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Below is a problem I encountered during an online course. There are a lot of similar questions asked on this, but I couldn't find a solution there as well as I believe my specific issue is a bit different from the rest.

Given a string s, find the first non-repeating character in the string If there is no such character print ‘@’. You are required to complete the function

Input Format: String s

Output Format: First non-repeating character or @

You are supposed to implement char char FirstNonRepeatedchar(char* str, int len): which takes a pointer to input string str and length of the string(including ‘0円’). The function should return either the first non repeating character or ‘@’

Input Constraint: All input characters will be either lower or upper case english alphabets.

 str[i] [a,z][A,Z]

There will not be any space in the input string.

 /************ Prefix Code Begins*******************/
 #include <iostream>
 #include <string>
using namespace std;
//function finds the first non-repeating character
 char FirstNonRepeatedchar(const char* , int);
int main(){
int len;
string line;
getline (cin, line);
len = int(line.length());
cout << FirstNonRepeatedchar(line.c_str(), len);
return 0;
}
/******************* Prefix Code Ends******************/

Also, note I cannot change anything in the prefix code above. I wrote 2 solutions: first one below is giving correct output.

char FirstNonRepeatedchar(const char* str, int len){
 int *count=0;
 for(int i=0;i<len;i++)
 {
 count[*(str+i)]++;
 }
 for(int i=0;i<len;i++)
 {
 if(count[*(str+i)]==1)
 return *(str+i);
 }
return '@';
} 

The above solution iterates the string twice. So, I tried to improve the complexity by implementing it using a hash.

char FirstNonRepeatedchar(const char* str, int len){
int count[26]={0};
int j;
char c;
if(isupper(*str)==1)
{ c='A';
}
else
c='a';
for(int i=0;i<len;i++)
{ j=*(str+i)-c;
 if(count[j]<0)
 ;
 else if(count[j]==0)
 count[j]=i+1;
 
 else if(count[j]>0)
 count[j]=-i;
 
}
int min=len+1;
for(int i=0;i<26;i++)
{ 
 if(count[i]>0 && count[i]<min)
 min=count[i];
 
}
if(min==len+1) 
return '@';
else
return *(str+min-1);
}

Is the second solution I gave the most optimized solution possible for this problem?