HackerEarth Beautiful Strings problem solution YASH PAL, 31 July 202415 February 2026 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;} coding problems solutions HackerEarth HackerEarth