CareerCup

Oh..
It will change the order.

MSGEEK, dont know who made you MS Geek. Thats the whole point buddy to push the elements to HashTable.

Think man, think

char * oddstring(char *s)
{
char *start = s;
char a[256];
for (int i=0;i<256;i++)
a[i] = 0;
while(*s)
a[*s++]++;
s = start;
char *left = start;
while(*s)
{
if (a[*s]%2 == 1)
{
*left++ = *s;
a[*s] = 0;
s++;
}
else
s++;
}
*left = '\0';
return start;
}

if memory is not a constraint, then create an array of size 26 where each index correspond to an alphabet[eg 0 ->a ,1->b..25-> z]. Intialize array with zero.Now iterate thru string and increment the correspondent array index.Now print all the alphabets whose count is odd.

Thinker... has gone nuts. MSGeek was right. Push elements in the HashTable means read each Character of the string and push it into Hashtable keeping its count as Value field. Keep on incrementing count as and when repitition occurs. In the end, scan the Hashtable, do value%2 check and create a string.

As simple as that!

use a bitset.. alternate the bit for the char which is found..

go through the bitset when the string ends.. chars which have their bits set to 1 appear odd number of times..

{{
//sort the string
void sort(char *c){
int length = strlen(c);
for(int i=1; i<length; i++){
int j = i;
int t = c[j];
while(j>0 && c[j-1]>t){
a[j]=a[j-1];
j--;
}
a[j]=t;
}
}
void deleteOddOccure(char *s){
//check for null
if(*s == '\0')
return;
sort(s);
int length = strlen(s);
char *k = s;
int j=0;
for(int i=0; i<length-1; ){
if( s[i]==s[i+1]){
i=i+2;
}else{
k[j]=a[i];
i++;
j++;
}
}
k[j]='\0';

}
//

}}

Simple solution would be to use Hash table.
1)Push all the elements in to the hashtable as keys along with count of their repetitions as values.
2)Now iterate through the elements in the hash table
3)Display only elements where (value%2==1)and these would be the elements with odd number of repetitions.

Lemme know your thoughts.