@va1ts7_100 i hope you are confused. By the way range of int is –2,147,483,648 to 2,147,483,647 .
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@hello_all Your code is runing fine. here is the linklink text i don’t see why it was giving tle.
Anyways u can place your strlen() out of for loop because in every iteration it will be called.
can i know the cases where my code fails.have been getting wrong answer even though i have tried a similar approach. here is my sol link text
Getting wrong answer but not sure why …Can anyone point out failing test case please
http://www.codechef.com/viewsolution/7021382
Help is appreciated !!
@Madhur . If you are saying changing System.out.print(new String(ans)) to System.out.println(new String(ans)) , I have already done that but still its not working .
Practice Link and Contest Link are swapped!
a test case where it fails?
Can someone look, where my solution is failing.
http://www.codechef.com/viewsolution/7145091
Thanks!!
can someone check this why is it getting error SIGSEGV
#include<bits/stdc++.h>
using namespace std;
int main()
{
int i,j,k,l,o,p,m,n,x;
cin>>x;
while(x–)
{
int sum=0;
string s;
cin>>s;
int arr[26]={0};
multimap<int,int> mymap;
multimap<int,int>::iterator it,itr;
for(i=0;i<s.length();i++)
arr[s[i]-‘a’]++;
int maxi=0;
for(i=0;i<26;i++)
{
maxi=max(maxi,arr[i]);
sum+=arr[i];
mymap.insert(make_pair(arr[i],i)); //count,alphabet
}
if(maxi<=((sum+1)/2))
{
string s1;
i=0;
it=mymap.end();
while(it!=mymap.begin())
{
it–;
n=it->first;
while(n!=0)
{
s1[i]=char(it->second+‘a’);
i+=2;
if(i>=s.length())
i=1;
n–;
}
}
for(i=0;i<s.length();i++)
{
cout<<s1[i];
}
cout<<endl;
}
else
cout<<"-1"<<endl;
}
}
The problem can be solved easily by using priority_queues. The queue stores the no. of occurrences(frequency) of a character and the character itself as a pair. Now you pop out the first two values from the queue. Decrease their occurrence frequency by 1 (until of course the frequency becomes 0) and add them to your final string and then again push the new pairs with updated frequencies. You do that till the priority queue becomes empty or it only has one element left, in which case you add the character of the remaining element its frequency number of times. Then just check your final string for equal adjacent characters. If so output -1 else you have your answer.
Here is the Link to my accepted code. View above solution.
Please someone help me find out for which case my solution gives wrong answer. I tried so many cases but didn’t find such case.
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