trie tree will be better to implement
On Thu, Feb 17, 2011 at 11:07 AM, Jammy <[email protected]> wrote:
> Greedy, always choose the remaining one that is the lexicographically
> smallest since choose any other way will yield a lexicographically
> greater one.
>
>
> void concante(char **strings, int n){
> qsort(strings,n,sizeof(char *),compareStr);
> }
>
> int compareStr(const void *a, const void *b){
> return strcmp(*(char **)a,*(char **)b);
> }
>
> On Feb 16, 10:05 pm, simplyamey <[email protected]> wrote:
> > Try this code.
> >
> > #include<iostream>
> > #include<string>
> > using namespace std;
> > int main()
> > {
> > string sArray[5] ={"aab","abc","bba","acc","bcc"};
> >
> > int n = 5;
> >
> > for(int i = 0;i < n; i++)
> > {
> > for( int j = 0; j < n - 1 ; j++)
> > {
> > if(sArray[j] > sArray[j+1])
> > {
> > string temp = sArray[j];
> > sArray[j] = sArray[j + 1];
> > sArray[j+1]= temp;
> > }
> > }
> > }
> >
> > string sTemp;
> > for(int k = 0; k < n ;k++)
> > {
> > sTemp = sTemp + sArray[k];}
> >
> > cout<<"printing result::"<< sTemp<<endl;
> >
> > }
> >
> > n = Number of strings.
> >
> > On Feb 17, 9:56 am, bittu <[email protected]> wrote:
> >
> >
> >
> >
> >
> >
> >
> > > Given a group of strings, find a string by concatenating all the
> > > strings in any order which produces the lexicographically smallest
> > > string.
> > > For e.g strings are acc bcc abb
> > > So the string formed should be abbaccbcc
> >
> > > Thanks
> > > Shashank
>
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