HackerEarth All Vowels problem solution YASH PAL, 31 July 2024 In this HackerEarth All Vowels problem solution Vowels are very essential characters to form any meaningful word in the English dictionary. There are 5 vowels in the English language – a, e, i, o u. You are given a random string containing only lowercase letters and you need to find if the string contains ALL the vowels. HackerEarth All Vowels problem solution. #include <bits/stdc++.h>#define rep(i,n) for(i=0;i<n;i++)#define ll long long#define elif else if#define pii pair< int, int>#define mp make_pair#define pb push_backusing namespace std;#define gc getchar_unlockedint randi(int l,int r){ if(r<l) return l; return rand()%(r-l+1) +l;}int done[5]={0};int main(){ freopen("in.txt","r",stdin); freopen("out","w",stdout); int i,j,n,mi=99999,ma=0; cin>>n; string st; cin>>st; rep(i,n) { if(st[i]=='a') done[0]++; if(st[i]=='e') done[1]++; if(st[i]=='i') done[2]++; if(st[i]=='o') done[3]++; if(st[i]=='u') done[4]++; } rep(i,5) { if(!done[i]) { cout<<"NO"; return 0; } } cout<<"YES"; return 0;} Second solution #include<bits/stdc++.h>using namespace std;int main(){ int N; string S; cin>>N>>S; assert(N>=1 && N<=10000); assert(S.length()==N); // a,e,i,o,u bool flag1=false,flag2=false,flag3=false,flag4=false,flag5=false; for(int i=0;i<N;i++){ if(S[i]=='a') flag1=true; else if(S[i]=='e') flag2=true; else if(S[i]=='i') flag3=true; else if(S[i]=='o') flag4=true; else if(S[i]=='u') flag5=true; if(flag1==true && flag2==true &&flag3==true && flag4==true && flag5==true) break; } if(flag1==true && flag2==true &&flag3==true && flag4==true && flag5==true) cout<<"YES"<<endl; else cout<<"NO"<<endl; } coding problems