Question

Efficient and O(n) code for this in c??
I know that solution of O(n*n)

stringCompare(str1, str2){
int freq1[100] = {0}, i;
int freq2[100]  = {0};

for(i=0; i<=strlen(str1); i++){
     freq1[str1[i]]+ = 1;
}

for(i=0; i<=strlen(str2); i++)

{
     freq2[str2[i]]+ = 1;
 }

for(i=0;i<26;i++){
     if(freq1[i]!=freq2[i])
      return 0;
   return 1;

}

}

Answer 1

I modified MAK‘s pseudocode slightly so it only uses one frequency count array. A positive value in the final freq array means a char in s1 is not in s2. A negative value signals extra chars in s2.

function same(s1,s2):
    freq=array of zeros

    for i=0 to length of s1:
       freq[s1[i]]+=1

    for i=0 to length of s2:
       freq[s2[i]]-=1

    for i=0 to alphabet_size:
        if not freq[i]=0
            return "no"
    return "yes"