Find the number of occurrences of a character upto preceding position

Given a string S of length N and an integer P(1?P?N) denoting the position of a character in the string. The task is to find the number of occurrences of the character present at position P up to P-1 index.

Examples:

Input : S = “ababababab”, P = 9 
Output : 4 
Character at P is ‘a’. Number of occurrences of ‘a’ upto 8th index is 4

Input : S = “geeksforgeeks”, P = 9 
Output : 1

Naive Approach:A naive approach is to iterate over the string till Position-1 searches for a similar character. Whenever a similar character occurs increment the counter variable by one. 

Below is the implementation of the above approach:  

C++

// CPP program to find the number of occurrences
// of a character at position P upto p-1
#include <bits/stdc++.h>
using namespace std;

// Function to find the number of occurrences
// of a character at position P upto p-1
int Occurrence(string s, int position)
{
    int count = 0;
    for (int i = 0; i < position - 1; i++)
        if (s[i] == s[position - 1])
            count++;

    // Return the required count
    return count;
}

// Driver code
int main()
{
    string s = "ababababab";

    int p = 9;

    // Function call
    cout << Occurrence(s, p);

    return 0;
}

Java

// Java program to find the number of occurrences
// of a character at position P upto p-1
class GFG
{

// Function to find the number of occurrences
// of a character at position P upto p-1
static int Occurrence(String s, int position)
{
    int count = 0;
    for (int i = 0; i < position - 1; i++)
        if (s.charAt(i) == s.charAt(position - 1))
            count++;

    // Return the required count
    return count;
}

// Driver code
public static void main(String[] args)
{
    String s = "ababababab";

    int p = 9;

    // Function call
    System.out.println(Occurrence(s, p));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to find the
# number of occurrences of
# a character at position P upto p-1

# Function to find the number of occurrences
# of a character at position P upto p-1
def Occurrence(s, position):
    count = 0
    for i in range(position - 1):
        if (s[i] == s[position - 1]):
            count += 1

    # Return the required count
    return count

# Driver code
s = "ababababab";

p = 9

# Function call
print(Occurrence(s, p))

# This code is contributed by Mohit Kumar

C#

// C# program to find the number of occurrences
// of a character at position P upto p-1
using System;

class GFG
{

// Function to find the number of occurrences
// of a character at position P upto p-1
static int Occurrence(String s, int position)
{
    int count = 0;
    for (int i = 0; i < position - 1; i++)
        if (s[i] == s[position - 1])
            count++;

    // Return the required count
    return count;
}

// Driver code
public static void Main(String[] args)
{
    String s = "ababababab";

    int p = 9;

    // Function call
    Console.WriteLine(Occurrence(s, p));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>
// Javascript program to find the number of occurrences
// of a character at position P upto p-1

// Function to find the number of occurrences
// of a character at position P upto p-1
function Occurrence(s,position)
{
    let count = 0;
    for (let i = 0; i < position - 1; i++)
        if (s[i] == s[position - 1])
            count++;

    // Return the required count
    return count;
}

// Driver code
let s = "ababababab";

let p = 9;

// Function call
document.write(Occurrence(s, p));


// This code is contributed by unknown2108
</script>

Output

4

Time Complexity: O(N) for each query.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Efficient Approach: In case, if there are multiple such queries and we are given a unique index P for every query then an efficient approach is to use a frequency array for storing the character count in each iteration of the string.

Below is the implementation of the above approach : 

C++

// CPP program to find the number of occurrences
// of a character at position P upto p-1
#include <bits/stdc++.h>
using namespace std;

// Function to find the number of occurrences
// of a character at position P upto p-1
int countOccurrence(string s, int position)
{
    int alpha[26] = { 0 }, b[s.size()] = { 0 };

    // Iterate over the string
    for (int i = 0; i < s.size(); i++) {
        // Store the Occurrence of same character
        b[i] = alpha[(int)s[i] - 97];

        // Increase its frequency
        alpha[(int)s[i] - 97]++;
    }

    // Return the required count
    return b[position - 1];
}

// Driver code
int main()
{
    string s = "ababababab";

    int p = 9;

    // Function call
    cout << countOccurrence(s, p);

    return 0;
}

Java

// Java program to find the number of occurrences
// of a character at position P upto p-1
import java.util.*;

class GFG
{

// Function to find the number of occurrences
// of a character at position P upto p-1
static int countOccurrence(String s, int position)
{
    int []alpha = new int[26];
    int []b = new int[s.length()];

    // Iterate over the string
    for (int i = 0; i < s.length(); i++)
    {
        // Store the Occurrence of same character
        b[i] = alpha[(int)s.charAt(i) - 97];

        // Increase its frequency
        alpha[(int)s.charAt(i) - 97]++;
    }

    // Return the required count
    return b[position - 1];
}

// Driver code
public static void main(String[] args)
{
    String s = "ababababab";

    int p = 9;

    // Function call
    System.out.println(countOccurrence(s, p));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to find the number of occurrences
# of a character at position P upto p-1

# Function to find the number of occurrences
# of a character at position P upto p-1
def countOccurrence(s, position):
    alpha = [0] * 26
    b = [0] * len(s)

    # Iterate over the string
    for i in range(0, len(s)):

        # Store the Occurrence of same character
        b[i] = alpha[ord(s[i]) - 97]

        # Increase its frequency
        alpha[ord(s[i]) - 97] = alpha[ord(s[i]) - 97] + 1

    # Return the required count
    return b[position - 1]

# Driver code
s = "ababababab"

p = 9

# Function call
print(countOccurrence(s, p))

# This code is contributed by Sanjit_Prasad

C#

// C# program to find the number of occurrences
// of a character at position P upto p-1
using System;

class GFG
{

// Function to find the number of occurrences
// of a character at position P upto p-1
static int countOccurrence(String s, int position)
{
    int []alpha = new int[26];
    int []b = new int[s.Length];

    // Iterate over the string
    for (int i = 0; i < s.Length; i++)
    {
        // Store the Occurrence of same character
        b[i] = alpha[(int)s[i] - 97];

        // Increase its frequency
        alpha[(int)s[i] - 97]++;
    }

    // Return the required count
    return b[position - 1];
}

// Driver code
public static void Main(String[] args)
{
    String s = "ababababab";

    int p = 9;

    // Function call
    Console.WriteLine(countOccurrence(s, p));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// Javascript program to find the number of occurrences
// of a character at position P upto p-1

// Function to find the number of occurrences
// of a character at position P upto p-1
function countOccurrence(s, position)
{
    let alpha = new Array(26);
    for(let i = 0; i < 26; i++)
    {
        alpha[i] = 0;
    }
    let b = new Array(s.length);

    // Iterate over the string
    for(let i = 0; i < s.length; i++)
    {

        // Store the Occurrence of same character
        b[i] = alpha[s[i].charCodeAt(0) - 97];

        // Increase its frequency
        alpha[s[i].charCodeAt(0) - 97]++;
    }

    // Return the required count
    return b[position - 1];
}

// Driver code
let s = "ababababab";

p = 9;

// Function call
document.write(countOccurrence(s, p));

// This code is contributed by patel2127

</script>

Output

4

Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(26) ? O(1), no extra space is required, so it is a constant.