# HackerEarth Kth Character problem solution

In this HackerEarth Kth Character problem solution, You are given a string S of lowercase alphabets. Now you need to remove all the instances of exactly one type of character such that the new string that is formed is lexicographically smallest across all the other strings that can be formed in a similar way.

For example: Let the string S be "appleap"

If you remove all instances of 'a' then the string formed will be : pplep
If you remove all isntances of 'p' then the string formed will be : alea
If you remove all intsances of 'l' then the string formed will be : appeap
If you remove all instances of 'e' then the string formed will be : applap
So among all the newly formed strings , alea is lexicographically smallest string.

## HackerEarth Kth Character problem solution.

`#include<bits/stdc++.h>#define LL long long int#define M 1000000007#define endl "\n"#define eps 0.00000001LL pow(LL a,LL b,LL m){LL x=1,y=a;while(b > 0){if(b%2 == 1){x=(x*y);if(x>m) x%=m;}y = (y*y);if(y>m) y%=m;b /= 2;}return x%m;}LL gcd(LL a,LL b){if(b==0) return a; else return gcd(b,a%b);}LL gen(LL start,LL end){LL diff = end-start;LL temp = rand()%start;return temp+diff;}using namespace std;vector<string> v;int main()    {      ios_base::sync_with_stdio(0);    string s;    cin >> s;    int mask = 0;    for(int i = 0; i < s.length(); i++)      {        mask = (mask | (1 << (s[i] - 'a')));      }    for(int i = 0; i < 26; i++)      {        if(mask & (1 << i))          {            string temp;            for(int j = 0; j < s.length(); j++)              {                if((s[j] - 'a') != i)                  {                    temp.push_back(s[j]);                  }              }            v.push_back(temp);          }      }    sort(v.begin() , v.end());    cout << v[0];  }`

### Second solution

`#include<bits/stdc++.h>using namespace std;int main(){  string s;  cin>>s;  int n=s.size();  vector<string>v;  for(char i='a';i<='z';i++)  {    string temp="";    for(int j=0;j<n;j++)    {      if(s[j]!=i)        temp+=s[j];    }    v.push_back(temp);  }  sort(v.begin(),v.end());  cout<<v[0]<<"\n";  return 0;}`