maintain an array indexed on alphabets 'a' to 'z'.
now count the occurence of each character taking *last character* of each
string
{"asdsb","csasaa", "bssssc", bddddc"}
in this case
arr[a]=1;
arr[b]=1;
arr[c]=2;
now start with each sting let first character be first and last character be
last
while (strings)
{
if(arr[first ]==0)
return false;
if(arr[first]==1)
{
if(first==last)
return false;
}
else{
arr[first--];
}
}/*end while
if at the end all elements of arr =0 then return true
else
return false;
On Sat, Aug 13, 2011 at 9:44 PM, Yasir <yasir....@gmail.com> wrote:
> An array of strings is given and you have to find out whether a circular
> chain with all character can be formed or not?
> Circular chain is formed in way that: if last char of a string is equal
> to first char of another string then it can be joined.:
> like axxxxb bxxxxxc ===> axxxxxbbxxxxxc (Notice that it got joined at
> b)
>
> example
> {"asdsb","csasaa", "bssssc"}
> Answer: TRUE
>
> {"asdsb","csasaa", "bssssc", bddddc"}
> Answer: FALSE
>
>
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--
Regards,
Kamakshi
kamakshi...@gmail.com
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