HackerEarth Beautiful Strings problem solution

In this HackerEarth Beautiful Strings problem solution, A string is beautiful if it has equal number of a,b,and c in it.

Example “abc” , “aabbcc” , “dabc” , “” are beautiful.

Given a string of alphabets containing only lowercas aplhabets (a-z), output the number of non-empty beautiful substring of the given string.

HackerEarth Beautiful Strings problem solution.

#include<bits/stdc++.h>
#define PB push_back
#define MP make_pair
#define F first
#define S second
#define SZ(a) (int)(a.size())
#define CLR(a) a.clear()
#define SET(a,b) memset(a,b,sizeof(a))
#define LET(x,a) __typeof(a) x(a)
#define TR(v,it) for( LET(it,v.begin()) ; it != v.end() ; it++)
#define FORi(i,a,b) for(LET(i,a) ; i<b; i++)
#define repi(i,n) FORi(i,(__typeof(n))0,n)
#define FOR(i,a,b) for(i=a ; i<b; i++)
#define rep(i,n) FOR(i,0,n)
#define si(n) scanf("%d",&n)
#define sll(n) scanf("%lld",&n)
#define pi(n) printf("%d",n)
#define piw(n) printf("%d ",n)
#define pin(n) printf("%dn",n)
using namespace std;

typedef long long LL;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef vector< PII > VPII;
string s;
PII a[100001];
int main()
{
    int t,n;
    cin>>t;
    while(t--)
    {
        cin>>s;
        n = SZ(s);
        a[0].F = a[0].S = 0;
        repi(i,n)
        {
            a[i+1]=a[i];
            if(s[i]=='a'){a[i+1].F--;a[i+1].S--;}
            if(s[i]=='b')a[i+1].F++;
            if(s[i]=='c')a[i+1].S++;
        }
        n++;
        sort(a,a+n);
        LL ans=0;
        LL c = 1;
        for(int i = 1; i < n; i++)
            if(a[i]==a[i-1])
                c++;
            else
            {
                ans += (c*(c-1))/2;
                c=1;
            }
        cout<<ans<<endl;
    }
    return 0;

}

Second solution

#include<bits/stdc++.h>
#define PB push_back
#define MP make_pair
#define F first
#define S second
#define SZ(a) (int)(a.size())
#define CLR(a) a.clear()
#define SET(a,b) memset(a,b,sizeof(a))
#define TR(v,it) for( typeof(v.begin()) it(v.begin()) ; it != v.end() ; it++)
#define FOR(i,a,b) for(i=(int)a;i<(int)b;i++)
#define si(n) scanf("%d",&n)
#define rep(i,n) FOR(i,0,n)
#define repi(i,n) for(int i=0; i<n; i++)
 
 
using namespace std;
 
typedef long long LL; 
typedef pair<int,int> PII;
typedef vector<int> VI; 
typedef vector< PII > VPII;
LL power(LL a, LL p)
{
  LL ret=1;
  while(p)
  {
    if(p&1) ret = (ret*a);
    p/=2; 
    a = (a*a);
  }
  return ret;
}
LL gcd(LL a, LL b)
{if(b) return gcd(b,a%b); return a;}
PII A[1000001];
int main() 
{
  int t;
  scanf("%d",&t);
  while(t--){
  
  A[0] = MP(0,0);
  string s; cin>>s; int n = SZ(s); 
  repi(i,n)
  {
    A[i+1] = A[i];
    if(s[i]=='a')
      A[i+1].F++;
    if(s[i]=='b')
    {
      A[i+1].F--;
      A[i+1].S++;
    }
    if(s[i]=='c') A[i+1].S--;
      
  }
  sort(A,A+n+1);
  LL ans=0;
  LL c=1;
  for(int i=1; i<=n;i++)
  {
    if(A[i]==A[i-1])c++;
    else
    {
      ans += (c*(c-1)) / 2;
      c=1;
    }
  }
  ans += (c*(c-1))/2;
  cout<<ans<<endl;
}
    return 0;
}

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