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 the position P upto P-1 index.

Examples:

Input : S = “ababababab”, P = 9
Output : 4
Character at P is ‘a’. Number of occurences 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 searching for a similar character. Whenever a similar character occurs increment counter variable by one.

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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 

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Output:

4

Time Complexity : O(N) for each query.

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

4


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