Determine if two strings are anagram? - c

Question

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
static int comp(const void *a, const void *b);
int main(int argc, char ** argv)
{
char s1[]="";
char s2[]="";
char *fmt = "%[^\n]%*c";
printf("Enter the first word? ");
scanf(fmt,&s1);
printf("Enter the second word? ");
scanf(fmt,&s2);
qsort(s1, strlen(s1), 1, comp);
qsort(s2, strlen(s2), 1, comp);
printf("%s : %s - %s\n", s1, s2, strcmp(s1, s2) ? "No" : "Yes");
return 0;
}
static int comp(const void *a, const void *b)
{
const char *pa = (char*)a;
const char *pb = (char*)b;
return
(*pa > *pb) ? 1 :
(*pa < *pb) ? -1 :
0;
}
When I key in 'smartest' as my first string and 'mattress' as my second string , my output becomes
emrsstt : aemrsstt - No
Why is the first character of the string missing?
Been sitting on this for hours. Gladly appreciate if someone could point out the silly mistake I may have made.

Answer 1

A smart way of doing this is to first check for a probable anagram. You do this by
Comparing the string lengths.
XORing every char from the total set; drawn from both strings.
If the result of (2) is not zero then the strings cannot possibly be anagrams. The advantage of doing this is that (2) is extremely fast in c and most strings are not anagrams, even if they are the same length. If your feeling particularly ambitious, you could combine (1) and (2) into a single O(N) algorithm!
Once you've done that, call your main code body. Your problem is in not allowing for the null terminator on each string. Your need to allocate enough space for the string characters plus one extra element for the null terminator.

Answer 2

Basically when you have to check whether any 2 set of values are same or not XOR operation is helpful.
Input Output
A B
0 0 0
0 1 1
1 0 1
1 1 0
So using this we can solve your problem as done below:
#include <stdio.h>
#include<stdlib.h>
#include <string.h>
int main()
{
int i=0,j=0,f;
char a[30] = "gooo";
char b[30] = "ooog";
if(strlen(a) == strlen(b))
{
while(a[i] != '\0')
{
j = 0;
f =0;
while(b[j] != '\0')
{
if( !(a[i] ^ b[j]) )
{
f =1;
break;
}
j++;
}
if(f)
i++;
else
break;
}
printf("%d\n",i);
if(i == strlen(a))
printf("Yes\n");
else
printf("No\n");
}
return 0;
}

Answer 3

Please check and the below solution for the anagram problem. This solution checks if two strings have identical counts for each unique char
bool anagram (char s[], char t[])
{
if (strlen(s) != strlen(t)){
return false;
}
int letters[256];
memset(letters,(int)0,sizeof(letters));
int numUniqueChars = 0;
int numCompletedT = 0;
int i;
for (i=0; i < strlen(s); i++){
if (letters[(int)s[i]]==0){
++numUniqueChars;
}
letters[s[i]]=letters[s[i]]+1;
}
for (i=0; i < strlen(t); i++){
int c = (int)t[i];
if (letters[c] == 0){
return false;
}
letters[c]=letters[c]-1;
if (letters[c] == 0){
numCompletedT++;
if (numUniqueChars==numCompletedT){
return i == strlen(t)-1;
}
}
}
return false;
}