HackerEarth Mancunian in Palindromia problem solution

In this HackerEarth Mancunian in Palindromia problem solution Mancunian lives in the magical far-away land of Palindromic. Due to his patriotic nature, he wants each and every one of his friends' names to be a palindrome. Note that Mancunian's patriotism does not extend towards his own name.

To please him, each of his friends decides to change their name so that it becomes a palindrome. They can do that by choosing at most two non-overlapping substrings of their own name and reversing them.

Given a list of Mancunian's friends' names which consist only of lowercase letters, count the total number of friends who will be successful.

HackerEarth Mancunian in Palindromia problem solution.

#include <iostream>
#include <algorithm>
#include <stack>
#include <vector>
#include <cassert>
#include <queue>
#include <cmath>
#define ff first
#define ss second
#define pb push_back
#define MOD (1000000007LL)
#define LEFT(n) (2*(n))
#define RIGHT(n) (2*(n)+1)

using namespace std;
typedef long long ll;
typedef pair<ll, ll> ii;
typedef pair<ll, ii> iii;

ll pwr(ll base, ll p, ll mod = MOD){
ll ans = 1;while(p){if(p&1)ans=(ans*base)%mod;base=(base*base)%mod;p/=2;}return ans;
}


bool is_palindrome(string str){
    string temp = str;
    reverse(temp.begin(), temp.end());
    return (str == temp);
}


int main(){

    ios_base::sync_with_stdio(0);

    int t, len, ans = 0;
    cin>>t>>len;
    assert(t <= 10);
    assert(len <= 50);
    while(t--){

        string str;
        cin>>str;

        int n = str.length();
        assert(n <= len);

        bool yes = false;
        if(is_palindrome(str))  yes = true;

        for(int l1=0;l1<n;l1++)
        for(int r1=l1;r1<n;r1++){
            string temp = str;
            reverse(temp.begin()+l1, temp.begin()+r1+1);
            if(is_palindrome(temp)){
                yes = true;
            }
        }

        for(int l1=0;l1<n;l1++)
        for(int r1=l1;r1<n;r1++)
            for(int l2=r1+1;l2<n;l2++)
            for(int r2=l2;r2<n;r2++){
                string temp = str;
                reverse(temp.begin()+l1, temp.begin()+r1+1);
                reverse(temp.begin()+l2, temp.begin()+r2+1);
                if(is_palindrome(temp))
                    yes = true;
            }

        if(yes){
            ans++;
        }
        else{
        }
    }

    cout<<ans;
    return 0;
}

Second solution

#include <string>
#include <vector>
#include <map>
#include <list>
#include <iterator>
#include <cassert>
#include <set>
#include <queue>
#include <iostream>
#include <sstream>
#include <stack>
#include <deque>
#include <cmath>
#include <memory.h>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <algorithm>
#include <utility>
#include <time.h>
#include <complex>
using namespace std;

#define FOR(i, a, b) for(int i=(a);i<(b);i++)
#define RFOR(i, b, a) for(int i=(b)-1;i>=(a);--i)
#define FILL(A,value) memset(A,value,sizeof(A))

#define ALL(V) V.begin(), V.end()
#define SZ(V) (int)V.size()
#define PB push_back
#define MP make_pair
#define Pi 3.14159265358979
#define x0 ikjnrmthklmnt
#define y0 lkrjhkltr
#define y1 ewrgrg

typedef long long Int;
typedef unsigned long long UInt;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef pair<Int, Int> PLL;
typedef pair<double, double> PDD;
typedef complex<double> base;

const int INF = 1000000000;
const int BASE = 1000000007;
const int MAX = 100007;
const int ADD = 1000000;
const int MOD = 1000000007;
const int CNT = 800;

int main()
{

  int res = 0;
  int n , l;
  cin >> n >> l;

  assert(n >= 1 && n <= 10);

  FOR(test,0,n)
  {
    string s;
    cin >> s;

    assert(SZ(s) >= 1 && SZ(s) <= l);

    bool ok = 0;
    int n = SZ(s);

    FOR(l1,0,n)
      FOR(r1,l1,n)
        FOR(l2,r1 + 1 , n)
          FOR(r2 , l2 , n)
          {
            string t = s;
            reverse(t.begin() + l1 , t.begin() + r1 + 1);
            reverse(t.begin() + l2 , t.begin() + r2 + 1);
            string tt = t;
            reverse(ALL(tt));
            if (tt == t)
              ok = 1;
          }

    res += ok;

  }

  cout << res << endl;

    return 0;
}