Two-gram | Editorial by khushalagarwal

A very good problem consisting concepts of mapping and strings.
To solve it we run a loop from starting character to s.length()-1 (penultimate) character and store every two consecutive characters in a temporary string variable x and push it to the 
vector.
Then we create a map of string and count how many times its present so <string,int> map helps us here.
Finally we store in max the maximium time a string is present by iterating over value of every string in map and do a search at last to find the first max key which has value as max. hence we print the key.
Hence, Kudos it is solved.

```
#include<bits/stdc++.h>
using namespace std;
int main() {
    int n;
    cin >> n;
    string s;
    cin >> s;
    map<string, int> m;
    vector<string> res;
    for (int i = 0; i < s.length() - 1; i++) {
        string x = "";
        x += s[i];
        x += s[i + 1];
        res.push_back(x);
    }
    for (auto j:res)
        m[j]++;
    int max = 0;
    string ans = "";
    for (auto k:m) {
        if (k.second > max) {
            max = k.second;
        }
    }
    for (auto k:m) {
        if (max == k.second) {
            cout << k.first;
            break;
        }
    }
}
```

Two-gram | Editorial by khushalagarwal

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