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implementStrStr
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implementStrStr
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/*Implement strStr().
Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
Clarification:
What should we return when needle is an empty string? This is a great question to ask during an interview.
For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C's strstr() and Java's indexOf().
*/
#include <bits/stdc++.h>
#define db1(x) cerr << #x << "=" << x << '\n'
#define db2(x, y) cerr << #x << "=" << x << "," << #y << "=" << y << '\n'
#define db3(x, y, z) cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << '\n'
#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define repA(i, a, n) for (ll i = a; i <= (n); ++i)
#define repD(i, a, n) for (ll i = a; i >= (n); --i)
#define mp make_pair
#define sp << ' ' <<
#define pb push_back
#define FastIO \
ios::sync_with_stdio(0); \
cin.tie(0); \
cout.tie(0);
using namespace std;
using ll = long long;
int strStr(string haystack, string needle)
{
if (needle.size() == 0)
return 0;
int i = 0, j = 0, pos = 0;
while (i != haystack.size())
{
int found = 0;
// gateway
if (needle[j] == haystack[i])
{
// assume that needle is found
found = 1;
pos = i;
// check if needle is found in haystack
while (j != needle.size() && i != haystack.size())
{
if (needle[j] == haystack[i])
{
i++;
j++;
}
else
{ // assumption was wrong, reset needle counter
found = 0;
j = 0;
i = pos + 1;
}
}
}
else
{ // if first character not same , then check ahead in haystack
i++;
}
if (found && j == needle.size())
return pos; // if found then return pos , else continue, needle should completely be there
}
return -1;
}
int main()
{
FastIO
ll t;
cin >> t;
while (t--)
{
string v, nee;
cin >> v >> nee;
cout << strStr(v, nee);
cout << v << "\n"
<< nee << "\n";
cout << '\n';
}
return 0;
}