Skip to content
Related Articles

Related Articles

Save Article
Improve Article
Save Article
Like Article

Find the number of occurrences of a character upto preceding position

  • Last Updated : 11 Jun, 2021

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: 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input : S = “ababababab”, P = 9 
Output :
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 searching for a similar character. Whenever a similar character occurs increment 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.

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

 




My Personal Notes arrow_drop_up
Recommended Articles
Page :