CareerCup

according to the question 'b' appears first in both s1 and s2
int fn(s1,s2)
{
char a,b;
int i=j=0;
while(s1)
{
a=s1[i++];
while(s2)
{
j=0;
b=s2[j++];
if (a==b)
retun j;
break;
}
}
return -1;
}

I agree with Pranav, as per the question input (S1 and S2) the answer is 'b'.
The time complexity of Pranav's solution is O(N^2) [ for each char of S1 traverse each char of S2].
Are we allowed to take any other data structure ? the question doesn't talk about it.
Let's say we are allowed to then we can take a HashTable and we will traverse S2 and the persist each char as key, and their position as value. [ in case of duplicate occurance the first one will get place in the HashTabe ].
And then we will traverse S1 and for each char we will do a lookup in the HashTabe ( which is O(1) ). so we will be able to solve this problem in O(N) time compleixty with extra memory[ the HashTable ].

1. Two strings s1[m], s2[n]. Take temp array A[26]

2. initialize all elements in A[26] to -1 ;

3. for each i=0 to n in S2[n]
if( A[ S2[i] - 'a' ] == -1 )
A[ S2[i] - 'a' ] = i;

4. Now temp array A[26] contains S2[] elements first indexes .

5. Take first element in S1[j] and look in Temp array (A[ S1[j]- 'a' ]), from temp array you will get first index in S[].

6. Time complexity O(n) & Space complexity O(1).