HackerEarth Substrings problem solution

In this HackerEarth Substrings problem solution You have a string S, but you like only special strings. So, you have to calculate the total number of special substrings in S.

A string T, of length L, is called special string, if either of the following property holds:

1. All characters of the string T are same. for example, aaa
2. The string has an odd length (i.e, L is odd) and all characters of T are same except the middle character, for example, aabaa.

Count the total number of special substrings in S.

#include<bits/stdc++.h>
using namespace std;

//Constants:
#define N 1000005
#define ll long long
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pritnf printf

ll le[N];
ll ri[N];

int main()
{
  ll t,i=0,j,k,x,y,z,count=0,p,flag=0,ans=0,sum=0,l,n,m,max1,min1,pos,tmp,q;
  
  //string s;
  
  t=1;
  
  while(t--)
  {
    string s;
      cin >> s;
      l=s.length();
      
      ans=0;
      //condition 1
      count=1;
      for(i=1;i<l;i++)
      {
        if(s[i]==s[i-1])
          count++;
        else
        {
          ans+=count*(count+1)/2;
          count=1;
        }
      }
      ans+=count*(count+1)/2;
      //printf("%lld ",ans);
      
      //condition 2
      count=1;
      for(i=1;i<l;i++)
      {
        if(s[i]==s[i-1])
          count++;
        else
        {
          le[i]=count;
          count=1;
        }
      }

      count=1;
    for(i=l-2;i>=0;i--)
      {
        if(s[i]==s[i+1])
          count++;
        else
        {
          ri[i]=count;
          count=1;
        }
      }     

      for(i=1;i<l-1;i++)
      {
        p=min(le[i],ri[i]);

        if(s[i-1]==s[i+1] && s[i]!=s[i+1])
          ans+=p;
        //printf("%lld ",ans);
      }

    printf("%lldn",ans);
    
  }
  
  
  return 0;
}