# HackerEarth String Operations problem solution

In this HackerEarth String Operations problem solution, You are given a string S.

Q number of operations are performed on string S.
In each of these Q operations, you are given an index ind and a character ch. For each operation, you have to update the character at index ind in string S to ch, that is, after this operation S[ind] = ch.
It is guaranteed that any index is updated at most once.
We call the final string after performing the Q number of operations as str.

After this, M number of operations are performed on string str.
In each of these M operations, you are given two indices a and b. For each operation, you have to reverse the substring that lies between the indices a and b (inclusive).
We call the final string after performing M operations as fin.

## HackerEarth String Operations problem solution.

`import java.util.*;import java.lang.*;class StringQue{  public static void main(String args[])  {    StringBuffer s= new StringBuffer();    Scanner in= new Scanner(System.in);    s.append(in.nextLine());    int q=in.nextInt();    for(int i=1;i<=q;i++)    {      int ind=in.nextInt();      char ch=in.next().charAt(0);      s.setCharAt(ind-1,ch);    }    StringBuffer str=new StringBuffer(s);    int m=in.nextInt();    for(int i=1;i<=m;i++)    {      int a=in.nextInt(),b=in.nextInt(),ans=0;      StringBuffer upd=new StringBuffer();      upd.append(s.substring(a-1,b));      upd.reverse();      s.replace(a-1,b,upd.toString());    }    int ans=0;    System.out.println(str);    System.out.println(s);    for(int i=0;i<s.length();i++)    {      if(str.charAt(i)==s.charAt(i))        ans++;    }    System.out.println(ans);  }}`