Queries to find the count of characters preceding the given location

Given a string S of length N containing only lowercase letters. Also, given Q queries where each query consists of an integer P such that 1? P ? N. The task is to find the count of occurrences of the same letter preceding the given location P.

Examples :

Input: S = "abacsddaa", Q[] = {9, 3}
Output:
3
1
For first query, P = 9, character at 9th location is 'a'.
The number of occurrences of 'a' before P i.e. 9 is three.

Similarly, for P = 3, 3rd character is 'a'.
The number of occurrences of 'a' before P. i.e. 3 is one.

Naive approach For each query Q_i which gives position ‘p’, run a loop from 1 to ‘p-1’ and check whether element present at that index is equal to element at index ‘p’, if it is true then increase the value of counter variable 1. After completion of loop print the value of the counter variable in a new line.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;

int Count(string s, int pos )
{
    // returns character at index pos - 1
    int c = s[pos - 1] ;
    int counter = 0 ;
    for (int i = 0; i < pos-1; i++ )
    {
        if(s[i] == c)
            counter = counter + 1;
    }
    return counter;
}

// Driver Code
int main()
{
    string s = "abacsddaa";
    int pos;
    int n = s.length();
    int query[] = {9, 3, 2};
    int Q = sizeof(query) / sizeof(query[0]);
    for(int i = 0; i < Q; i++ )
    {
        pos = query[i] ;
        cout << Count( s, pos ) << endl;
    }
    return 0;
}

// This code is contributed by
// divyamohan123

Java

// Java implementation of the approach
class GFG
{
    static int Count(String s, int pos )
    {
        // returns character at index pos - 1
        int c = s.charAt(pos - 1);
        int counter = 0;
        for (int i = 0; i < pos - 1; i++ )
        {
            if(s.charAt(i) == c)
                counter = counter + 1;
        }
        return counter;
    }

    // Driver Code
    public static void main (String[] args)
    {
        String s = "abacsddaa";
        int pos;
        int n = s.length();

        int query[] = {9, 3, 2};
        int Q = query.length;

        for(int i = 0; i < Q; i++)
        {
            pos = query[i];
            System.out.println(Count(s, pos));
        }
    }
}

// This code is contributed by AnkitRai01

Python3

# Python 3 implementation of the approach
def Count( s, pos ):
    # returns character at index pos - 1
    c = s[pos - 1]
    counter = 0
    for i in range( pos - 1 ):
        if s[i] == c:
              counter = counter + 1
    return counter

# Driver Code
if __name__ == "__main__" :
    s = "abacsddaa"
    n = len(s)
    query = [9, 3, 2]
    Q = len(query)
    for i in range( Q ):
        pos = query[i]
        print(Count( s, pos ))

C#

// C# implementation of the approach
using System;

class GFG
{
    static int Count(String s, int pos )
    {
        // returns character at index pos - 1
        int c = s[pos - 1];
        int counter = 0;
        for (int i = 0; i < pos - 1; i++ )
        {
            if(s[i] == c)
                counter = counter + 1;
        }
        return counter;
    }

    // Driver Code
    public static void Main (String[] args)
    {
        String s = "abacsddaa";
        int pos;
        int n = s.Length;

        int []query = {9, 3, 2};
        int Q = query.Length;

        for(int i = 0; i < Q; i++)
        {
            pos = query[i];
            Console.WriteLine(Count(s, pos));
        }
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>

    // Javascript implementation of the approach

    function Count(s, pos )
    {
        // returns character at index pos - 1
        let c = s[pos - 1];
        let counter = 0;
        for (let i = 0; i < pos - 1; i++ )
        {
            if(s[i] == c)
                counter = counter + 1;
        }
        return counter;
    }

    // Driver code

        let s = "abacsddaa";
        let pos;
        let n = s.length;

        let query = [9, 3, 2];
        let Q = query.length;

        for(let i = 0; i < Q; i++)
        {
            pos = query[i];
            document.write(Count(s, pos) + "<br/>");
        }

// This code is contributed by susmitakundugoaldanga.
</script>

Output

3
1
0

Time Complexity: O(N) 
Auxiliary Space: O(1)

Better approach: Using Hashing or Dictionary and auxiliary space. 
Create an auxiliary array say ‘temp’ which initialize to zero and hash table or dictionary ‘d’
initially empty. For pre-processing purposes, that is to store the previous count of elements at index ‘pos’ at index pos in temp. 

