Finding a Pair of character in a string [duplicate] - c++

Question

This question already has answers here:
Counting the number of occurrences of a string within a string
(5 answers)
Closed 2 years ago.
#include<bits/stdc++.h>
using namespace std;
int main(){
int t;
cin >> t;
while(t--){
int count = 0;
vector<string> v;
string resp;
cin >> resp;
v.push_back(resp);
for(int i = 0; i < v.size(); i++){
if(find(v.begin(), v.end(), "xy") != v.end()){
count++;
}
cout << count << endl;
}
return 0;
}
I want to find the character "xy" in the string for multiple test cases.
For input xy, my count value outputs correctly as 1.
But for the input xyxxy instead of 2 it gives the value as 0
It is only finding the value once but i want to check the count of xy in whole string
I tried to use the substring function as well but it failed to work

Answer 1

I don't get the idea of while loop but that worked for me.
#include <iostream>
#include <vector>
int main()
{
std::string str;
std::cin >> str;
int count = 0;
for (int i(0); i < str.size()-1; ++i)
{
if ((str[i] == 'x') && (str[i + 1] == 'y'))
{
++count;
}
}
std::cout << count;
}

Answer 2

You're looking for "xy" within a vector of strings, which in your example, has a single element, "xyxxy". Since "xy" is not equal to "xyxxy", you're not finding any matches.
But even if you tried to std::find "xy" within "xyxxy" itself - that would fail too, since std::find looks for a single element within a range (or rather, iterator pair).
Instead, you can use the string::find() method, as described here; or, as the case may be, std::string_view::find():
#include <string>
#include <vector>
#include <iostream>
#include <string_view>
int main() {
const std::string needle{"xy"};
std::string haystack;
std::cin >> haystack;
std::size_t count{0};
std::string_view remainder{haystack};
while(true) {
auto first_pos = remainder.find(needle);
if (first_pos == std::string_view::npos) { break; }
count++;
remainder = remainder.substr(first_pos+needle.length());
}
std::cout << "Found " << count << " occurrences of \"" << needle << "\"\n";
}
Note: This does not account for overlapping occurrences. If you want those, you could either always increase the starting position by just 1; or make your solution more complex by employing something Boyer-Moore or Knuth-Morris-Pratt search (see String search algorithms), and resuming it at the correct state after each occurrence found.