be marked as Moderate
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