# HackerEarth One String No Trouble problem solution

In this HackerEarth One String, No Trouble problem solution A string S is called a good string if and only if two consecutive letters are not the same. For example, abcab and cda are good while abaa and accba are not. You are given a string S. Among all the good substrings of S, print the size of the longest one.

## HackerEarth One String No Trouble problem solution.

`#include<bits/stdc++.h>using namespace std;#define ll long long#define pb push_backconst int maxn = 2e5 + 20;int a[maxn];int main(){  ios_base::sync_with_stdio(false);  cin.tie(0);  cout.tie(0);  string s;  cin >> s;    char last = 'a' - 1;  int len = 0 , res = 0;  for(auto ch : s)  {      if(ch == last)          len = 0;      len++;      last = ch;      res = max(res , len);  }    cout << res << endl;}`

### second solution

`#ifndef BZ#pragma GCC optimize "-O3"#endif#include <bits/stdc++.h>#define FASTIO#define ALL(v) (v).begin(), (v).end()#define rep(i, l, r) for (int i = (l); i < (r); ++i)#ifdef FASTIO#define scanf abacaba#define printf abacaba#endiftypedef long long ll;typedef long double ld;typedef unsigned long long ull;using namespace std;template<typename T> T mo(T x, T y) { x %= y; return x <= 0 ? x + y : x; }int main() {#ifdef FASTIO    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);#endif    string s;    cin >> s;    int n = s.size();    int len = 0;    int ans = 0;    for (int i = 0; i < n; i++) {        len++;        if (i < n && s[i] == s[i + 1]) {            ans = max(ans, len);            len = 0;        }    }    ans = max(ans, len);    cout << ans << "\n";    return 0;}`