Question

- Problem Description
Tahir and Mamta are woking in a project in TCS. Tahir being a problem solver came up with an interesting problem for his friend Mamta.
Problem consists of a string of length N and contains only small case alphabets.
It will be followed by Q queries, in which each query will contain an integer P (1<=P<=N) denoting a position within the string.
Mamta's task is to find the alphabet present at that location and determine the number of occurrence of same alphabet preceding the g
location P.
Mamta is busy with her office work. Therefore, she asked you to help her.
Constraints
1<= N<= 500000
S consisting of small case alphabets
1 <= Q<= 10000
1<= P <=N
- Input Format
First line contains an integer N, denoting the length of string.
Second line contains string S itself consists of small case alphabets only ('a' - 'z).
Third line contains an integer Q denoting number of queries that will be asked.
Next Q lines contains an integer P (1 <= P <= N) for which you need to find the number occurrence of character present at the Pth locatio
preceeding P.
- Output
For each query, print an integer denoting the answer on single line.​

Answer 1

Answer:

class solve

{

public static void main(String args[])

{

Scanner sc= new Scanner(System . in);

int arr [ ];

System . out . println("Enter the length of string");

int len=sc . nextInt();

arr =new int [ len ];

System . out . println("Enter the string");

string str=sc . next();

System . out . println("Enter number of queries to be made");

int q=sc . nextInt();

for ( int i = 0; i< q; i++)

{

int count =0;

System . out . println( "Enter the query" );

int p=sc . nextInt();

for (int j=0; j<p-1 ; j++)

{  

if( str . charAt ( j ) = = str . charAt ( p-1 ) )

{

count + + ;

}

}

arr [ i ] = count;

}

for ( int i= 0; i< q; i++)

{

System .  out .  println (arr [ i ]);

}

}

}