HackerEarth Vowel Query problem solution

In this HackerEarth Vowel Query problem solution, You are given a string S consisting of lowercase alphabets and an integer array val consisting of N integers. Using the given string S and array val, you need to create another string X according to the code snippet below:

Initially string X is empty

Let len be the length of string S

for i := 0 to N-1

    pos := absolute value of val[i]

    if(val[i] >= 0)

      X := X + S[0,pos] // append substring of string S from index 0 to pos (both including) into X

    else

      X := X + S[pos,len-1] // append substring of string S from index pos to len-1 (both including) into X

You have to answer Q tasks. For each task, you are given an integer K and you have to print the Kth vowel in the string X. If the Kth vowel doesn't exist print -1.

HackerEarth Vowel Query problem solution.

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
string s;
ll q,n,val[100005],m,pre[100005],suf[100005],x;
vector<char>pref,suff;
vector<ll>cum;
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    ll i,j,k;
    cin>>s;
    cum.push_back(0);
    n=s.size();
    cin>>m;
    for(i=1;i<=m;i++)
    {
        cin>>val[i];
    }
    for(i=0;i<n;i++)
    {
        if(i>0)
        pre[i]=pre[i-1];
        if(s[i]=='a'||s[i]=='o'||s[i]=='e'||s[i]=='i'||s[i]=='u')
        {
            pre[i]++;
            pref.push_back(s[i]);
        }
    }
    suff=pref;
    reverse(suff.begin(),suff.end());
    for(i=n-1;i>=0;i--)
    {
        suf[i]=suf[i+1];
        if(s[i]=='a'||s[i]=='o'||s[i]=='e'||s[i]=='i'||s[i]=='u')
        {
            suf[i]++;
        }
    }
    for(i=1;i<=m;i++)
    {
        if(val[i]<0)
        {
            cum.push_back(cum[cum.size()-1]+suf[-1*val[i]]);
        }
        else
        {
            cum.push_back(cum[cum.size()-1]+pre[val[i]]);
        }
    }
    cin>>q;
    while(q--)
    {
        cin>>x;
        if(cum[cum.size()-1]<x)
        {
            cout<<"-1\n";
        }
        else
        {
            j=lower_bound(cum.begin(),cum.end(),x)-cum.begin();
            x-=cum[j-1];
            j=val[j];
            if(j<0)
            {
                j=suf[-j]-x;
                cout<<suff[j]<<"\n";
            }
            else
            {
                x--;
                cout<<pref[x]<<"\n";
            }
        }
    }
    return 0;
}