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# ERROR - Editorial

Author: Vivek Hamirwasia
Tester: Mahbub
Editorialist: Jingbo Shang

Cakewalk

# PREREQUISITES:

Programming Language.

# PROBLEM:

Determine whether the given binary string contains the substring (consecutive) "010" or "101".

# EXPLANATION:

You can use a loop and condition clauses to check directly. Also, you can use some built-in functions to solve this problem too.

For example, C++, we can use

int position = strstr(s, "010") - s;


to get the first occurrence of "010" and check whether this position is in range. Or, if string is used in C++, then

int position = s.find("010");


will work for string.

Similarly, Java, Python, etc... a lot of languages have such functions. Post your accepted solution and let's find which one is the shortest! I think it will be interesting :P

# AUTHOR'S AND TESTER'S SOLUTIONS:

Author's solution can be found here.
Tester's solution can be found here.

This question is marked "community wiki".

asked 13 Jan '14, 15:11

161446375
accept rate: 0%

19.8k350498541

4

I foolishly tried to implement "Z- algo".. :P. Someone kick me on my ass...

(14 Jan '14, 11:14) 2★

 2 @ rd13123013 When 010 or 101 comes at the last of the string ..then d=='\n' .Therefore ,it doesn't enters into While loop...so the count doesn't increasing .....and there lies the bug.  So , you can overcome this by using "do() ...while()..loop" :)  answered 11 Jul '14, 16:29 3★neal95 318●7●12 accept rate: 10%
 1 Using Python's elegance http://www.codechef.com/viewsolution/3250941 for _ in xrange(input()): s = raw_input() if "010" in s or "101" in s: print "Good" else: print "Bad"  P.S.: Waiting for a one liner in Perl or GolfScript :P UPDATE: If that's the case @shangjingbo, then how about a Perl regex check $s =~ m/"101"/  answered 13 Jan '14, 15:23 8.7k●19●48●98 accept rate: 9% 1 sound great, but you have benn beaten by @kuruma, have another try :D (13 Jan '14, 18:53) @shangjingbo >> I was not trying to enter any competition for shortest code. I was just demonstrating Python's elegance. :D (13 Jan '14, 19:26) Wow, great! (13 Jan '14, 22:40)  1 Well, when I'm good with Monads I can return and fully solve this one in Haskell, but, the built-in functions sure are a charm: isInfixOf "010" "10010"  :D answered 13 Jan '14, 16:29 3★kuruma 17.7k●72●143●209 accept rate: 8% I think it may be the shortest one :) (13 Jan '14, 18:52) The problem with this check is that it doesn't receive input and produce output, instead, it can be used in an interactive fashion, but, I'm loving Haskell so far :D (13 Jan '14, 21:09) kuruma3★  1 link text @s_pandey01 It is in the checking stage. if '010' in j and '101' in j: it should use 'or' instead of 'and'. If 'and' is used, then 'Good' will only print out when both '010' and '101' are in the line. answered 31 Jan '17, 09:28 11●1 accept rate: 0% @all: May i know what is wrong with this approach. http://www.codechef.com/viewsolution/3177973 I was checking characters at i, i-1, i-2 in a loop. Thanks :) answered 14 Jan '14, 16:49 4★squal 1.2k41533 accept rate: 14% 1 You missed corner case when length < 3, for example for 1 1  your code returns 'Good'... (14 Jan '14, 17:00) feeling bad to have missed this case :( thanks (14 Jan '14, 17:07) 4★ don't be sad, I can write a book about stupid bugs I did in contests :-D My advice here is, do not use some side effects (like if I scanned whole string), use additional boolean flag that is set to false at the beginning and when you really find what you are looking for, set it to true. Also you do not need to use such difficult input