Preprocessing function
Pseudo Code:
function Preprocess(s):
      for i=1 to length(s),
           if s(i) not in hash_table,
                hash_table(s(i)) := i,
                end;
           else,
                auxiliary_array(i) = hash_table(s(i)) + 1,
                hash_table(s(i)) =  i,
                end;                

It place all the required value at respective index of auxiliary array. For each query Q_i print val at position ‘pos’ in temp.

C++

// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;

void Count(vector<int> temp)
{
    int query[] = {9, 3, 2};
    int Q = sizeof(query) /
            sizeof(query[0]);
    for (int i = 0; i < Q; i++)
    {
        int pos = query[i];
        cout << (temp[pos - 1]) << endl;
    }
}

vector<int> processing(string s, int len)
{
    vector<int> temp(len);
    map<char, int> d;

    for (int i = 0; i < len; i++)
    {
        if (d.find(s[i]) == d.end())
        {
            d[s[i]] = i;
        }
        else
        {
            temp[i] = temp[d[s[i]]] + 1;
            d[s[i]] = i;
        }
    }
    return temp;
}

// Driver Code
int main()
{
    string s = "abacsddaa";
    int n = s.length();
    vector<int> temp = processing(s, n);
    Count(temp);
}

// This code is contributed
// by Surendra_Gangwar

Java

// Java implementation of the approach
import java.util.*;

class GFG
{
    static void Count(int[] temp)
    {
        int[] query = {9, 3, 2};
        int Q = query.length;
        for (int i = 0; i < Q; i++)
        {
            int pos = query[i];
            System.out.println(temp[pos - 1]);
        }
    }

    static int[] processing(String s, int len)
    {
        int[] temp = new int[len];
        HashMap<Character,
                Integer> d = new HashMap<Character,
                                         Integer>();
        for (int i = 0; i < len; i++)
        {
            if (!d.containsKey(s.charAt(i)))
            {
                d.put(s.charAt(i), i);
            }
            else
            {
                temp[i] = temp[d.get(s.charAt(i))] + 1;
                d.put(s.charAt(i), i);
            }
        }
        return temp;
    }

    // Driver Code
    public static void main(String[] args)
    {
        String s = "abacsddaa";
        int n = s.length();
        int[] temp = processing(s, n);
        Count(temp);
    }
}

// This code is contributed by Rajput-Ji

Python3

# Python 3 implementation of the approach
def Count( temp ):
    query = [9, 3, 2]
    Q = len(query)
    for i in range( Q ):
        pos = query[i]
        print( temp[pos-1] )
def processing( s ):
    temp = [ 0 ] * len( s )
    d = dict( )
    for i in range( len( s ) ):
        if s[i] not in d:
            d[ s[i] ] = i
        else:
            temp[i] = temp[ d[ s[i] ] ] + 1
            d[ s[i] ] = i
    return temp

# Driver Code
if __name__ == "__main__" :
    s = "abacsddaa"
    n = len(s)
    temp = processing( s )
    Count( temp )

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG
{
    static void Count(int[] temp)
    {
        int[] query = {9, 3, 2};
        int Q = query.Length;
        for (int i = 0; i < Q; i++)
        {
            int pos = query[i];
            Console.WriteLine(temp[pos - 1]);
        }
    }

    static int[] processing(String s, int len)
    {
        int[] temp = new int[len];
        Dictionary<int,
                   int> d = new Dictionary<int,
                                           int>();
        for (int i = 0; i < len; i++)
        {
            if (!d.ContainsKey(s[i]))
            {
                d.Add(s[i], i);
            }
            else
            {
                temp[i] = temp[d[s[i]]] + 1;
                d[s[i]] = i;
            }
        }
        return temp;
    }

    // Driver Code
    public static void Main(String[] args)
    {
        String s = "abacsddaa";
        int n = s.Length;
        int[] temp = processing(s, n);
        Count(temp);
    }
}

// This code is contributed by PrinciRaj1992

Javascript

const Count = (temp) => {
  const query = [9, 3, 2];
  const Q = query.length;
  for (let i = 0; i < Q; i++) {
    const pos = query[i];
    console.log(temp[pos - 1]);
  }
};

const processing = (s, len) => {
  const temp = Array(len).fill(0);
  const d = new Map();
  for (let i = 0; i < len; i++) {
    if (!d.has(s[i])) {
      d.set(s[i], i);
    } else {
      temp[i] = temp[d.get(s[i])] + 1;
      d.set(s[i], i);
    }
  }
  return temp;
};

const s = "abacsddaa";
const n = s.length;
const temp = processing(s, n);
Count(temp);

// This code is contributed by anskalyan3.

Output

3
1
0

Time Complexity: O(N),  where N is the length of the given string.
Auxiliary Space: O(N)