# Strings in C++ : Find the first non-repeated character in a string – 3

APPROACH 3

We can approach this problem using a lookup table. Using a lookup table, the running time can be significantly reduced.

For simplicity, let us assume that our input string can only contain characters from the English alphabet i.e. only 26 characters are possible, though each may be repeated any number of times.

The idea is to construct a lookup table having space for 26 entries. Each entry in the lookup table corresponds to a character in the input string.

We will proceed as follows— Create a lookup table of size 26. Initialize each entry in the lookup table to be 0. The entries in the lookup table correspond to the 26 characters in the English alphabet, and are equal to the number of times that particular character appears in the input string.

Traverse the input string. Say the string that you are given is “elephant”. The first character is ‘e’. Go to the lookup table entry corresponding  to ‘e’ and increment the count in that entry by 1. But which is the entry in the lookup table corresponding to ‘e’?

Let us call the lookup table L.

L[0] will store the count of ‘a’. L[1] will store the count of ‘b’. … L[25] will store the count of ‘z’.

So, for ‘e’, we go to L[4] and increment that count by 1.

Now, once we have finished traversing the string, and have the counts of all characters available to us in L, we need to output the first non-repeated character.

For this, we proceed as follows:

Traverse the input string again. Take the first character in the input string. Go to the entry corresponding to that character in L. (Note that we can do this in O(1) time.) Check if the count is 1. If so, output it. Else, go to the next character in the input string, and repeat the same procedure.

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Now, what would be the running time of this procedure?

Initializing the look up table takes O(1) [since the size of the lookup table is a constant.]

Traversing the string to populate the lookup table with appropriate counts takes O(n) time. (n is the length of the input string.)

Traversing the string again, to output the first non-repeated character takes O(n) time.

Hence, the total running time is O(n).

The memory requirement is that of a lookup table of constant size, which is O(1).

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In C++, we can code it as follows:

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#include <iostream.h>

//func f returns the first non-repeated character in s.

char f ( char *s)

{

//check for null string.

if ( ! s)

return ”;

int L [26] = {0}; //lookup table, initially all entries are 0.

//we now traverse the string to populate L.

char *t = s;

while (  *t != ”)

{

L [ *t – ‘a’]++; // e.g. if *t is ‘e’, we need to increment L[4] which is L [‘e’ – ‘a’].

t++;

}

//we have now populated L with counts.

//we now traverse s and using L, output the first non-repeated char.

t = s;

while ( *t != ”)

{

if ( L [ *t – ‘a’ ] == 1)

return *t;

}

//no non-repeated char in s.

return ”; //we may also write: return 0; (using the ASCII value of NUL)

}

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Now, what if remove the simplifying assumption made earlier and say  that s can contain any character. In this case, the lookup table needs to have 128 entries (one for each ASCII character).

For the preceding program, we use L [ ch – ‘a’ ] to access the lookup table entry corresponding to ch. In the new lookup table having 128 entries, we can use the ASCII values of the characters as direct indices into the lookup table, i.e. we can use L[ch] to access the count of character ch.

The remaining part of the code would remain more or less the same as above.

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