reading, see my code (http://www.codechef.com/viewsolution/3153331 ) for this problem if you want or this one (http://www.codechef.com/viewsolution/3250870 ), to see how to use StringTokenizer. (14 Jan '14, 17:46) what is wrong with this... # include <stdio.h> # include <string.h> int main(void) { int t; char a , b , c ,d ,count ; scanf("%d" , &t); while(t--) { count = 0; a = getchar(); b = getchar(); c = getchar(); while( (d = getchar()) != '\n') ( (a=='0' && b=='1' && c=='0') || (a == '1' && b=='0' && c=='1')) ? ( (count++) , (a = b) , ( b = c) , (c = d )): ((a = b) , ( b = c), (c = d )); (count > 0)?(printf("Good\n")):(printf("Bad\n")); } return 0; } (15 Jan '14, 02:10) Try pasting ur code after pressing CTRL+K (01 Aug '15, 21:35) what is wrong wid this... # include <stdio.h> # include <conio.h> # include <string.h> int main(void) { int t; char a , b , c ,d ,count ; scanf("%d" , &t); while(t--) { count = 0; a = getchar(); b = getchar(); c = getchar(); while( (d = getchar()) != '\n') ( (a=='0' && b=='1' && c=='0') || (a == '1' && b=='0' && c=='1')) ? ( (count++) , (a = b) , ( b = c) , (c = d )): ((a = b) , ( b = c), (c = d )); (count > 0)?(printf("Good\n")):(printf("Bad\n")); } getch(); return 0; } answered 15 Jan '14, 02:09 0 accept rate: 0%  0 int main() { int t; cin>>t; while(t--) { std::string str(""); cin>>str; if((str.find("010")==!std::string::npos) || (str.find("101")==!std::string::npos)) { cout<  0 What is wrong with this code? Similar code accepted in java but this is not working. class Sample7 @t=0 @str def initialize t=gets.to_i for i in 0...t str=gets if str.include?"101"||"010" puts "Good" else puts "Bad" end end  end end obj=Sample7.new answered 10 Jun '16, 17:19 1★s9w311 1 accept rate: 0% WHAT IS WRONG IN MY CODE? # include<iostream> # include<string.h> using namespace std; int main() { int t; long int l,i,c,d,j,e; cin>>t; while(t--) { char s[100005]; cin>>s; l=strlen(s); for(i=0;i<l-2;i++) { e=2; c=0; j=i; while(e--) { if(s[j]!=s[j+1]) c++; else break; j++; } if(c==2) { cout<<"good"<<endl; break; } } if(c!=2) cout<<"bad"<<endl; } return 0; } answered 21 Jul '16, 13:12 1 accept rate: 0%  0 Guys can anyone tell what is wrong with this python code for the problem, according to me it is correct..... lst=[] a=int(input()) i=0 while i  0 Guys can anyone tell what is wrong with this python code for the problem, according to me it is correct..... lst=[] a=int(input()) i=0 while i # include<iostream> # include<string> using namespace std; int main(){ int a; cin>>a; while(a--){ string s;int c=0; cin>>s; for(int i=0;i<(s.length()-1)&&c<2;i++){ if(s[i]!=s[i+1])c++; else c=0; } if(c==2)cout<<"Good"<<endl; else cout<<"Bad"<<endl; } return 0; } answered 30 Jan '17, 22:49 1 accept rate: 0%  0 import java.util.Scanner; class untitled { public static void main(String[] args) { Scanner kb=new Scanner(System.in); int test=kb.nextInt(); while(test-->0) { String s1=kb.next(); String s2="010"; String s3="101"; if(s1.contains(s2)||s1.contains(s3)) System.out.println("Good"); else System.out.println("Bad"); } } }  benefit of java!..... :-P answered 06 Apr '17, 16:58 15●3 accept rate: 0%  toggle preview community wiki: Preview ### Follow this question By Email: Once you sign in you will be able to subscribe for any updates here By RSS: Answers Answers and Comments Markdown Basics • *italic* or _italic_ • **bold** or __bold__ • link:[text](http://url.com/ "title") • image?![alt text](/path/img.jpg "title") • numbered list: 1. 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question asked: 13 Jan '14, 15:11

question was seen: 5,534 times

last updated: 06 Apr '17, 16:58